ML Aggarwal Solutions – Page 6

ML Aggarwal Maths for Class 12 Solutions Pdf – Understanding ISC Mathematics Class 12 Solutions

November 28, 2023

ISC Mathematics Class 12 Solutions – ML Aggarwal Class 12 ISC Solutions

Section A

ML Aggarwal Class 12 Solutions ISC Pdf Chapter 1 Relations and Functions

ML Aggarwal Class 12 Solutions Chapter 2 Inverse Trigonometric Functions

ML Aggarwal Maths for Class 12 Solutions Pdf Chapter 3 Matrices

Understanding ISC Mathematics Class 12 Solutions Chapter 4 Determinants

ML Aggarwal Class 12 ISC Solutions Chapter 5 Continuity and Differentiability

ML Aggarwal Class 12 ISC Solutions Chapter 6 Indeterminate Forms

Class 12 ML Aggarwal Solutions Chapter 7 Applications of Derivatives

ISC Mathematics Class 12 Solutions Chapter 8 Integrals

ISC Maths Class 12 Solutions Chapter 9 Differential Equations

Class 12 ISC Maths Solutions Chapter 10 Probability

Section B

ISC Class 12 Maths Solutions Chapter 1 Vectors

ISC Class 12th Maths Solutions Chapter 2 Three Dimensional Geometry

Maths Class 12 ISC Solutions Chapter 3 Applications of Integrals

Section C

ISC 12th Maths Solutions Chapter 1 Application of Calculus in Commerce and Economics

Maths ISC Class 12 Solutions Chapter 2 Linear Regression

ISC Class 12 Mathematics Solutions Chapter 3 Linear Programming

Case Study Based Questions

Case Study Based Questions Section A Chapter 1-7
Case Study Based Questions Section A Chapter 8-10
Case Study Based Questions Section B
Case Study Based Questions Section C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Chapter Test

November 27, 2023

Access to comprehensive ML Aggarwal Maths for Class 12 Solutions Chapter 7 Applications of Derivatives Chapter Test encourages independent learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Chapter Test

Question 1.
A circular disc is being heated. Due to heat its radius increases at the rate of 005 cm / sec. Find the rate at which its area is increasing when the radius is 3.2 cm. (NCERT)
Solution:
Let r be the radius of circle
Then A = area of circle = πr²
Differentiating w.r.t. t, we have
\(\frac{d A}{d t}\) = 2πr \(\frac{d r}{d t}\) ……….(1)
given, \(\frac{d r}{d t}\) = 0.05 cm/sec ;
r = 3.2 cm
∴ from (1); we have
\(\frac{d A}{d t}\) = (2π × 3.2 × 0.05) cm²/sec
= 0.32 π cm²/sec
Hence, the required rate at which area is increasing be 0.32 π cm²/sec.

Question 2.
The radius of a circular blot of ink is increasing at 0.5 cm per minute. Find the rate of increase of the area when the radius is equal to 5 cm. Find also the approximate change in the area when the radius increases from 5 cm to 5.001 cm.
Solution:
LetA be the area of circular blot of radius r
Then A = πr² ……………(1)
On differentiating eqn. (1) w.r.t. t, we have
\(\frac{d A}{d t}\) = 2πr \(\frac{d r}{d t}\) ……….(2)
given \(\frac{d r}{d t}\) = 0.5 cm / mm ; r = 5 cm
∴ from eqn. (2) ; we have
\(\frac{d A}{d t}\) = (2π × 5 × 0.5) cm²/min
= 5π cm²/min.
Hence the required rate at which area is increasing be 5π cm²/min.
Diff. (1) w.r.t. r, we have
\(\frac{d A}{d r}\) = 2πr
Take r = 5 ;
r + δr = 5.001
∴ δr = dr = 5.001 – 5 = 0.001
Now, dA = \(\frac{d A}{d r}\) dr
= 2πr × δr
= 2π × 5 × 0.001 = 0.01π
∴ δA = 0.01π [∵ δA = dA]
Thus approximate change in area = δA = 0.01 π cm².

Question 3.
A balloon in the form of a right circular cone surmounted by a hemi-sphere, having a diameter equal to the height of the cone, is being inflated by pumping air into it. How fast is its volume changing with respect to its total height h, when h = 9 cm?
Solution:
Let r be the radius of hemisphere and base radius of cone.
It is given that diameter of hemisphere is equal to height of cone.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives chapter Test 7

Let x be the height of cone.
∴ x = 2r
given total height h = x + r
= 2r + r = 3r
∴ r = \(\frac{h}{3}\)
∴ V = Volume of solid
= \(\frac{\pi}{3}\) r²x + \(\frac{2}{3}\) πr³
V = \(\frac{\pi}{3}\left(\frac{h}{3}\right)^2 \times 2 \times \frac{h}{3}+\frac{2}{3} \pi\left(\frac{h}{3}\right)^3\)
= \(\frac{4 \pi}{3}\left(\frac{h}{3}\right)^3\)
= \(\frac{4 \pi}{81} h^3\)
∴ \(\frac{4 \pi}{81}\) 3h²
= \(\frac{4 \pi}{27}\) h²
at h = 9 cm ;
\(\frac{4 \pi}{27}\) × 9²
= 12π cm³ / cm.

Question 4.
Find the equation of the normal to the curve 2y + x² = 3 at the point (1, 1).
Solution:
Given equation of curve be, 2y + x² = 3 …………(1)
Diff. eqn. (1) w.r.t. x; we have
2 \(\frac{d y}{d x}\) + 2x = 0
\(\frac{d y}{d x}\) = – x
∴ slope of Normal to curve at (1, 1)
= \(-\frac{1}{\left(\frac{d y}{d x}\right)_{(1,1)}}=\frac{-1}{-1}\) = – 1
Thus, the required eqn. of normal to given curve at point (1, 1) is given by
y – 1 = 1 (x – 1)
⇒ x = y
⇒ x – y = 0.

Question 5.
Find the equation of the tangent to the curve y = x³ – x² at the point (2, 4). Also find the equation of tangents to the curve which are parallel to x-axis.
Solution:
Given equation of curve be,
y = x³ – x²
Diff. eqn. (1) both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 3x² – 2x
∴ slope of tangent to curve at point (2, 4) = (\(\frac{d y}{d x}\))_{(2, 4)}
Thus, the eqn. of tangent to curve at point (2, 4) is given by
y – 4 = 8 (x – 2)
⇒ 8x – y – 12 = 0
Let (x₁, y₁) be any point on curve (1).
∴ y₁ = x₁³ – x₁² ………….(2)
Since the tangent to curve is parallel to x – axis.
∴ (\(\frac{d y}{d x}\))_{(x₁, y₁)} = 0
⇒ 3x₁² – 2x₁ = 0
x₁ = 0, 2/3
∴ from (2);
y₁ = 0, – \(\frac{4}{27}\)
Thus, the points on curve at which tangent is parallel to x-axis are (0, 0) and \(\left(\frac{2}{3},-\frac{4}{27}\right)\).
Now, eqn. of langent to given curve at (0, 0) is given by
y – 0 = 0 (x – 0)
⇒ y = 0
and eqn. of tangent to given curve at point \(\left(\frac{2}{3},-\frac{4}{27}\right)\) is given by
y + \(\frac{4}{27}\) = 0
⇒ 27y + 4 = 0.

Question 6.
Find the equations of the tangents to the curve y = (x³ – 1) (x – 2) at the points where the curve cuts the x-axis.
Solution:
Given curve be,
y = (x³ – 1) (x – 2) …………….(1)
It cuts x-axis
∴ y = 0
⇒ (x³ – 1) (x – 2) = 0
⇒ x = 1, 2
eqn. (1) cuts x-axis at (1, 0) and (2, 0).
Diff. (1) w.r.t. x, we get
\(\frac{d y}{d x}\) = 4x³ – 6x² – 1
∴ slope of tangent to given curve at (1, 0) = \(\left.\frac{d y}{d x}\right]_{(1,0)}\)
= 4 – 6 – 1 = – 3
and slope of to given curve tangent at (2, 0) = \(\left.\frac{d y}{d x}\right]_{(2,0)}\)
= 32 – 24 – 1 = 7
eqn. of tangent at (1, 0) is given by
y – 0 = – 3 (x – 1)
⇒ 3x + y – 3 = 0
and eqn. of tangent at (2, 0) is given by y – 0 = 7 (x – 2)
⇒ 7x – y – 14 = 0.

Question 7.
Find the equation of the tangent to the curve y = 3x³ + 4x² + 2x at the point whose abscissa is – 1. Also find the coordinates of another point on the curve at which the tangent is parallel to that already obtained.
Solution:
Given eqn. of curve be,
y = 3x³ + 4x² + 2x ………..(1)
Duff. eqn. (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = 9x² + 8x + 2
slope of tangent to given curve (1) at x = – 1
= (\(\frac{d y}{d x}\))_{x = – 1}
= 9 – 8 + 2 = 3
When x = – 1
∴ from (1);
y – 3 + 4 – 2 = – 1
Thus the point on curve (1) be (- 1, – 1).
Hence the eqn. of tangent to given curve (1) at p (- 1, – 1) is given by
y + 1 = 3 (x + 1)
⇒ 3x – y + 2 = 0 ………….(2)
Let (x1, y) be any point on curve (1).
∴ y₁ = 3x₁³ + 4x₁² + 2x₁ ………..(3)
Thus slope of tangent to curve (1) at (x₁, y₁) = (\(\frac{d y}{d x}\))_{(x₁, y₁)}
= 9x₁² + 8x₁ + 2
Since the tangent is parallel to eqn. (2) whose slope be \(\frac{-3}{-1}\) = 3
Thus, 9x₁² + 8x₁ + 2 = 3
⇒ 9x₁² + 8x₁ – 1 = 0
⇒ (x₁ + 1) (9x₁ – 1) = 0
⇒ x₁ = – 1, \(\frac{1}{9}\)
when x₁ = \(\frac{1}{9}\)
∴ from (2) ; we have
y₁ = \(\frac{1}{243}+\frac{4}{81}+\frac{2}{9}\)
= \(\frac{1+12+54}{243}\)
⇒ y₁ = \(\frac{67}{243}\)
Hence the required point on given curve be (\(\frac{1}{9}\), \(\frac{67}{243}\)).

Question 8.
Find the slope of the tangent to the curve x = t² + 31 – 8, y = 2t² – 2t – 5 at the point (2, – 1). (NCERT)
Solution:
Given eqn. of curve be;
x = t² + 3t – 8 …………(1)
and y = 2t² – 2t – 5 ………….(2)
Duff. eqn. (1) and eqn. (2) ; w.r.t. r, we have
∴ \(\frac{d x}{d t}\) = 2t + 3;
\(\frac{d y}{d t}\) = 4t – 2
Thus, \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{4 t-2}{2 t+3}\) …………..(3)
at given point (2, – 1);
⇒ t² + 3t – 8 = 2
⇒ t² + 3t – 10 = 0
⇒ (t – 2) (t + 5) = 0
⇒ t = 2, – 5 and y = 1
⇒ – 1 = 2t² – 2t – 5
⇒ 2t² – 2t – 4 = 0
⇒ t² – t – 2 = 0
⇒ (t + 1) (t – 2) = 0
⇒ t = – 1 , 2
Hence the common value of be 2.
∴ slope of tangent to given curve at t = 2 = (\(\frac{d y}{d x}\))_{t = 2}
= \(\frac{8-2}{4+3}=\frac{6}{7}\)

Question 9.
Find the equation of the tangent to the curve x = a sin³ t, y = b cos³ t at any point ‘t’.
Solution:
Curve eqn. of curve be
x = a sin³ t
and y = b cos³ t
Diff. eqn. (1) and eqn. (2) w.r.t. t, we have
\(\frac{d x}{d t}\) = 3a sin² t cos t;
\(\frac{d y}{d t}\) = 3b cos² t (- sin t)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{-3 b \cos ^2 t \sin t}{3 a \sin ^2 t \cos t}\)
= – \(\frac{b}{a}\) cot t
Now at any point t, we mean the point (a sin³ t, b cos³ t) on given curve.
Thus, required equation of tangent to given curve at (a sin³ t, b cos³ t) is given by
y – b cos³ t = – \(\frac{b}{a}\) cot t (x – a sin³ t)
⇒ ay – ab cos³ t = – bx cot t + ba sin³ t cot t
⇒ ay + bx cot t = ab cos t (sin² t + cos² t)
⇒ ay + bx cot t = ab cos t
⇒ \(\frac{a y}{a b \cos t}+\frac{b x \cot t}{a b \cos t}\) = 1
⇒ \(\frac{y}{b \cos t}+\frac{x}{a \sin t}\) = 1

Question 10.
Find the condition for the curves \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 and xy = c² to intersect orthogonally. (NCERT Exampler)
Solution:
Given eqns. of curves be
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 ………(2)
and xy = c²
Let (x₁, y₁) be the point of intersection of two given curves (1) and (2).
Diff. (1) both sides w.r.t. x; we get
\(\frac{2 x}{a^2}-\frac{2 y}{b^2} \frac{d y}{d x}\) = 0
∴ \(\frac{d y}{d x}=\frac{b^2 x}{a^2 y}\)
Thus, m₁ = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{b^2 x_1}{a^2 y_1}\)
Diff. (2) both sides w.r.t. x; we get
x \(\frac{d y}{d x}\) + y = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{y}{x}\)
Now two curves intersect orthogonally iff m₁m₂ = – 1
⇒ \(\left(\frac{b^2 x_1}{a^2 y_1}\right)\left(-\frac{y_1}{x_1}\right)\) = – 1
⇒ b² = a²

Question 11.
Prove that the curves y² = 4ax and xy = c² cut at right angles if c⁴ = 32a⁴.
Solution:
Given curves are y² = 4ax ……….(1)
and xy = c² …………(2)
From (1) and (2) ; we have
\(\frac{c^4}{x^2}\) = 4
⇒ c⁴ – ax³ = 0
⇒ x = \(\left(\frac{c^4}{4 a}\right)^{1 / 3}\)
∴ From (2) ; we have
y = \(\frac{c^2}{x}=\frac{c^2}{c^{4 / 3}}(4 a)^{1 / 3}\)
= \(\frac{(4 a)^{1 / 3}}{c^{-2 / 3}}\)
Thus both curves intersect at \(\mathrm{P}\left[\left(\frac{c^4}{4 a}\right)^{1 / 3}, \frac{(4 a)^{1 / 3}}{c^{-2 / 3}}\right]\)
Diff. (1) w.r.t. x, we have
\(\frac{d y}{d x}=\frac{2 a}{y}\)
∴ m₁ = \(\left.\frac{d y}{d x}\right]_{\mathrm{at} P}=\frac{(2 a) \cdot c^{-2 / 3}}{(4 a)^{1 / 3}}\)
Diff. (2) w.r.t. x, we have
∴ m₂ = \(\frac{d y}{d x}\)]_{at P}
= – \(\frac{(4 a)^{1 / 3}}{c^{-2 / 3}} \times \frac{(4 a)^{1 / 3}}{c^{4 / 3}}\)
= – \(\frac{(4 a)^{2 / 3}}{c^{2 / 3}}\)
Since the given curves cut right angles at P
∴ m₁m₂ = – 1
⇒ \(\frac{(2 a) c^{-2 / 3}}{(4 a)^{1 / 3}} \cdot\left(-\frac{(4 a)^{2 / 3}}{c^{2 / 3}}\right)\) = – 1
⇒ c^4/3 = 2 . 4^1/3 a^4/3
On cubing ; we have
c⁴ = 32 a⁴.

Question 12.
Using differentials, find the approximate values of:
(i) (81.5)^1/4 (NCERT)
(ii) \(\left(\frac{17}{81}\right)^{1 / 4}\) (NCERT)
(iii) (33)^-1/5 (NCERT)
Solution:
(i) Let y = f(x) = x^1/4
Take x = 81,
x + ∆x = 81.5
⇒ ∆x = 0.5
when x = 81, then y (81)^1/4 = 3,
Let dx = ∆x = 0.5
Now y = x^1/4
∴ \(\frac{d y}{d x}=\frac{1}{4} x^{\frac{-3}{4}}\)
∴ \(\left.\frac{d y}{d x}\right]_{x=81}=\frac{1}{4}(81)^{-3 / 4}=\frac{1}{108}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{108}\) × 0.5
= \(\frac{0.5}{108}\)
∴ (81.5)^1/4 = y + ∆y
= 3 + \(\frac{0.5}{108}\) = 3.004

(ii) Let y = f(x) = x^1/4
Take x = \(\frac{16}{81}\)
⇒ x + ∆x = \(\frac{17}{81}\)
∴ ∆x = \(\frac{1}{81}\)
when x = \(\frac{16}{81}\)
then y = \(\left(\frac{16}{81}\right)^{1 / 4}=\frac{2}{3}\)
Let ∆x = dx = \(\frac{1}{81}\)
Now y = x^1/4
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{4}\) x^{– 3/4}
⇒ \(\left.\frac{d y}{d x}\right]_{x=\frac{16}{81}}=\frac{1}{4}\left(\frac{16}{81}\right)^{-3 / 4}\)
= \(\frac{1}{4}\left(\frac{2}{3}\right)^{-3}\)
= \(\frac{1}{4}\left(\frac{3}{2}\right)^3=\frac{27}{32}\)
∴ dy = \(\frac{d y}{d x}\) ∆x
= \(\frac{27}{32} \times \frac{1}{81}=\frac{1}{96}\)
= ∆y [∵ dy ≅ ∆y]
∴ \(\left(\frac{17}{81}\right)^{1 / 4}\) = y + ∆y
= \(\frac{2}{3}+\frac{1}{96}\) = 0.677.

(iii) Let y = f(x) = x^{– 1/5}
Take x = 32,
x + ∆x = 33
∆x = 1
When x = 32 then y = (32)^{– 1/5} = \(\frac{1}{2}\)
Let dx = ∆x = 1
Now y = x^{– 1/5}
⇒ \(\frac{d y}{d x}\) = – \(\frac{1}{5}\) x^{– 6/5}
∴ \(\frac{d y}{d x}\)]_{x = 32} = – \(\frac{1}{5}\) x^{– 6/5}
= – \(\frac{1}{320}\)
∴ dy = \(\frac{d y}{d x}\) dx
= – \(\frac{1}{320}\) × 1
= – \(\frac{1}{320}\) = ∆y
∴ (33)^{– 1/5} = y + ∆y
= \(\frac{1}{2}-\frac{1}{320}\) = 0.497.

Question 13.
Find the approximate change in the volume of a cube ofsidex metres caused by increasing the side by 3%.(NCERT)
Solution:
Given the length of each side of cube be x m.
Then volume of cube = V = x³
∴ \(\frac{d V}{d x}\) = 3x²
Let ∆x be the change in x and the corresponding change in V be ∆V.
∴ ∆V = \(\frac{d V}{d x}\) ∆x [∵ ∆V = dV ; ∆x = dx]
= 3x² ∆x
also it is given that \(\frac{\Delta x}{x}\) × 100 = 3
∆x = \(\frac{3 x}{100}\) = 0.03x
∴ ∆V = 3x² × 003x = 0.09 x³ m³.

Question 14.
If f(x) = 3x² + 15x + 5, then find the approximate value of f (3.02). (NCERT)
Solution:
Given f(x) = y = 3x² + 15x + 5
Take x = 3, x + ∆x = 3.02
⇒ ∆x = 0.02
When x = 3 then y = 3.3² + 153 + 5
⇒ y = 27 + 45 + 5 = 77,
Let ∆x = dx = 0.02
Now y = 3x² + 15x + 3
∴ \(\frac{d y}{d x}\) = 6x + 15
⇒ \(\frac{d y}{d x}\)]_{x = 3} = 6 × 3 + 15 = 33
∴ dy = \(\frac{d y}{d x}\) ∆x
= 33 × 0.02
= 0.66 = ∆y
[∵ ∆y ≅ dy]
∴ f (3.02) = y + ∆y
= 77 + 0.66 = 77.66.

Question 15.
Show that the relative error in computing the volume of a sphere, due to an error in measuring its radius, is approximately equal to three times the relative error in the radius.
Solution:
Let r be the radius of sphere
Then V = volume of sphere = \(\frac{4 \Pi}{3}\) r³
Taking logarithm on both sides, we get
log V = log \(\frac{4 \Pi}{3}\) + log r³
log V = log \(\frac{4 \Pi}{3}\) + 3 log r
On differentiating ; we get
\(\frac{1}{V}\) dV = 0 + \(\frac{3}{r}\) dr
[∵ dV = ∆V and dr = ∆r]
\(\frac{\Delta \mathrm{V}}{\mathrm{V}}=3 \times\left(\frac{\Delta r}{r}\right)\)
∴relative error in V = 3 × (relative error in r)
Hence, the relative error in computing the volume of sphere is equal to 3 times the relative error in radius of sphere.

Question 16.
Determine for which values of x, the function f(x) = \(\frac{x}{x^2+1}\) is increasing and for which values of x, it is decreasing. Find also the points on the graph of the function at which the tangent is parallel to x-axis.
Solution:
Given f(x) = \(\frac{x}{x^2+1}\)
Diff. both sides w.r.t. x, we get
f'(x) = \(=\frac{\left(x^2+1\right) \cdot 1-x \cdot 2 x}{\left(x^2+1\right)^2}\)
= \(\frac{1-x^2}{\left(1+x^2\right)^2}\)
Now f'(x) ≥ o iff \(\frac{1-x^2}{\left(1+x^2\right)^2}\) ≥ 0
⇒ 1 – x² ≥ 0
[∵ (1 + x²)² > 0 ∀ x ∈ R]
⇒ x² ≤ 1
⇒ |x| ≤ 1
⇒ – 1 ≤ x ≤ 1
Thus, f(x) is increasing in [- 1, 1]
Now f'(x) ≤ 0 iff \(\frac{1-x^2}{\left(1+x^2\right)^2}\) > 0
⇒ 1 – x² ≤ 0
[∵ (1 + x²)² > 0 ∀ x ∈ R]
⇒ x² ≥ 1
⇒ |x| ≥ 1
⇒ x ≥ 1 or x ≤ – 1
Thus, the function f(x) is decreasing in (- ∞, – 1] ∪ [1, ∞).
Let (x₁, y₁) be any point on given curve
∴ y₁ = \(\frac{x_1}{x_1^2+1}\) …………(2)
Now slope of tangent to given curve at (x₁, y₁) = [f'(x)]_{(x₁, y₁)}
= \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= \(\frac{1-x_1^2}{\left(1+x_1^2\right)^2}\)
Also tangent to given curve is || to x-axis.
∴ (\(\frac{d y}{d x}\))_{(x₁, y₁)} = 0
⇒ \(\frac{1-x_1^2}{\left(1+x_1^2\right)^2}\) = 0
⇒ 1 – x₁² = 0
⇒ x₁ = ± 1
When x₁ = 1
∴ from (2) ;
y₁ = \(\frac{1}{1+1}=\frac{1}{2}\)
When x₁ = – 1
∴ from (2) ;
y₁ = \(\frac{-1}{1+1}=-\frac{1}{2}\)
Hence the required points on graph of function at which tangent is || to x-axis be (1, \(\frac{1}{2}\)) and (- 1, – \(\frac{1}{2}\)).

Question 17.
Find the points at which the function f given by f(x) = (x – 2)⁴ (x + 1)³ has
(i) local maxima
(ii) local minima
(iii) point of inflexion. (NCERT)
Solution:
Given, f(x) = (x – 2)⁴ (x + 1)³
Diff. both sides w.r.t. x ; we have
f'(x) = (x – 2)⁴ 3 (x + 1)² + (x + 1)³ 4 (x – 2)³
= (x + 1)² (x – 2)³ [3 (x – 2) + 4 (x + 1)]
= (x + 1)² (x – 2)³ (7x – 2)
For critical points, f’ (x) = 0
⇒ (x + 1)² (x – 2)³ (7x – 2) = 0
⇒ x = – 1, 2, \(\frac{2}{7}\)

Case – I:
at x = – 1
When x slightly < – 1
⇒ x + 1 < 0 also x < – 1 < 2
⇒ x – 2 < 0
∴ f’ (x) = (+ve) (- ve) (- ve) = + ve
When x slightly > – 1
⇒ x + 1 > 0, x – 2 < 0
f’ (x) = (+ ve) (- ve) (- ve) = + ve
So f’ (x) does not changes its sign as we move from slightly < – 1 to slightly > – 1.
∴ x = – 1 be a point of neither maxima nor minima.
Hence x = – 1 be a point of inflexion.

Case-II :
at x = 2
When x slightly < 2
⇒ x – 2 < 0 but 7x – 2 > 0
∴f’ (x) = (+ve) (- ve) (+ve) = – ve
When x slightly > 2
⇒ x – 2 > 0 and 7x – 2 > 0
∴ f’ (x) = (+ ve) (+ ve) (+ ve) = + ve
Thus, f ‘(x) changes its sign from -ve to + ve as we move from slightly < 2 to slightly > 2
∴ x = 2 is a point of minima.

Case – III :
at x = \(\frac{2}{7}\)
When x slightly < \(\frac{2}{7}\)
⇒ 7x – 2 < 0 and x – 2 < 0 ∴ f'(x) = (+ ve) (- ve) (- ve) = + ve When x slightly > \(\frac{2}{7}\)
⇒ 7x – 2 > 0 and x – 2 < 0
∴ f’ (x) = (+ ve) (- ve) (+ ve) = – ve
Thus, f’ (x) changes its sign from + ve to – ve as we move from slightly < \(\frac{2}{7}\) to 2 slightly > \(\frac{2}{7}\).
∴ x = \(\frac{2}{7}\) be a point of local maxima.

Question 18.
Find the points of local maximum and minima (if any) of the function :
f(x) = 2 cos x + x in [0, π].
Find also the absolute maximum and minimum values.
Solution:
Given f(x) = 2 cos x + x
Diff. both sides w.r.t. x, we have
f’ (x) = – 2 sin x + 1
f” (x) = – 2 cos x
For maxima/minima, f’ (x) = 0
⇒ – 2 sin x + 1 = 0
⇒ sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
⇒ x = nπ + (- 1)ⁿ \(\frac{\pi}{6}\) ∀ n ∈ I
but x ∈ [0, π]
∴ x = \(\frac{\pi}{6}\), \(\frac{5 \pi}{6}\) ∈ [0, π]
Thus, the critical points are 7t 571
x = 0, \(\frac{\pi}{6}\), \(\frac{5 \pi}{6}\), π
Now at x = \(\frac{\pi}{6}\)
f” (x) = – 2cos \(\frac{\pi}{6}\)
= – 2 × \(\frac{\sqrt{3}}{2}\)
= – √3 < 0
∴ x = \(\frac{\pi}{6}\) be a point of local maxima.
and f” (\(\frac{5 \pi}{6}\)) = – 2 cos (π – \(\frac{\pi}{6}\))
∴ x = \(\frac{5 \pi}{6}\) be a point of local minima.
Also f” (0) = – 2 cos 0 = – 2 < 0 ∴ x = 0 be a point of local maxima, and f” (π) = – 2 cos π = 2 > 0
∴ x = π be a point of local minima.
∴ f(0) = 2 cos 0 + 0 = 2
f(π) = 2 cos π + π = – 2 + π
f(\(\frac{\pi}{6}\)) = 2 cos \(\frac{\pi}{6}\) + \(\frac{\pi}{6}\)
= √3 + \(\frac{\pi}{6}\)
f(\(\frac{5 \pi}{6}\)) = 2 cos \(\frac{5 \pi}{6}\) + \(\frac{5 \pi}{6}\)
= – √3 + \(\frac{5 \pi}{6}\)
∴ absolute minimum value = – √3 + \(\frac{5 \pi}{6}\)
and absolute maximum value = √3 + \(\frac{\pi}{6}\).

Question 19.
Find the maximum value of the function 2x³ – 24x + 107 in the interval [1, 3]. Also find the maximum value of the same function in the interval [- 3, – 1]. (NCERT)
Solution:
Given f(x) = 2x³ – 24x + 107
∴ f (x) = 6x² – 24
= 6 (x – 2) (x + 2)
For critical points we have f’ (x) = 0
⇒ 6 (x – 2) (x + 2) = 0
⇒ x = 2, – 2
if f(x) be defined on [1, 3]
∴ f’ (x) = 0 at x = 2
Now let us compute the values of f(x) at critical point x = 2 and also at the end point of given interval [1, 3].
∴ f(2) = 16 – 48 + 107 = 123 – 48 = 75
f(1) = 2 – 24 + 107 = 109 – 24 = 85
f(3) = 54 – 72 + 107 = 161 – 72 = 89
out of these values, maximum value of f(x) be f(3) = 89
∴ absolute max. value = 89 at x = 3.
if f(x) be defined in [- 3, – 1]
Then f’ (x) = 0 at x = – 2.
Thus we compute the values of f (x) at critical point x = – 2 and also at the end points of given interval [- 3, – 1]
∴ f(- 2) = – 16 + 48 + 107 = 139
f (- 1) = -2 + 24 + 107 = 129
f(- 3) = – 54 + 72 + 107 = 125
Out of these values, maximum value of f (x) be f(- 2) = 139.
Hence, absolute maximum value = 139 at x = – 2.

Question 20.
OABC is a square of side 1 m. A point X in AB and a point Y in BC are such that AX = x m and BY = k x m and BY is longer than AX. For a given value of k, show that the minimum area of AOXY is ((4k – 1)/8k) m².
Solution:
Given OABC be a square s.t OA = AB = BC = OC = 1 m.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Chapter Test 1

Given AX = x m and BY = kx
and BY > AX
From figure, it is clear that
Let S = area of ∆OXY
= area of square OABC – area of ∆OAX – area of ∆XBY – area of ∆OCY
⇒ S = 1² – \(\frac{1}{2}\) × OA × AX – \(\frac{1}{2}\) × BY × BX – \(\frac{1}{2}\) × OC × CY
⇒ S = 1 – \(\frac{1}{2}\) × 1 × x – \(\frac{1}{2}\) × kx (1 – x) – \(\frac{1}{2}\) × 1 × (1 – kx)
⇒ S = 1 – \(\frac{x}{2}-\frac{k x}{2}+\frac{k x^2}{2}-\frac{1}{2}+\frac{k x}{2}\)
S = \(\frac{1}{2}-\frac{x}{2}+\frac{k x^2}{2}\)
On differentiating w.r.t. x, we have
\(\frac{d \mathrm{~S}}{d x}\) = – \(\frac{1}{2}\) + kx ;
\(\frac{d^2 S}{d x^2}\) = k
For maxima/minima, \(\frac{d \mathrm{~S}}{d x}\) = 0
⇒ – \(\frac{1}{2}\) + kx = 0
⇒ x = \(\frac{1}{2 k}\)
∴ \(\left(\frac{d^2 \mathrm{~S}}{d x^2}\right)_{x=\frac{1}{2 k}}\) = k > 0
Hence S is minimise for x = \(\frac{1}{2 k}\)
∴ Minimum area of ∆OXY = S = \(\frac{1}{2}\left[1-\frac{1}{2 k}+k \times \frac{1}{4 k^2}\right]\)
= \(\frac{1}{2}\left[1-\frac{1}{4 k}\right]\)
= \(\left(\frac{4 k-1}{8 k}\right)\) m².

Question 21.
Find the area of the largest isosceles triangle having perimeter 18 metres.
Solution:
Let x metre be the lengths of equal side of isosceles triangle.
Since the perimeter of an isosceles ∆ be 18 cm.
∴ length of base of isosceles A be (18 – 2x) m
Then by Heron’s formula, we have
A = area of isosceles triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) ………..(1)
given s = \(\frac{a+b+c}{2}=\frac{18}{2}\) = 9
from (1) ; we have
A = \(\sqrt{9(9-x)(9-x)(9-18+2 x)}\)
= \(\sqrt{9(9-x)^2(2 x-9)}\)
Since the sum of two sides of ∆ is greater than third side.
x + x > 18 – 2x
⇒ 4x > 18
⇒ x > \(\frac{9}{2}\)
and x – x< 18 – 2x
⇒ 0 < 18 – 2x
⇒ 2x < 18
⇒ x < 9
Thus \(\frac{9}{2}\) < x < 9.
∴ A = 3 (9 – x) \(\sqrt{2 x-9}\)
Now A is maximum when A² is maximum.
Let P = A² = 9 (9 – x)² (2x – 9)
Diff. both sides w.r.t. x, we have
\(\frac{d P}{d x}\) = 9 [(9 – x)² . 2 + (2x – 9) 2 (9 – x) (- 1)]
= 18 (9 – x) [9 – x – 2x + 9]
= 18 (9 – x) (18 – 3x)
= 54 (9 – x) (6 – x)
= 54 (x² – 15x + 54)
and \(\frac{d^2 P}{d x^2}\) = 54 (2x – 15)
For maxima/minima, \(\frac{d P}{d x}\) = 0
⇒ 54 (x² – 15x + 54) = 0
⇒ (9 – x) (6 – x) = 0
⇒ x = 6, 9
But \(\frac{9}{2}\) < x < 9
∴ x = 6
∴ \(\left(\frac{d^2 \mathrm{P}}{d x^2}\right)_{x=6}\) = 54 (12 – 15) = – 162 < 0
Thus x = 6 is a point of maxima.
∴ P is maximum for x = 6
Thus A is maximum for x = 6.
and Maximum area of isosceles triangle = 3 (9 – 6) \(\sqrt{12-9}\) = 9√3 m².

Question 22.
Divide the number 4 into two positive numbers such that the sum of square of one and the cube of other is minimum.
Solution:
Let the required two positive numbers bex and 4 – x
since the sum of two positive numbers be 4.
Let P = (4 – x)² + x³ ;
Diff. both sides w.r.t. x, we get
\(\frac{d P}{d x}\) = 2 (4 – x) (- 1) + 3x²
∴ \(\frac{d^2 P}{d x^2}\) = 2 + 6x
For maxima/minima, \(\frac{d P}{d x}\) = 0
⇒ – 8 + 2x + 3x² = 0
⇒ (x + 2) (3x – 4) = 0
⇒ x = – 2, \(\frac{4}{3}\)
but x > 0
∴ x = \(\frac{4}{3}\)
Thus, \(\left(\frac{d^2 \mathrm{P}}{d x^2}\right)_{x=\frac{4}{3}}\) = 2 + 6 × \(\frac{4}{3}\)
= 2 + 8
= 10 > 0
∴ x = \(\frac{4}{3}\) be a point of minima.
∴ P is minimise at x = \(\frac{4}{3}\).
Hence, the required numbers are \(\frac{4}{3}\) and 4 – \(\frac{4}{3}\).
i.e. \(\frac{4}{3}\) and \(\frac{8}{3}\).

Question 23.
Find the point on the curve y² = 2x which is nearest to the point A (1, – 4).
Solution:
Let P (x, y) be any point on the curve
y² = 2x ………….(1)
and let the given point be A (1, – 4).
∴ AP² = (x – 1)² + (y + 4)²
and let S = AP².
Then S is max/min. according as AP is maximum/minimum.
Now S = \(\left(\frac{y^2}{2}-1\right)^2\) + (y + 4)²
∴ \(\frac{d S}{d y}\) = 2 \(\left(\frac{y^2}{2}-1\right)\) y + 2 (y + 4)
= y³ + 8
For Max/Min. \(\frac{d S}{d y}\) = 0
∴ y³ = – 8
= (- 2)³
⇒ y = – 2
∴from (1) ; x = 2
Now \(\frac{d^2 S}{d y^2}\) = 3y²
∴ \(\left(\frac{d^2 \mathrm{~S}}{d y^2}\right)_{y=-2}\) = 12 > 0
∴ S is minimise for y = – 2 and x = 2 and hence the required point be (2, – 2).

Question 24.
A wire of length 25 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the square and the circle Is minimum ?
Solution:
Let the length of one piece of wire be x m
∴ Length of other piece must be (25 – x) m.
Let ‘x’ metres be made into circle and
(25 – x) metres be made into square.
So perimeter of circle = x = 2πr,
where r = radius of circle
∴ r = \(\frac{x}{2 \pi}\)
thus area of circle = πr²
Further perimeter of square = 4 × side
∴ side of square = \(\frac{25-x}{4}\)
Thus area of square = \(\left(\frac{25-x}{4}\right)^2\)
Let A = combined area of circle and square
= π \(\frac{x^2}{4 \pi^2}+\left(\frac{25-x}{4}\right)^2\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Chapter Test 2

∴ A is manimise for x = \(\frac{25 \pi}{\pi+4}\)
Hence the length of two pieces are \(\), 25 – \(\frac{25 \pi}{\pi+4}\)
i.e. \(\frac{25 \pi}{\pi+4}\) m and \(\frac{100}{\pi+4}\) m.

Question 25.
Find the dimensions of the rectangle of maximum area that can be inscribed in the portion of the parabola y² = 4px intercepted by the line x = a.
Solution:
Let ABDC be he rectangle that is inscribed in the portion of given parabola y² = 4px intercepted by the line x = a.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Chapter Test 3

Let (x, y) be the coordinates of point A.
So AC = 2y and CD = a – x
∴ S = area of rectangle
= (a – x) 2y
= (a – \(\frac{y^2}{4 p}\)) 2y
Diff., both sides w.r.t. y ; we have
\(\) ;
\(\frac{d^2 \mathrm{~S}}{d y^2}=-\frac{12 y}{4 p}=-\frac{3 y}{p}\)
For maxima / minima, \(\frac{d S}{d y}\) = 0
⇒ 2a – \(\frac{6 y^2}{4 p}\) = 0
⇒ 6y² = 8ap
⇒ y = ± \(\sqrt{\frac{4 a p}{3}}\)
= ± \(\frac{2}{\sqrt{3}} \sqrt{a p}\)
∴ \(\left(\frac{d^2 \mathrm{~S}}{d y^2}\right)_{y=\frac{2}{\sqrt{3}} \sqrt{a p}}=-\frac{3}{p} \times \frac{2}{\sqrt{3}} \sqrt{a p}\) < 0
Thus S is maximise for
y = \(\frac{2}{\sqrt{3}} \sqrt{a p}\)
= \(\frac{2}{3} \sqrt{3 a p}\)
∴ x = \(\frac{y^2}{4 p}\)
= \(\frac{1}{4 p} \times \frac{4}{9}\) × 3ap = \(\frac{a}{3}\)
Thus, the dimensions of the rectangle are a – x and 2y
i.e. a – \(\frac{a}{3}\) and 2 × \(\frac{2}{3} \sqrt{3 a p}\)
i.e. \(\frac{2 a}{3}\) and \(\frac{4}{3} \sqrt{3 a p}\).

Question 26.
An open topped box is to be constructed by removing equal squares from each corner of a 3 metres by 8 metres rectangular sheet of aluminimum and folding up the sides. Find the volume of the largest such box.(NCERT)
Solution:
Given sides of rectangular sheet be 3 m and 8 m.
Let x units be the length of each side of squares of same size removed from each corner of the sheet.
Thus the dimensions of the open box formed by folding up the flaps are;

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Chapter Test 4

Length = 8 – 2x;
breadth = 3 – 2x
and height = x
Let V = volume of cuboid = (8 – 2x)(3 – 2x) x
∴ V = (8 – 2x) (3x – 2x²)
= 4x³ – 22x² + 24x
Diff. both sides w.r.t. x, we have
\(\frac{d V}{d x}\) = 12x² – 44x + 24 ;
\(\frac{d^2 V}{d x^2}\) = 24x – 44
For maxima / minima, \(\frac{d V}{d x}\) = 0
⇒ 4 (3x² – 11x + 6) = 0
⇒ (x – 3) (3x – 2) = 0
⇒ x = 3, \(\frac{2}{3}\)
Further x ≠ 3 if x = 3,
breadth = 3 – 6 = – 3, which is not possible
Thus, \(\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=\frac{2}{3}}\) = 24 × \(\frac{2}{3}\) – 44
= 16 – 44 = – 28 > 0
∴ x = \(\frac{2}{3}\) be a point of maxima.
Thus V is maximum when x = \(\frac{2}{3}\).
∴ Maximum volume of the box = \(\left(8-\frac{4}{3}\right)\left(3-\frac{4}{3}\right) \frac{2}{3}\)
= \(\frac{20}{3} \times \frac{5}{3} \times \frac{2}{3}\)
= \(\frac{200}{27}\) m³.

Question 27.
Show that the maximum volume of a cylinder which can be inscribed in a cone of height h, and semi-vertical angle 30° is \(\frac{4}{81}\) πh³.
Solution:
Let x be the radius of cylinder andy be the height of cylinder
Then, AF = h – y
and FG = x
∴ tan 30° = \(\frac{x}{h-y}\)
⇒ x = (h – y) \(\frac{1}{\sqrt{3}}\) ……….(1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives chapter Test 5

Let V = volume of cylinder = \(\frac{\pi}{3}\) (h – y)² . y
∴ \(\frac{d V}{d y}\) = \(\frac{\pi}{3}\) [(h – y)² + 2y (h – y) (- 1)]
∴ \(\frac{d^2 V}{d y^2}\) = \(\frac{\pi}{3}\) [- 2 (h – y) – 2 (h – 2y)]
= \(\frac{\pi}{3}\) [- 4h + 6y]
For max/min, \(\frac{d V}{d y}\) = 0
⇒ (h – y) [(h – y) – 2y] = 0
⇒ (h – y) (h – 3y) = 0
⇒ y = h, \(\frac{h}{3}\) [but y ≠ h as cylinder is inscribed in cone]
and \(\left(\frac{d^2 \mathrm{~V}}{d y^2}\right)_{y=\frac{h}{3}}\) = \(\frac{\pi}{3}\) [- 4h + 2h]
= \(\frac{-2 \pi h}{3}\) < 0
∴ V is maximise for y = \(\frac{h}{3}\)
and Required Max. value = \(\frac{\pi}{3}\left(\frac{2 h}{3}\right)^2 \times \frac{h}{3}=\frac{4 \pi}{81}\) h³.

Question 28.
A cylinder is such that the sum of its height and circumference of its base is 10 metres. Find the maximum volume of the cylinder.
Solution:
Let h metres be the height and r metres be the radius of cylinder.
Then h + 2πr = 10 …………(1)
Let V = volume of cylinder = πr²h
V = πr² (10 – 2πr)
Diff, both sides w.r.t. r we have
\(\frac{d V}{d r}\) = π (20r – 6πr²)
∴ \(\frac{d^2 V}{d r^2}\) = π (20 – 12πr)
For maxima / minima, \(\frac{d V}{d r}\) = 0
⇒ π (20r – 6πr²) = 0
⇒ (20 – 6πr) r = 0
⇒ r = 0, \(\frac{10}{3 \pi}\)
Since r > 0
∴ r = \(\frac{10}{3 \pi}\) m
∴ \(\left(20-12 \pi \times \frac{10}{3 \pi}\right)\) = – 20π < 0
Thus, V is maximise at r = \(\frac{10}{3 \pi}\) metre
and h = 10 – \(\frac{20 \pi}{3 \pi}\) = \(\frac{10}{3}\) m
and Maximum volume of cylinder = πr²h
= \(\pi \times\left(\frac{10}{3 \pi}\right)^2 \times \frac{10}{3}\) m³
= \(\frac{1000}{27 \pi}\) m³

Question 29.
Find the area of the largest rectangle in the first quadrant with two sides on x-axis and y- axis and one vertex on the curve)’ = 12 – x².
Solution:

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Chapter Test 6

Let P (x, y) be any point on the given curve y = 12 – x² in the first quadrant.
Let M and N be the foot of perpendiculars drawn from P on y-axis and x-axis respectively.
Then, A area of rectangle = xy = x (12 – x²)
Diff. both sides w.r.t. x, we have
\(\frac{d A}{d x}\) = 12 – 3x²
For maxima / minima, \(\frac{d A}{d x}\) = 0
⇒ 12 – 3x² = 0
⇒ x² = 4
⇒ x = ± 2
at x = 2,
\(\frac{d^2 A}{d x^2}\) = – 12 < 0
∴ x = 2 be a point of maxima.
Hence A will be maximum at x = 2.
∴ maximum area of rectangle = 2 (12 – 2²) = 2 (12 – 4) = 16 units.

Question 30.
A tank with rectangular base and rectangular sides, open at the top Is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs ₹ 70 per sq. metre for the base and ₹ 45 per m² for sides, what is the cost of least expensive tank?
Solution:
Letx, x, y be the length, width and height of the tank
∴ area of square base = x2
and area of four walls = 4xy
Given V = volume of open tank = 8x²y ………….(1)
∴ E = 70x² + 45 (4 xy)
= 70x² + 180x . \(\frac{8}{x^2}\) [using (1)]
⇒ E = 70x² + \(\frac{180 \times 8}{x}\)
Given y = depth of tank = 2m
∴ from (1);
8 = x² × 2
⇒ x = 2[∵ x > 0]
Now we check whether x = 2 is point of maximum or minima.
\(\frac{d E}{d x}\) = 140x – \(\frac{180 \times 8}{x^2}\)
\(\frac{d^2 E}{d x^2}\) = 140 + \(\frac{180 \times 16}{x^2}\)
at x = 2,
\(\frac{d^2 E}{d x^2}\) > 0
∴ E is minimum for x = 2.
∴ from (2) ;
least value of E = 70 (2)² + \(\frac{180 \times 8}{2}\) = ₹ 1000.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8

November 24, 2023

Access to comprehensive ML Aggarwal Class 12 Solutions ISC Chapter 7 Applications of Derivatives Ex 7.8 encourages independent learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8

Question 1.
Find two numbers whose sum is 24 and whose product is as large as possible. (NCERT)
Solution:
Let the one number be x, (0 < x < 24)
∴ other number be 24 – x because sum of two numbers be 24.
Let P = x (24 – x) ……..(1)
Duff. (1) w.r.t. x; we have
\(\frac{d P}{d x}\) = 24 – 2x
For maximum/minima, \(\frac{d P}{d x}\) = 0
⇒ 24 – 2x = 0
⇒ x = 12
∴ \(\frac{d^2 P}{d x^2}\) = – 2
⇒ (\(\frac{d^2 P}{d x^2}\))_{x = 12} = – 2 < 0
Thus, x = 12 be a point of local maxima and x = 12 be the only point of local maxima.
So the local maximum value be the absolute maximum value.
Thus, required numbers are 12 and (24 – 12) i.e. 12.

Question 2.
The sum of two positive numbers is 20. Find the numbers
(i) if the sum of their squares is minimum.
(ii) if the product of the square of one and the cube of the other is maximum.
Solution:
(i) Let the required numbers be x, 20 – x,
since the sum of two positive integers is 20.
Let S = x² + (20 – x)²
Diff. both sides w.r.t. x, we have
\(\frac{d S}{d x}\) = 2x + 2 (20 – x) (- 1)
= 2x – 40 + 2x
= 4x – 40
For maxima/minima, \(\frac{d S}{d x}\) = 0
⇒ 4x – 40 = 0
⇒ x = 10
∴ (\(\frac{d^2 P}{d x^2}\))_{x = 10} = 4 > 0
Thus x = 10 be a point of local minima.
∴ required numbers are 10 and (20 – 10) i.e. 10 and 10.

(ii) Let the required numbers be x and 20 – x
Since the sum of two numbers be 20.
Let P = x² (20 – x)³ ;
Diff. both sides w.r.t. x, we have
\(\frac{d P}{d x}\) = 3x² (20 – x)² (- 1) + (20 – x)³ 2x
= – x (20 – x)² [3x – 2 (20 – x)]
= – x (20 – x)² (5x – 40)
For maxima / minima, \(\frac{d P}{d x}\) = 0
⇒ x (20 – x)² (5x – 40) = 0
⇒ x = 0, 20, 8
But 0 < x < 20 ∴ x = 8
When x slightly < 8,
\(\frac{d P}{d x}\) = (- ve) (+ ve) (+ ve) = – ve
∴ \(\frac{d P}{d x}\) changes its sign from +ve to – ve.
Hence x = 8 be a point of local maxima.
Thus the required numbers are 8 and 20 – 8 i.e. 8 and 12.

Question 3.
The product of two positive numbers is 16. Find the numbers
(i) if their sum is least.
(ii) if the sum of one and the square of the other is least.
Solution:
(i) Let the two numbers be x and \(\frac{16}{x}\),
since the product of two numbers be 16.
Let S = x + \(\frac{16}{x}\)
∴ \(\frac{d S}{d x}\) = 1 – \(\frac{16}{x^2}\)
For maxima/minima, \(\frac{d P}{d x}\) = 0
⇒ 1 – \(\frac{16}{x^2}\) = 0
⇒ x² = 16
⇒ x = ± 4 but x, y ∈ N
∴ x = 4
and \(\frac{d^2 S}{d x^2}\) = \(\frac{32}{x^3}\)
∴ \(\left(\frac{d^2 S}{d x^2}\right)_{x=4}=\frac{32}{64}=\frac{1}{2}\) > 0
∴ x = 4 be a point local minima.
Thus S is least for x = 4
∴ required numbers are 4 and \(\frac{16}{4}\) i.e. 4 and 4.

(ii) Let the required two numbers be x and \(\frac{16}{x}\),
Since the product of two numbers be 16.
Let S = x² + 16/x
\(\frac{d S}{d x}\) = \(\frac{-16}{x^2}\) + 2x
∴ \(\frac{d^2 S}{d x^2}\) = \(\frac{32}{x^3}\) + 2
For maxima/minimum, \(\frac{d S}{d x}\) = 0
⇒ \(\frac{-16}{x^2}\) + 2x = 0
⇒ x³ = 8
⇒ x = 2
at x = 2;
\(\frac{d^2 S}{d x^2}\) = \(\frac{32}{8}\) + 2
= 4 + 2 = 6 > 0
∴ x = 2 be a point of local minima.
Thus S is minimum for x = 2.
Hence the required numbers are 2 and \(\frac{16}{2}\) i.e. 2 and 8.

Question 4.
Find the positive numbers x and y such that x + y = 60 and xy is maximum. (NCERT)
Solution:
Given x + y = 60 …………(1)
where x, y > 0
Let P = xy³
= y³ (60 – y)
∴ \(\frac{d P}{d y}\) = 180y² – 4y³
= 4y² [45 – y]
For maxima/minina, \(\frac{d P}{d y}\) = 0
⇒ 4y² (45 – y) = 0
⇒ y = 0, 45
but y > 0
∴ y = 45
and \(\frac{d^2 P}{d y^2}\) = 12 × 45 × (- 15) = – 8100 < 0
∴ P is maximise for y = 45
∴ from (1) ; x = 15.
Hence the required two numbers be 15, 45.

Question 5.
Find two positive numbers x andy such that their sum is 35 and the product x²y⁵ is maximum. (NCERT)
Solution:
Let the two positive numbers be x and y.
Given x + y = 35 ………..(1)
Let P product = x²y⁵
⇒ P = (35 – y²) y⁵
∴ \(\frac{d P}{d y}\) = (35 – y)² 5y⁴ ± y⁵ 2(35 – y) (- 1)
= y⁴ (35 – y) [5 (35 – y) – 2y]
= y⁴ (35 – y) (175 – 7y)
For maxima/minima, \(\frac{d P}{d y}\) = 0
⇒ y = 0, 35, 25
but y ≠ 0, 35
[∵ y > 0 and further if y = 35 then x = 0]
∴ possible value of y = 25.
When y slightly < 25 ⇒ \(\frac{d P}{d y}\) = (+ve) (+ve) (+ve) = +ve When y slightly > 25
⇒ \(\frac{d P}{d y}\) = (+ve) (+ve) (-ve) = -ve
∴ \(\frac{d P}{d x}\) varies from +ve to -ve
∴ P is maximise for y = 25
∴ from (1); x = 10
∴ Required two numbers are 10, 25.

Question 6.
(i) Find two positive numbers whose sum is 16 and the sum of whose cube is minimum.
Solution:
(i) Let the two numbers be x andy s.t. x, y > 0
also given, x + y = 16
Let S = x<sup3 + y³
= x³ + (16 – x)³
∴ \(\frac{d S}{d x}\) = 3x² + 3 (16 – x)² (- 1)
∴ \(\frac{d^2 S}{d x^2}\) = 3 [2x + 2 (16 – x)] = 96 > 0
For maximaJminima, \(\frac{d S}{d x}\) = 0
⇒ 3 [x² – (16 – x)²] = 0
⇒ x² – (256 + x² – 32x) = 0
⇒ x = \(\frac{256}{32}\) = 8
∴ (\(\frac{d^2 S}{d x^2}\))_{x = 8} = 3 (16 + 16) = 96 > 0
∴ S is minimum for x = 8 and
from (1) ; y = 8
Thus the required two numbers are 8 and 8.

Question 6 (old).
(i) Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. (NCERT)
Solution:
Let the two numbers be x and y s.t. x, y > 0
Given x + y = 15
Let S = x² + y²
= x² + (15 – x)² …………….(2)
∴ \(\frac{d S}{d x}\) = 2x + 2 (15 – x) (- 1)
= 2 [x – 15 + x]
= 2 (2x – 15);
\(\frac{d^2 S}{d x^2}\) = 4
For maxima/minima
∴ \(\frac{d S}{d x}\) = 0
⇒ x = \(\frac{15}{2}\)
∴ \(\frac{d^2 S}{d x^2}\) = 4 > 0
Thus S is minimise for x = \(\frac{15}{2}\)
∴ from (1) ;
y = 15 – \(\frac{15}{2}\) = \(\frac{15}{2}\)
Thus required two positive numbers are \(\frac{15}{2}\) and \(\frac{15}{2}\).

Question 7.
Three numbers are given whose sum is 180 and the ratio of first two of them is 1 : 2. If the product of the numbers is greatest, find the numbers. (ISC 2003)
Solution:
Let the first two numbers be x and 2x, since the ratio of first two of them be 1 : 2.
Also the s um of first three numbers be 180. required numbers are x, 2x, 180 – 3x.
Let P = x × 2x × (180 – 3x)
= 2x² (180 – 3x)
= 6 (60x² – x3)
Diff. both sides w.r.t. x, we have
\(\frac{d P}{d x}\) = 6 [120x – 3x²] ;
\(\frac{d^2 P}{d x^2}\) = 6 (120 – 6x)
For maxima/minima, \(\frac{d P}{d x}\) = 0
⇒ 120x – 3x² = 0
⇒ – 3x [x – 40] = 0
⇒ x = 0, 40
But x > 0
∴ x = 40
∴ (\(\frac{d^2 P}{d x^2}\))_{x = 40} = 6 (120 – 240)
= – 720 < 0
∴ x = 40 be a point of maxima.
Thus product P is maximum for x = 40.
Hence the required numbers are, 40, 2 × 40 and 180 – 120 i.e. 40, 80 and 60.

Question 8.
Find the maximum profit that a company can make, if the profit function is given by p (x) = 41 + 72x – 18x². (NCERT)
Solution:
Given profit function p(x) = 41 + 72x – 18x²
Diff. both sides w.r.t. x, we have
\(\frac{d P}{d x}\) = 72 – 36x
For maxima/minima, \(\frac{d P}{d x}\) = 0
⇒ 72 – 36x = 0
⇒ x = 2
Also, \(\frac{d^2 y}{d x^2}\) = – 36
∴ (\(\frac{d^2 P}{d x^2}\))_{x = 2} = – 36 < 0
Thus x = 2 be a point of maxima.
∴ profit P(x) will be maximum when x = 2.
and Maximum profit = p (2) = 41 + 144 – 72 = 113 units.

Question 9.
The cost (in Rs.) per article C, of manufacturing a certain article is given by the formula C = 5 + \(\frac{48}{x}\) + 3x², where x is the number of articles manufactured per hour. Find the minimum value of C.
Solution:
Given C = 5 + \(\frac{48}{x}\) + 3x²
Diff. both sides w.r.t. x, we have
\(\frac{d C}{d x}\) = – \(\frac{48}{x^2}\) + 6x
∴ \(\frac{d^2 C}{d x^2}\) = \(\frac{96}{x^3}\) + 6
For maxima/minima, we have
\(\frac{d C}{d x}\) = 0
⇒ – \(\frac{48}{x^2}\) + 6x = 0
⇒ 6x³ – 48 = 0
⇒ x = 2
∴ (\(\frac{d^2 P}{d x^2}\))_{x = 2} = \(\frac{96}{8}\) + 6
= 12 + 6 = 18 > 0
Thus x = 2 be a point of minima.
and Minimum value of C = 5 + \(\frac{48}{2}\) + 3 × 2²
= 5 + 24 + 12 = Rs. 41

Question 10.
Prove that of all rectangles with given area, the square has the smallest perimeter.
Solution:
Let x and y be the length and breadth of rectangle.
∴ A = area of rectangle = xy [given] ………..(1)
Let P = perimeter = 2 (x + y)
= 2 [x + \(\frac{A}{x}\)] [using (1)]
∴ \(\frac{d P}{dx}\) = 2 (1 – \(\frac{\mathrm{A}}{x^2}\))
⇒ \(\frac{d^2 \mathrm{P}}{d x^2}=\frac{4 \mathrm{~A}}{x^3}\)
For maxima or minima, \(\frac{d P}{dx}\) = 0
⇒ 1 – \(\frac{\mathrm{A}}{x^2}\) = 0
⇒ x = √A [∵ x > 0]
∴ \(\frac{d^2 \mathrm{P}}{d x^2}=\frac{4 \mathrm{~A}}{\mathrm{~A}^{3 / 2}}=\frac{4}{\sqrt{\mathrm{A}}}\) > 0
∴ P is minimise for x = √A
∴ From (1) ;
y = \(\frac{\mathrm{A}}{\sqrt{\mathrm{A}}}\) = √A
∴ x = y
Hence the rectangle becomes square.
Thus, the rectangle with given area, the square has the smallest perimeter.

Question 11.
Of all rectangles each of which has perimeter 24 cm, find the one having maximum area. Also find that area.
Solution:
Let x cm be the length and y cm be the breadth of rectangle
s.t 2(x + y) = 24
[∵ perimeter of rectangle = 24]
⇒ x + y = 12
Let A = area of rectangle = xy = x (12 – x) [using (1)]
Differentiating w.r.t. x, we have
\(\frac{d A}{d x}\) = 12 – 2x;
\(\frac{d^2 A}{d x^2}\) = – 2
For maxima/minima, \(\frac{d A}{d x}\) = 0
⇒ 12 – 2x = 0
⇒ x = 6
∴ (\(\frac{d^2 A}{d x^2}\))_{x = 6} = – 2 < 0
∴ x = 6 be a point of maxima.
Thus Area A is maximum for x = 6
∴ from (1);
y = 12 – 6 = 6
Therefore x = y = 6 cm.
Thus the square has the maximum area.
and Maximum area = xy = (6 × 6) cm² = 36 cm².

Question 12.
Prove that of all rectangles with given perimeter, the square has the largest area.
Solution:
Let x cm and y cm be the length and breadth of the rectangle.
Let P be the perimeter of rectangle
∴ P = 2x + 2y …………(1)
Let A = area of rectangle = xy
= x (\(\frac{P}{2}\) – x)
⇒ A = \(\frac{P x}{2}\) – x²
Diff, both sides w.r.t. x; we have
\(\frac{d A}{d x}\) = \(\frac{P}{2}\) – 2x ;
\(\frac{d^2 A}{d x^2}\) = – 2
For maxima/minima, \(\frac{d A}{d x}\) = 0
⇒ \(\frac{P}{2}\) – 2x = 0
⇒ x = \(\frac{P}{4}\)
∴ (\(\frac{d^2 A}{d x^2}\))_{x = \(\frac{P}{4}\)} = – 2 < 0
Thus, x = \(\frac{P}{4}\) be a point of maxima
∴ from (1) ;
2y = P – \(\frac{P}{2}\) = \(\frac{P}{2}\)
⇒ y = \(\frac{P}{4}\)
Thus area of the rectangle is maximum when x = y = \(\frac{P}{4}\)
∴ rectangle becomes square.
Thus, Maximum area = \(\frac{P}{4} \times \frac{P}{4}=\frac{P^2}{16}\)
Hence, the rectangle with given perimeter, the square has the largest area.

Question 13.
Prove that the perimeter of a right- angled triangle of given hypotenuse is maximum when the triangle is isosceles.
Solution:
Let ΔABC be right angled Δ at angle B with hypotenuse h.
Let AB = x
∴ BC = \(\sqrt{h^2-x^2}\)
[using pythagoras theorem]

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 1

Thus P = perimetcr of ∆ABC = AB + BC + AC
⇒ P = x + \(\sqrt{h^2-x^2}\) + h
Diff. (1) both sides w.r.t. x; we have
\(\frac{d \mathrm{P}}{d x}=1+\frac{1}{2} \frac{1}{\sqrt{h^2-x^2}}(-2 x)\)
= 1 – \(\frac{x}{\sqrt{h^2-x^2}}\)
For maxima / minima, \(\frac{d P}{d x}\) = 0
⇒ 1 – \(\frac{x}{\sqrt{h^2-x^2}}\) = 0
⇒ \(\sqrt{h^2-x^2}\) = x
⇒ h² – x² = x²
⇒ 2x² = h²
⇒ x = ± \(\frac{h}{\sqrt{2}}\)
Since x, h > 0
∴ x = \(\frac{h}{\sqrt{2}}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 2

∴ AB = BC
Hence the perimeter of right angle triangle is maximum when the triangle is isosceles.

Question 14.
Two sides of a triangle have lengths a and b, and the angle between them is 6. What value of 9 will maximize the area of the triangle ? Also find the maximum area of the triangle.
Solution:
Let ABC be the given triangle with AC = b
and BC = a be given two sides.
Let θ be the angle between them
Let A = area of ∆ABC
= \(\frac{1}{2}\) ab sin θ

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 3

Clearly A is maximum when sin θ is maximum i.e. sin θ = 1
⇒ θ = \(\frac{\pi}{2}\)
and required maximum area = \(\frac{1}{2}\) ab × 1 = \(\frac{1}{2}\) ab

Question 15.
A right-angled triangle with constant area is given. Prove that the hypotenuse of the triangle is least when the triangle is-isosceles.
Solution:
Let ∆ABC be right angled triangle with constant area P.
Let l be the length of the hypotenuse of ∆ABC
∴ AB = l sin θ
and BC = l cos θ
Also ∠ACB = θ.
∴ area of triangle ∆ABC = \(\frac{1}{2}\) × BC × AB
⇒ P = \(\frac{1}{2}\) (l cos θ) (l sin θ) = \(\frac{l^2}{4}\) sin 2θ
⇒ l² = 4P cosec 2θ
Now l is minimise when l² is minimise, i.e. we want to find the value of 0 for which l² is least.
Let L = l² = 4P cosec 2θ
Diff. both sides w.r.t. θ, we have
\(\frac{d \mathrm{~L}}{d \theta}\) = – 8P cosec 2θ cot 2θ
and \(\frac{d^2 \mathrm{~L}}{d \theta^2}\) = – 8P [- 2 cosec³ 2θ + cot 2θ (- 2 cosec 2θ cot 2θ)]
= 16P [cosec³ 2θ + cosec 2θ cot² 2θ]
For maxima/minima \(\frac{d \mathrm{~L}}{d \theta}\) = 0
⇒ cosec 2θ cot 2θ = 0
⇒ \(\frac{\cos 2 \theta}{\sin ^2 2 \theta}\) = 0
⇒ cos 2θ = 0
⇒ 2θ = \(\frac{\pi}{2}\)
⇒ θ = \(\frac{\pi}{4}\)
∵ θ ∈ (0, \(\frac{\pi}{2}\))
At θ = \(\frac{\pi}{4}\),
\(\frac{d^2 \mathrm{~L}}{d \theta^2}\) = 16P [1³ + 1 × 0] = 16P > 0
∴ Lis minimum for θ = \(\frac{\pi}{4}\)
Thus l is least for θ = \(\frac{\pi}{4}\)
∴ AB = l sin \(\frac{\pi}{4}\)
= \(\frac{l}{\sqrt{2}}\)
and BC = l cos θ
= l cos \(\frac{\pi}{4}\)
= \(\frac{l}{\sqrt{2}}\)
∴ AB = BC.
Hence ∆ABC be an isosceles triangle.

Question 16.
The lengths of the sides of a triangle are 9 + x², 9 + x² and 18 – 2x². Calculate:
(i) the area of the triangle in terms of x.
(ii) the value of x for which this area is maximum.
Solution:
(i) Here given lengths of sides of triangle are 9 + x², 9 + x² and 18 – 2x².
Let a = 9 + x² ;
b = 9 + x²;
c = 18 – 2x²
∴ s = \(\frac{a+b+c}{2}\)
= \(\frac{9+x^2+9+x^2+18-2 x^2}{2}\) = 18
Let A = area of ∆ABC
= \(\sqrt{s(s-a)(s-b)(s-c)}\) [By Heron’s formula]
= \(\sqrt{18\left(18-9-x^2\right)\left(18-9-x^2\right)\left(18-18+2 x^2\right)}\)
= \(\sqrt{18\left(9-x^2\right)^2 \times 2 x^2}\)
= 6 (9 – x²) x
= 54x – 6x³

(ii) Diff, both sides w.r.t. x, we have
\(\frac{d A}{d x}\) = 6 [9 – 3x²] ;
\(\frac{d^2 A}{d x^2}\) = 6 (0 – 6x) = – 36x
For maxima / minima, \(\frac{d A}{d x}\) = 0
⇒ \(\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\sqrt{3}}\) = – 36√3 < 0
Thus, x = √3 be a point of maxima
and maximum value of area A = 54 × √3 – (6√3)³
= 54√3 – 18√3 = 36√3 sq. units.

Question 17.
The perimeter of a triangle is 8 cm. If one of the sides is 3 cm, what are the lengths of the other sides for maximum area of the triangle ?
Solution:
Let a, b, c be the lengths of sides of triangle then a + b + c = 8
∴ s = \(\frac{a+b+c}{2}=\frac{8}{2}\) = 4
given one of its sides be of length 3 cm,
let a = 3
b + c = 5
⇒ b = 5 – c
Let A = area of the ∆ABC
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{4(4-3)(4-5+c)(4-c)}\)
= \(\sqrt{4(c-1)(4-c)}\)
⇒ A² = 4 (c – 1) (4 – c) = 4 (- c² + 5c – 4)
Now A is maximise when A2 is maximise.
Let P = A² = 4 (- c² + 5c – 4)
On differentiating w.r.t. c, we get
\(\frac{d P}{d c}\) = 4 (- 2c + 5) ;
∴ (\(\frac{d^2 P}{d c^2}\))_{c = \(\frac{5}{2}\)} = – 8 < 0
Thus P is maximum for c = \(\frac{5}{2}\)
Hence Area A is maximum for c = \(\frac{5}{2}\)
Thus the required lengths of sides of triangle are 3, 5 – \(\frac{5}{2}\) and \(\frac{5}{2}\) i.e. 3 cm, \(\frac{5}{2}\) cm and \(\frac{5}{2}\) cm.

Question 18.
A sheet of paper is to contain 18 cm² of printed matter. The margins at the top and bottom are 2 cm each, and at the sides 1 cm each. Find the dimensions of the sheet which require the least amount of paper.
Solution:
Let x cm (x > 0) be the one dimension of the page then the other dimension be \(\frac{18}{x}\) cm, since the area of given page be 18 sq. cm.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 4

Let A cm² be the area of printed page.
Then, A = (x – 2) (\(\frac{18}{x}\) – 4)
= 18 – 4x – \(\frac{36}{x}\) + 8
= 26 – 4x – \(\frac{36}{x}\)
On differentiating w.r.t. x, we have
\(\frac{d \mathrm{~A}}{d x}=-4+\frac{36}{x^2}\)
\(\frac{d^2 \mathrm{~A}}{d x^2}=-\frac{72}{x^3}\)
For maxima / minima, \(\frac{d A}{d x}\) = 0
⇒ – 4 + \(\frac{36}{x^2}\) = 0
⇒ x² = 9
⇒ x = 3
∴ (\(\frac{d^2 A}{d x^2}\))_{x = 3} = \(-\frac{72}{27}=-\frac{8}{3}\) < 0
Thus A is maximum for x = 3
∴ the dimensions of the printed page be x and \(\frac{18}{x}\) i.e. 3 cm and 6 cm
Hence, the required dimensions of the sheet that requires the least amount of paper be (x + 2) and (\(\frac{18}{x}\) + 4) cm i.e. 5 cm and 10 cm.

Question 19.
Show that of all rectnagles inscribed in a given fixed circle the square has the maximum area.
Solution:
Let x and y be the length and breadth of the rectangle respectively.
Which is inscribed in a circle of radius r
∴ x² + y² = (2r)²
x² + y² = 4r² …………(1)
Thus A = area of rectangle = xy
= x \(\sqrt{4 r^2-x^2}\) [using (1)]

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 5

For max/ minima, \(\frac{d A}{d x}\) = 0
⇒ 4r² – 2x² = 0
⇒ x = √2r
[∵ x > 0]
∴ \(\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\sqrt{2} r}=\frac{-\sqrt{2} r\left(8 r^2\right)}{\left(2 r^2\right)^{3 / 2}}\) < 0
∴ A is maximum for x = √2r
∴ From (1) ;
2r² + y² = 4r²
⇒ y = √2r
Hence A is maximise for x = y = √2r
i.e. rectangle becomes square.

Question 20.
(i) The sum of perimeters of a circle and a square is k, where k is some constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle. (NCERT)
(ii) Given the sum of perimeters of a circle and a square, show that the sum of areas is least when the diameter of the circle is equal to the side of the square.
Solution:
(i) Let x be the side of square and y be the radius of circle.
Let P be the sum of their perimeters.
Then P = 4x + 2πy
Let A combined area of square and circle
⇒ A = x² + πy² [using (1)]
= x² + π \(\left(\frac{\mathrm{P}-4 x}{2 \pi}\right)^2\) [using (1)]
⇒ A = x² + \(\frac{1}{4 \pi}\) (P – 4x)²
∴ \(\frac{d A}{d x}\) = 2x + \(\frac{1}{2 \pi}\) (P – 4x) (- 4)
For maxima / minima, we have \(\frac{d A}{d x}\) = 0
⇒ 4r² – 2x² = 0
⇒ x = √2r [∵ x > 0]
∴ \(\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\sqrt{2} r}=\frac{-\sqrt{2} r\left(8 r^2\right)}{\left(2 r^2\right)^{3 / 2}}\) < 0
∴ A is minimum for x = √2r
∴ From (1) ;
2r² + y = 4r²
⇒ y = √2r
Hence A is maximise for x = y = √2r
i.e. rectangle becomes square.

Question 20 (old).
(i) The sum of perimeters of a circle and a square is k, where A is some constant. Prove that the sum of their arcas is least when the side of the square is double the radius of the circle. (NCERT)
(ii) Given the sum of perimeters of a circle and a square, show that the sum of areas is least when the diameter of the circle is equal to the side of the square.
Solution:
(i) Let x be the side of square and y be the radius of circle.
Let P be the sum of their perimeters.
Then P = 4x + 2πy …………..(1)
Let A = combined area of square and circle
⇒ A = x² + πy²
= x² + π \(\left(\frac{P-4 x}{2 \pi}\right)^2\) [using (1)]
⇒ A = x² + \(\frac{1}{4 \pi}\) (P – 4x)²
∴ \(\frac{d A}{d x}\) = 2x + \(\frac{1}{2 \pi}\) (P – 4x) (- 4)
For maxima/minima, we have \(\frac{d A}{d x}\) = 0
2x – \(\frac{2}{\pi}\) (P – 4x) = 0
⇒ 2πx – 2P + 8x = 0
⇒ x = \(\frac{2 P}{2 \pi+8}=\frac{P}{\pi+4}\)
Also \(\frac{d^2 P}{d x^2}\) = 2 + \(\frac{16}{2 \pi}\) > 0
∴ P is minimum for x = \(\frac{9}{\pi+4}\)
∴ from (1) ;
2πy = P – \(\frac{4 \mathrm{P}}{\pi+4}\)
⇒ 2πy = \(\frac{\pi \mathrm{P}}{\pi+4}\)
⇒ y = \(\frac{\mathrm{P}}{2(\pi+4)}=\frac{x}{2}\)
⇒ x = 2y
i.e. side of square = diameter of circle.
Hence A is minimum when side of the square is equal to the diameter of the circle.

(ii) Let x be the length of each side of square and r be the radius of circle.
Then P = sum of perimeters of a circle and as square = 2πr + 4x …………(1)
Let A = combined area of square and circle
⇒ A = x² + r²
= x² + π \(\left(\frac{P-4 x}{2 \pi}\right)^2\) [using (1)]
⇒ A = x² + \(\frac{1}{4 \pi}\) (P – 4x)² ;
Diff. both sides w.r.t. x, we have
\(\frac{d A}{d x}\) = 2x + \(\frac{1}{4 \pi}\) × 2 (P – 4x) (- 4)
and \(\frac{d^2 A}{d x^2}\) = 2 + (- \(\frac{2}{\pi}\)) (- 4)
= 2 + \(\frac{8}{\pi}\)
For maxima and minima, \(\frac{d A}{d x}\) = 0
⇒ 2x – \(\frac{2}{\pi}\) (P – 4x) = 0
⇒ 2πx – 2P + 8x = 0
⇒ (π + 4) x = P
⇒ x = \(\frac{P}{\pi+4}\)
Now, \(\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\frac{\mathrm{P}}{\pi+4}}\) = 2 + \(\frac{8}{\pi}\)
Hence A is least when x = \(\frac{P}{\pi+4}\)
⇒ (π + 4) x = 2πr + 4x [using (1)]
⇒ πx = 2πr
⇒ x = 2r
Thus, diameter of circle is equal to side of square.

Question 21.
A wire 10 metres long is cut into two parts. One part is bent into the shape of a circle and the other into the shape of an equilateral triangle. How should the wire be cut so that the combined area of the two figures is as small as possible?
Solution:
Let r metre be the radius of circle and a metres be the side of equilateral triangle.
Then 2πr + 3a = 10 ………….( 1)
Let A combined area of circle and equilateral triangle
⇒ A = πr² + \(\frac{\sqrt{3}}{4}\) a²
= πr² + \(\frac{\sqrt{3}}{4}\left(\frac{10-2 \pi r}{3}\right)^2\) [using (1)]
On differentiating both sides w.r.t. r, we have
\(\frac{d A}{d r}\) = 2πr + \(\frac{\sqrt{3}}{36}\) 2 (10 – 2πr) (- 2π)
= 2πr – \(\frac{\sqrt{3}}{9}\) π (10 – 2πr)
∴ \(\frac{d^2 A}{d x^2}\) = 2π – \(\frac{\sqrt{3} \pi}{9}\) (0 – 2π)
= 2π + \(\frac{2 \sqrt{3}}{9}\) π²
For maxima / minima, \(\frac{d A}{d r}\) = 0
⇒ 2πr – \(\frac{\sqrt{3} \pi}{9}\) (10 – 2πr) = 0
⇒ 18πr – 10√3π + 2√3π²r = 0
⇒ (18 + 2√3π) r = 10√3
⇒ r = \(\frac{10 \sqrt{3}}{18+2 \sqrt{3} \pi}\)
⇒ r = \(\frac{10 \sqrt{3}}{2 \sqrt{3}(\pi+3 \sqrt{3})}\)
= \(\frac{5}{\pi+2 \sqrt{3}}\) m
Also, at r = \(\frac{5}{\pi+3 \sqrt{3}}\) ;
\(\frac{d^2 A}{d x^2}\) = 2π + \(\frac{2 \sqrt{3}}{9}\) π² > 0
∴ A is least for r = \(\frac{5}{\pi+3 \sqrt{3}}\) m
Thus, length of piece of wire bent into the form of circle = 2πr = \(\frac{10 \pi}{\pi+3 \sqrt{3}}\) m
and length of piece of wire bent into form of equilateral triangle = \(\left(10-\frac{10 \pi}{\pi+3 \sqrt{3}}\right)\) m
= \(\frac{30 \sqrt{3}}{\pi+3 \sqrt{3}}\) m

Question 22.
A wire of length 36 cm is cut into two pieces. One of the piece is turned into the form of a square and the other in the form of an equilateral triangle. Find the lengh of each piece so that the sum of the areas of the two figures be minimum.
Solution:
Let the length of one piece of wire be x m
∴ other piece must be (36 – x) m.
Let x metres be made into square and (36 – x) m be made into equilateral ∆.
So perimeter of square = 4 × side = x
∴ side of square = \(\frac{x}{4}\)
∴ area of square = \(\left(\frac{x}{4}\right)^2\)
Further perimeter of equilateral ∆ = 3 × side
= 36 – x
∴ side of equilateral ∆ = 12 – \(\frac{x}{3}\)
∴ area of equilateral ∆ = \(\frac{\sqrt{3}}{4}\) (side)²
= \(\frac{\sqrt{3}}{4}\left(12-\frac{x}{3}\right)^2\)
∴ A = combined area of square and equilateral ∆ = \(\frac{x^2}{16}+\frac{\sqrt{3}}{4}\left(12-\frac{x}{3}\right)^2\)
∴ \(\frac{d A}{d x}\) = \(\frac{x}{8}+\frac{\sqrt{3}}{2}\left(12-\frac{x}{3}\right)\left(-\frac{1}{3}\right)\)
\(\frac{d A}{d x}\) = \(\frac{x}{8}+\frac{x}{6 \sqrt{3}}\) – 2
For maxima / minima, \(\frac{d A}{d x}\) = 0

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 6

Hence the length of two pieces are \(\frac{144}{4+3 \sqrt{3}}\) m and \(\left(36-\frac{144}{x+3 \sqrt{3}}\right)\) m i.e. \(\frac{108 \sqrt{3}}{4+3 \sqrt{3}}\) m.

Question 22.
A window is in the form of a rectangle surrounded by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to adniit maximum light through the whole opening.
Solution:
Let the width and height of the window be 2x metre and y metre respectively.
Given total perimeter of window = 10 m
⇒ 2x + 2y + πx = 10 ………..(1)
Let A = total area of window
⇒ A = 2xy + \(\frac{\pi x^2}{2}\)
⇒ A = x (10 – πx – 2x) + \(\frac{\pi x^2}{2}\)
∴ \(\frac{d A}{d x}\) = 10 – (π + 2) 2x + πx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 7

For maxima/minima, \(\frac{d A}{d x}\) = 0
⇒ 10 – 2πx – 4x + πx = 0
⇒ 10 – 4x – πx = 0
⇒ x = \(\frac{10}{\pi+4}\)
∴ \(\frac{d^2 A}{d x^2}\) = – 2 (π + 2)
= – π – 4
at x = \(\frac{10}{\pi+4}\);
\(\frac{d^2 A}{d x^2}\) = – π – 4 < 0
Hence A is maximum when x = \(\frac{10}{\pi+4}\)
∴ from (1) ;
2y = 10 – \(\frac{10}{\pi+4}\)
= \(\frac{10 \pi+40-10 \pi-20}{\pi+4}\)
⇒ y = \(\frac{10}{\pi+4}\)
Thus, dimensions of the window are 2x and y
i.e. \(\frac{2 \times 10}{\pi+4}\) and \(\frac{10}{\pi+4}\)
i.e. \(\frac{10}{\pi+4}\) and \(\frac{10}{\pi+4}\).

Question 23.
A window is in the form of a rectangle above which there is a semicircle. If the perimeter of the window is p cm, show that the window will admit maximum possible light only when the radius of semicircle is \(\frac{p}{\pi+4}\) cm.
Solution:
Let r cm be the radius of semi-circle and x cm be the side BC of rectangle as shown in figure shown alongside.
Then p perimeter of combined figure
⇒ p = πr + 2x + 2r ……….(1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 8

Let A = area of the window.
⇒ A = \(\frac{1}{2}\) πr² + 2rx
= \(\frac{\pi r^2}{2}+2 r \frac{(p-\pi r-2 r)}{2}\) ………(2) [using (1)]
Now maximum light wil be admitted through the window if the area of the window is maximum.
For this we have to maximise A.
Diff. eqn. (2) both sides w.r.t. r, we have
\(\frac{d A}{d r}\) = πr
For maxima/minima, \(\frac{d A}{d r}\) = 0
⇒ πr + p – 2πr – 4r = 0
⇒ p – πr – 4r = 0
⇒ r = \(\frac{p}{\pi+4}\)
∴ at r = \(\frac{p}{\pi+4}\),
\(\frac{d^2 A}{d x^2}\) = – (π + 4) < 0
Thus A is least for r = \(\frac{p}{\pi+4}\)
Thus, from eqn. (1); we have
2x = p – \(\frac{(\pi+2) p}{\pi+4}\)
= \(\frac{p \pi+4 p-\pi p-2 p}{\pi+4}\)
⇒ x = \(\frac{p}{\pi+4}\)
Hence, maximum light is admitted when radius of semi-circle be r = \(\frac{p}{\pi+4}\) cm.

Question 24.
A rectangular area of 9000 m² is to be surrounded by a fence with two opposite sides being made of brick and the other two of wood. One metre of wooden fencing costs 25 while one metre of brick walling costs 10. What is the least amount of money that must be alloted for the construction of such a fence?
Solution:
Let x be the length of rectangular area and
y be the breadth of rectangula area
∴ xy = 9000 m²
Let C = amount of money that must be allotted for the construction of such a fence.
∴ C = 25 × 2x + 2y × 10
= 50x + 20 × \(\frac{9000}{x}\) [using (1)]
Diff. both sides w.r.t. x; we have
\(\frac{d C}{d x}\) = 50 – \(\frac{180000}{x^2}\) ;
\(\frac{d^2 \mathrm{C}}{d x^2}=\frac{360000}{x^3}\)
For maxima / minima, \(\frac{d C}{d x}\) = 0
⇒ 50 – \(\frac{180000}{x^2}\) = 0
⇒ x² = \(\frac{18000}{50}\) = 3600
⇒ x = 60 (∵ x > 0)
∴ from (1) ;
⇒ y = \(\frac{9000}{60}\) = 150
Now \(\frac{d^2 \mathrm{C}}{d x^2}=\frac{360000}{x^3}\)
∴ \(\left(\frac{d^2 \mathrm{C}}{d x^2}\right)_{x=60}=\frac{360000}{60 \times 60 \times 60}\)
= \(\frac{360}{216}=\frac{5}{3}\) > 0
Thus C is minimise for x = 60 and y = 150
∴ required amount of money = C
= ₹ (50 × 60 + 20 × 150)
= ₹ 6000.

Question 25.
(i) Find the point on the parabola y² = 4x which is nearest to the point (2, – 8).
(ii) Find the point on the curve x² = 8y which is nearest to the point (2, 4).
Sol.
(I) Let P(x, y) be any point on the curve
y² = 4x …………(1)
Let A (2, – 8) be the given point.
∴ |AP| = \(\sqrt{(x-2)^2+(y+8)^2}\)
⇒ AP² = (x – 2)² + (y + 8)²
= (\(\frac{y^2}{4}\) – 4)² + (y + 8)² [using (1)]
Let S = AP²,
Then S is maximum or minimum according as AP is maximum or minimum.
Now, S = (\(\frac{y^2}{4}\) – 4)² + + (y + 8)²
∴ \(\frac{d S}{d y}\) = 2 (\(\frac{y^2}{4}\) – 2) (\(\frac{y}{2}\)) + 2 (y + 8)
∴ \(\frac{d S}{d y}\) = \(\frac{y^3}{4}\) – 2y + 2y + 16
= \(\frac{y^3}{4}\) + 16
∴ \(\frac{d^2 \mathrm{~S}}{d y^2}=\frac{3 y^2}{4}\)
For max./minima, \(\frac{d S}{d y}\) = 0
⇒ y³ = – 64
⇒ y³ = – 64
⇒ y = – 4
∴ (\(\frac{d^2 S}{d y^2}\))_{y = – 4} = \(\frac{3}{4}\) (16) = 12 > 0
∴ S is minimise for y = – 4
and from (1); x = 4
Hence the point (4, – 4) on curve y³² = 4x is nearest to the given point (2, – 8).

(ii) Let P(x, y) be any point on the curve
x² = 8y …………………..(1)
and let the given point be A (2, 4).
∴ AP² = (x – 2)² + (y – 4)²
= (x – 2)² + \(\left(\frac{x^2}{8}-4\right)^2\) [using (1)]
Let S = AP²,
Now S is max./min.
according as AP is max./min.
∴ S = (x – 2)² + \(\left(\frac{x^2}{8}-4\right)^2\)
∴ \(\frac{d S}{d x}\) = 2 (x – 2) + 2 \(\left(\frac{x^2}{8}-4\right) \frac{2 x}{8}\)
= 2x – 4 + \(\frac{x^3}{16}\) – 2x
∴ \(\frac{d \mathrm{~S}}{d x}=\frac{x^3}{16}\) – 4
∴ \(\frac{d^2 \mathrm{~S}}{d x^2}=\frac{3 x^2}{16}\)
For max./min., \(\frac{d S}{d x}\) = 0
⇒ x³ = 8
⇒ x = 4
when x = 4,
\(\left(\frac{d^2 \mathrm{~S}}{d x^2}\right)=\frac{3 \times 16}{16}\) = 3 > 0
∴ S is minimize for x = 4
∴ from (1) ; y = 2
∴ Required point on given curve be (4, 2).

Question 26.
Find the maximum area of an isosceles triangle inscribed in the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}\) = 1 with its vertex at one end of the major axis.
Solution:
Let any point P on the ellipse

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 9

\(\frac{x^2}{16}+\frac{y^2}{9}\) = 1 is (4 cos θ, 3 sin θ)
∴ A = Area of ∆ \(\frac{1}{2}\) (PP’) (AB)
= \(\frac{1}{2}\) (6 sin θ) (4 – 4 cos θ)
= 12 sin θ (1 – cos θ)
= 12 [sin θ – \(\frac{1}{2}\) sin 2θ]
∴ \(\frac{d \mathrm{~A}}{d \theta}\) = 12 [cos θ – cos 2θ]
For Maxima / Minima, \(\frac{d \mathrm{~A}}{d \theta}\) = 0
⇒ cos 2θ – cos θ = 0
⇒ 2 cos θ – cos θ – 1 = 0
⇒ (cos θ – 1) (2 cos θ + 1) = 0
⇒ cos θ = 1 or
cos θ = – \(\frac{1}{2}\) = cos (π – \(\frac{\pi}{3}\))
⇒ θ = 0 or θ = \(\frac{2 \pi}{3}\)
When θ = 0, the point P coincide with A (4, 0)
When θ = \(\frac{2 \pi}{3}\),
Now \(\frac{d^2 \mathrm{~A}}{d \theta^2}\) = 12 (- sin θ + 2 sin 2θ)
⇒ \(\left(\frac{d^2 \mathrm{~A}}{d \theta^2}\right)_{\theta=\frac{2 \pi}{3}}=12\left(-\sin \frac{2 \pi}{3}+2 \sin \frac{4 \pi}{3}\right)\)
= \(\left(-\frac{\sqrt{3}}{2}+2\left(\frac{-\sqrt{3}}{2}\right)\right)\)
= – 18√3 < 0
∴ A is maximise at θ = \(\frac{\pi}{3}\)
Hence maximum area = A
= 12 \(\left[\sin \frac{2 \pi}{3}-\frac{1}{2} \sin \frac{4 \pi}{3}\right]\)
= 12 \(\left[\frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{\sqrt{3}}{2}\right]\)
= 9√3 sq. units

Question 27.
Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.
Solution:
Let x, x, y are the length, breadth and height of closed cuboid.
∴ area of base square = x² ;
area of four walls = 4xy
Let S = area of cuboid = 2x² + 4xy ……….(1)
Further, V = volume of cuboid = x²y ………..(2)
∴ S = 2x² + 4x \(\left(\frac{\mathrm{V}}{x^2}\right)\) [using (1)]
∴ \(\frac{d \mathrm{~S}}{d x}=4 x-\frac{4 \mathrm{~V}}{x^2}\)
∴ \(\frac{d^2 \mathrm{~S}}{d x^2}=4+\frac{8 \mathrm{~V}}{x^3}\)
For Max/Min, \(\frac{d S}{d x}\) = 0
⇒ 4x = \(\frac{4 V}{x^2}\)
⇒ x³ = V
⇒ x = (V)^1/3
at x = (V)^1/3
⇒ \(\frac{d^2 S}{d x^2}\) = 4 + \(\frac{8 V}{V}\) = 12 > 0
∴ S is minimise for x = \(\sqrt[3]{v}\)
∴ y = \(\frac{\mathrm{V}}{x^2}=\frac{\mathrm{V}}{\mathrm{V}^{2 / 3}}\)
= V^1/3
Hence all the dimensions of cuboid are equal.
∴ Cuboid because cube.
Hence the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.

Question 28.
An open box with a square base is to be made of given iron sheet of area 27 sq.m. Show that the maximum volume of the box is 13.5 cu. m .
Solution:
Let x, x and y are the length, breadth and height of square box.
It is given that
x² + 4xy = 27 sq. m
and V = volume of box = x²y
⇒ V = x² \(\left(\frac{27-x^2}{4 x}\right)\)
= \(\frac{27}{4} x-\frac{x^3}{4}\)
∴ \(\frac{d V}{d x}=\frac{27}{4}-\frac{3 x^2}{4}\)
For maximaJminima,
⇒ \(\frac{d V}{d x}\) = 0
⇒ \(\frac{27}{4}-\frac{3 x^2}{4}\) = 0
⇒ 27 = 3x²
⇒ at x = 3; (∵ x > 0)
Thus V is maximise volume x = 3
∴ y = \(\frac{27-9}{4 \times 3}=\frac{3}{2}\)
∴ Maximum volume of box = x²y
=3² × \(\frac{3}{2}\)
= \(\frac{27}{2}\) cu.m = 13.5 cu.m.

Question 28 (old).
An open box with a square base is to be made of given ¡ron sheet of area 27 sq. m. Show that the maximum volume of the box is 13.5 cu. m.
Solution:
Let x be the side of the square base and y be the height of the box.
Thus area of square base = x²
and area of four walls = 4 xy
given area of square box + area of four val1s = 27 m²
⇒ x² + 4xy = 27
Let V = volume of box = x²y = x² \(\left(\frac{27-x^2}{4 x}\right)\) [using (1)]
V = \(\frac{1}{4}\) (27x – x³)
Diff, both sides w.r,t. x, we have
\(\frac{d V}{d x}\) = \(\frac{1}{4}\) (27 – 3x²) ;
\(\frac{d^2 V}{d x^2}\) = \(\frac{1}{4}\) (- 6x)
For maxima / minima, \(\frac{d V}{d x}\) = 0
⇒ 27 – 3x² = 0
⇒ x² = 9
⇒ x = 3 (∵ x > 0)
∴ \(\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=3}=-\frac{3}{2} \times 3=-\frac{9}{2}\) < 0
Thus V is minimise for x = 3
∴ from (1) ; we have
y = \(\frac{27-x^2}{4 x}\)
= \(\frac{27-9}{12}\)
= \(\frac{18}{12}\)
= \(\frac{3}{2}\)
Hence, the required maximum volume of box = x²y
= (3² × \(\frac{3}{2}\)) cu.m = 13.5 cu.m.

Question 29.
The volume of a closed metal box with a square base is 4096 cm³. The cost of polishing the outer surface of the box is Rs. 4 per cm². Find the dimensions of the box for the minimum cost of polishing it. (ISC 2019)
Solution:
Let the dimensions of closed metal box are x cm, x cm and y cm respectively.
Then volume of closed metal box = x²y
and given volume of box = 4096 cm³
∴ x²y = 4096
Let S = surface area of closed metal box = 2x² + 4xy.
Given, the cost of polishing the outer surface of box be Rs. 4 per cm².
Let C be the cost of polishing the whole metal box
Then C = 4(2x² + 4xy) = 8(x² + 2xy)
C = 8(x² + 2x × \(\frac{4096}{x^2}\)) [using eqn. (1)]
∴ \(\frac{d \mathrm{C}}{d x}=8\left(2 x-\frac{2 \times 4096}{x^2}\right)\)
= 16 \(\left(x-\frac{4096}{x^2}\right)\)
For maximum / minima, \(\frac{d C}{d x}\) = 0
⇒ x³ = 4096
⇒ x = 16 (∵ x > 0)
Now \(\frac{d^2 \mathrm{C}}{d x^2}=16\left(1+\frac{2 \times 4096}{x^3}\right)\)
∴ \(\left(\frac{d^2 c}{d x^2}\right)_{x=16}=16\left(1+\frac{2 \times 4096}{16^3}\right)\)
= 48 > 0
Thus x = 16 be a point of minima
4096
∴ from (1) ;
y = \(\frac{4096}{16^2}\) = 16
Thus, cost is minimum when x = 16 cm and y = 16 cm.

Question 30.
An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when the depth of the tank is half of its width.
Solution:
Let x, x and y are the length, width and height of an open tank
∴ Volume of tank V = x²y (given) …………..(1)
Now it is given that, the tank to hold a given quantity of water
∴ V = constant.
To minimise the cost of material. we have to minimise the surface area of the tank.
Let S = x² + 4xy
= x² + 4x (\(\frac{\mathrm{~V}}{x^2}\)) (using (1))
⇒ \(\frac{d \mathrm{~S}}{d x}=2 x-\frac{4 \mathrm{~V}}{x^2}\)
∴ \(\frac{d^2 \mathrm{~S}}{d x^2}=2+\frac{8 \mathrm{~V}}{x^3}\)
For max./minima, \(\frac{d S}{d x}\) = 0
⇒ x = (2V)^1/3
and \(\frac{d^2 S}{d x^2}\) = 2 + \(\frac{8 V}{2 V}\) = 6 > 0
∴ S is minimise for x³ = 2V.
i.e. x³ = 2x²y
⇒ x = 2y
⇒ y = \(\frac{x}{2}\)
i.e. S is minimum when depth of the tank is half of its width.

Question 31.
A tank with rectangular sides, open at the top, is to be constructed so that its depth is 2 m and volume is 8 m3. 1f building of tank costs Rs. 70 per square metre for the base and Rs. 45 per square metre for the sides, what is the cost of least expensive tank ?
Solution:
Let x, x, y be the length, width and height of the tank
∴ area of square base = x² and
area of four walls = 4xy
Given V = volume of open tank = 8 x²y …………(1)
∴ E = 70x² + 45 (4xy)
= 70x² + 180x . \(\frac{8}{x^2}\) [using (1)]
⇒ E = 70x² + \(\frac{180 \times 8}{x}\) ……….(2)
Given y = depth of tank = 2m
∴ from (1) ;
8 = x² × 2
⇒ x = 2[∵ x > 0]
Now we check whether x = 2 is point of maximum or minima.
\(\frac{d E}{d x}\) = 140x – \(\frac{180 \times 8}{x^2}\)
\(\frac{d^2 \mathrm{E}}{d x^2}=140+\frac{180 \times 16}{x^2}\)
at x = 2,
\(\frac{d^2 E}{d x^2}\) > 0
∴ E is minimum for x = 2.
∴ from (2) ; we have
least value of E = 70 (2)² + \(\frac{180 \times 8}{2}\) = ₹ 1000.

Question 32.
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top by cutting off squares from the corners and folding up the flaps. What should be the side of the in square in order that the volume of the box is maximum ? (NCERT)
Solution:
Let x be the side of the square that is cut off from each comer of the plate.
Then sides of the base are 45 – 2x, 24 – 2x, x cm.
∴ Volume of the box = V = (45 – 2x) (24 – 2x) x
∴ V = (45 – 2x) (24x – 2x²)
= 2x (12 – x) (45 – 2x)
∴ V = 2x [540 – 24x – 45x + 2x²]
= 2x [2x² – 69x + 540]
∴ \(\frac{d V}{d x}\) = 2 [6x² – 138x + 540]
= 4 [3x² – 69x + 270]

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 10

For max/minima, \(\frac{d V}{d x}\) = 0
3x² – 69x + 270 = 0
∴ x = \(\frac{69 \pm 39}{6}\) = 5, 18
But x ≠ 18.
Thus, x = 5,
\(\frac{d^2 V}{d x^2}\) = 4 (6x – 69)
∴ (\(\frac{d^2 V}{d x^2}\))_{x = 5} = 4 (- 39)
= – 156 < 0
∴ V is maximise for x = 5.
Hence the volume of the box is maximum when the side of square is 5 cm and max. volume = 35 × 14 × 5 = 2450 cm³.

Question 33.
Show that the semi-vertical angle of a cone of the maximum volume and of given slant height is cos^-1 \(\frac{1}{\sqrt{3}}\).
Solution:
Let θ be the semi-vertical angle of cone and / be the given slant height of cone.
∴ r = radius of cone = l sin θ
and h = height of cone = l cos θ
Let V = volume of cone = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) π (l sin θ)² (l cos θ)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 11

⇒ \(\frac{d V}{d \theta}\) = \(\frac{\pi l^3}{3}\) [sin² θ (- sin θ) + 2 cos² θ sin θ]
= \(\frac{\pi l^3}{3}\) sin θ [2 cos² θ – sin² θ]
∴ \(\frac{d^2 \mathrm{~V}}{d \theta^2}\) = \(\frac{\pi l^3}{3}\) [- 3 sin² θ cos θ + 2 cos³ θ – 4 cos θ sin² θ]
For Max/minima, \(\frac{d V}{d \theta}\) = 0
⇒ sin θ (2 cos² θ – sin² θ) = 0
∴ sin θ = 0
⇒ θ = 0, π which is not possible, (∵ in this case, r = 0)
∴ tan² θ = 2
⇒ tan θ = √2 [∵ θ lies between 0 and π/2]
∴ θ = tan^-1 (√2)
at tan θ = √2,

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 12

sin θ = \(\frac{\sqrt{2}}{\sqrt{3}}\)
and cos θ = \(\frac{1}{\sqrt{3}}\)
\(\frac{d^2 \mathrm{~V}}{d \theta^2}=\frac{\pi l^3}{3}\left[2 \times \frac{1}{3 \sqrt{3}}-7 \times \frac{1}{\sqrt{3}} \times \frac{2}{3}\right]\)
= \(\frac{\pi l^3}{3} \times\left(-\frac{12}{3 \sqrt{3}}\right)\) < 0
∴ V is maximise for θ = tan^-1 (√2) i.e. cos^-1 \(\frac{1}{\sqrt{3}}\)

Question 34.
Show that a right circular cylinder of given volume, which is open at the top, has minimum total surface area if its height is equal to radius of its base.
Solution:
Let h be the height and r be the radius of cylinder
V = πr²h (given) ………….(1)
Let S = Total surface area
= πr² + 2πrh
= πr² + 2πr \(\frac{\mathrm{V}}{\pi r^2}\) [using (1)]
∴ S = πr² + \(\frac{2 V}{r}\)
∴ \(\frac{d S}{d r}\) = 2πr – \(\frac{2 V}{r^{2}}\)
For Maxima / Minima, \(\frac{d S}{d r}\) = 0
⇒ 2πr = \(\frac{2 V}{r^{2}}\)
⇒ r = \(\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}\)
Now \(\frac{d^2 S}{d r^2}\) = 2π + \(\frac{4 V}{r^{3}}\)
∴ at r = \(\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}\) ;
\(\frac{d^2 S}{d r^2}\) = 2π + \(\frac{4 \mathrm{~V} \times \pi}{\mathrm{V}}\) = 6π > 0
∴ S is minimise for r = \(\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}\)
∴ From (1) ; we have
h = \(\frac{\mathrm{V}}{\pi\left(\frac{\mathrm{V}}{\pi}\right)^{2 / 3}}\)
= \(\frac{\mathrm{V}^{1 / 3}}{\pi^{1 / 3}}=\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}\)
Hence S is minimise when radius of cylinder is equal to height of cylinder.

Question 34 (old).
Show that the height of a closed right circular cylinder of given surface and maximum volume is equal to the diameter of base.
Solution:
Let h be the height and r be the radius of closed cylinder.
Let S = surface area of cylinder
⇒ S = 2πr + 2πrh ……………(1) (given)
Let V = volume of cylinder = πr²h
= πr² \(\left[\frac{\mathrm{S}-2 \pi r^2}{2 \pi r}\right]\)
⇒ V = \(\frac{r}{2}\) [S – 2πr²]
⇒ \(\frac{d V}{d r}\) = \(\frac{1}{2}\) [S – 6πr²]
For MaximaJMinima, \(\frac{d V}{d r}\) = 0
⇒ S = 6πr²
∴ \(\frac{d^2 V}{d r^2}\) = \(\frac{1}{2}\) (- 12πr)
= – 6πr
= – 6π\(\sqrt{\frac{\mathrm{S}}{6 \pi}}\) < 0 [∵ r > 0]
Thus V is maximise when S = 6πr²
2πr² + 2πrh = 6πr² [using (1)]
⇒ 2πrh = 4πr²
⇒ h = 2r
Thus height of cylinder = diameter of closed cylinder.

Question 35.
(i) Show that a closed cylindrical vessel of given volume has the least (total) surface area when its height is twice its radius.
(ii) A closed right circular cylinder has a volume of 2156 cu. cm. What will be the radius of its base so that its total surface area is minimum ? Find the height of the cylinder when its total surface area is minimum. Take π = \(\frac{22}{7}\). (ISC 2004)
Solution:
(1) Let h be the height and r be the radius of closed cylinder
V = πr²h (given) …………(1)
and Let S = surface area of cylinder = 2πr² + 2πrh
⇒ S = 2πr² + 2πr \(\left(\frac{\mathrm{V}}{\pi r^2}\right)\)
= 2πr² + \(\frac{2 V}{r}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 13

∴ S is minimise for r³ = \(\frac{V}{2 \pi}\)
i.e. V = 2πr³
⇒ 2πr³ = πr²h
⇒ h = 2r [using (1)]
i.e. height of cylinder diameter of cylinder.

(ii) Let r be the radius of the base of cylinder
and h be the height of cylinder.
Then volume of cylinder = 2156 cm³
⇒ πr²h = 2156
⇒ h = \(\frac{2156}{\pi r^2}\) ……….(1)
Let S = Total surface area of cylinder
∴ S = 2πr² + 2πrh
S = 2πr (r + \(\frac{2156}{\pi r^2}\)) [using (1)]
∴ \(\frac{d \mathrm{~S}}{d r}=2 \pi\left(2 r-\frac{2156}{\pi r^2}\right)\)
For maxima/minima, \(\frac{d S}{d r}\) = 0
⇒ 2r – \(\frac{2156}{\pi r^2}\) = 0
⇒ r³ = \(\frac{2156}{2 \pi}=\frac{1078}{22}\) × 7
= 49 × t = 7³
⇒ r = 7 cm
∴ \(\left(\frac{d^2 \mathrm{~S}}{d r^2}\right)_{r=7}=2 \pi\left(2+\frac{4312}{\pi \times 7^3}\right)\)
= 2π \(\left(2+\frac{4312}{22 \times 49}\right)\)
= 12π > 0
Thus, S is minimise for r = 7
Hence the required radius of base of cylinder = 7 cm.

Question 36.
Show that the radius of a closed right circular cylinder of given surface area and maximum volume is equal to half of its height. (ISC 2020)
Solution:
Let r be the radius and h be the height of right circular cylinder.
Let S be given surface area of cylinder.
∴ S = 2πr² + 2πrh
Let V = volume of cylinder = πr²h
⇒ V = πr² \(\left(\frac{\mathrm{S}-2 \pi r^2}{2 \pi r}\right)\)
[∵ from (1) ;
h = \(\frac{S-2 \pi r^2}{2 \pi r}\)]
⇒ V = \(\frac{r}{2}\) (S – 2πr²)
= \(\frac{ S r}{2}\) – πr³
∴ \(\frac{d V}{d r}\) = \(\frac{S}{2}\) – 3πr²
For maxima/minima, \(\frac{d V}{d r}\) = 0
⇒ \(\frac{S}{2}\) = 3πr²
⇒ S = 6πr²
⇒ 2πr² + 2πrh = 6πr²
⇒ 2πrh = πr²
⇒ h = 2r
⇒ r = \(\frac{h}{2}\)
∴ \(\frac{d^2 V}{d x^2}\) = – 6πr
⇒ \(\frac{d^2 V}{d x^2}\) < 0
Thus V is maximise when r = \(\frac{h}{2}\)
i.e. V is maximum when radius of right circular cylinder is half of its height.

Question 36 (old).
Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.
Solution:
Let ABC be the cone of largest volume that can be inscribed in the sphere, it is understandable that for maximum value, the axis of the cone must be along the diameter of sphere. Let DE = x,
∴ radius of cone = \(\sqrt{r^2-x^2}\) =BE
and height of cane = r + x
Then volume of cone
⇒ V = \(\frac{1}{3}\) π (r² – x²) (r + x)
⇒ \(\frac{d \mathrm{~V}}{d x}=\frac{\pi}{3}\) [r² – 2rx – 3x²]
⇒ \(\frac{d^2 V}{d x^2}\) = \(\frac{\pi}{3}\) (- 6x – 2r)
For Max./minima, \(\frac{d V}{d x}\) = 0
⇒ 3x² + 2rx – r² = 0
⇒ (x + r) (3x – r) = 0
⇒ x = \(\frac{r}{3}\) [∵ r ≠ x]
at x = \(\frac{r}{3}\),
⇒ \(\frac{d^2 V}{d x^2}\) = \(\frac{\pi}{3}\) (- 4r) < 0
∴ V is maximum for x = \(\frac{r}{3}\)
∴ Max. volume of cone = \(\frac{\pi}{3}\left(r^2-\frac{r^2}{9}\right)\left(r+\frac{r}{3}\right)\)
= \(\frac{32 \pi r^3}{81}\)
Thus Max. value of cone = \(\frac{8}{27}\left(\frac{4}{3} \pi r^3\right)\)
= \(\frac{8}{27}\) (volume of sphere)
∴ height of cone = x + r
= \(\frac{r}{3}\) + r = \(\frac{4r}{3}\)
= \(\frac{4 \times 12}{3}\) = 16 cm then.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 15

[Since given radius of sphere = r = 12 cm]

Question 37.
Find the maximum volume of the cylinder which can be inscribed in a sphere of radius k 3√3 cm. (Leave the answer in terms of it). (ISC 2013)
Solution:
Let h be the height and R be the radius of the cylinder that is inscribed in a sphere of radius r.
In ∆EFA,
R² = r² – (\(\frac{h}{2}\))² ………..(1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 14

Let V = volume of under
= πr²h
= πh [r² – \(\left(\frac{h}{2}\right)^2\)]
∴ \(\frac{d V}{d h}\) = π \(\left[r^2-\frac{3}{4} h^2\right]\)
∴ \(\frac{d^2 V}{d h^2}\) = – \(\frac{3}{2}\) πh
For maxima / minima, \(\frac{d V}{d h}\) = 0
⇒ r² – \(\frac{3}{4}\) h² = 0
⇒ h = \(\frac{2 r}{\sqrt{3}}\)
∴ \(\left(\frac{d^2 \mathrm{~V}}{d h^2}\right)_{h=\frac{2 r}{\sqrt{3}}}=-\frac{3}{2} \pi \times \frac{2 r}{\sqrt{3}}\)
= – \(\frac{3 \pi r}{\sqrt{3}}\) < 0
∴ V is maximise h = \(\frac{2 r}{\sqrt{3}}\)
∴ Max. volume = π × \(\frac{2 r}{\sqrt{3}}\left[r^2-\frac{1}{4} \times \frac{4 r^2}{3}\right]\)
= \(\frac{2 r \pi}{\sqrt{3}} \times \frac{2 r^2}{3}=\frac{4 \pi r^3}{3 \sqrt{3}}\)
Given radius of sphere r = 3√3 cm
∴ Max. value = \(\frac{4 \pi}{3 \sqrt{3}} \times(3 \sqrt{3})^3\)
= 108π cm³.

Question 38.
A cone is inscribed in a sphere of radius 12 cm. If the volume of the cone Is maximum, find its height.
Solution:
Let ABC be the cone of largest volume that can be inscribed in the sphere, it is understandable that for maximum value, the axis of the cone must be along the diameter of sphere.
Let DE = x,
radius of cone = \(\sqrt{r^2-x^2}\) = BE
and height of cane = r + x
Then volume of cone
⇒ V = \(\frac{1}{3}\) π (r² – x²)
⇒ \(\frac{d V}{d x}\) = \(\frac{\pi}{3}\) [r² – 2rx – 3x²]
⇒ \(\frac{d^2 V}{d x^2}\) = \(\frac{\pi}{3}\) [- 6x – 2r]
For Max./minima, \(\frac{d V}{d x}\) = 0
⇒ 3x² + 2rx – r² = 0
⇒ (x + r) (3x – r) = 0
⇒ x = \(\frac{r}{3}\) [∵ r ≠ x]
at x = \(\frac{r}{3}\),
\(\frac{d^2 V}{d x^2}\) = \(\frac{\pi}{3}\) (- 4r) < 0
∴ V is maximise for x = \(\frac{r}{3}\)
∴ Max. volume of cone = \(\frac{\pi}{3}\left(r^2-\frac{r^2}{9}\right)\left(r+\frac{r}{3}\right)=\frac{32 \pi r^3}{81}\)
Thus Max. volume of cone = \(\frac{8}{27}\left(\frac{4}{3} \pi r^3\right)\)
= \(\frac{8}{27}\) (volume of the sphere)
∴ height of cone = x + r
= \(\frac{r}{3}\) + r
= \(\frac{4 r}{3}=\frac{4 \times 12}{3}\)
= 16 cm then.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.7 16

[Since given radius of sphere = r = 12 cm].

Question 39.
Show that the altitude of a right circular cone of maximum curved surface area which can be inscribed in a sphere of radius r is \(\frac{4 r}{3}\).
Solution:
Let PAB be a cone of maximum curved area that can be inscribed in a sphere of radius r.
Now it is obvious that. For maximum volume, axis of cone must lie along diameter of sphere.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 17

∴ PM height of cone = OP + OM = r + x
In right angled ∆OAM,
AM² = OA² – OM²
= r² – x²
Let S = curved surface area of cone
= π (AM) l
= π \(\sqrt{r^2-x^2}\left[\sqrt{(r+x)^2+r^2-x^2}\right]\)
⇒ V = S² = π² (r² – x²) (2r² + 2rx)
To maximise S it is convenient to maximise S² = V
∴ \(\frac{d V}{d x}\) = π² [2r³ – 4r²x – 6rx²]
and \(\frac{d^2 V}{d x^2}\) = π² [- 4r² – 12rx]
For maxima / minima, \(\frac{d V}{d x}\) = 0
⇒ 2r³ – 4r²x – 6rx² = 0
⇒ – 2r (3x² + 2rx – r²) = 0
⇒ (x + r) (3x – r) = 0
⇒ x = – r, \(\frac{r}{3}\) [but x > 0]
∴ x = \(\frac{r}{3}\)
∴ \(\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=\frac{r}{3}}\)
= – 8π²r² < 0
Thus curved surface area of cone is maximise when x = \(\frac{r}{3}\)
:. altitude of cone = r + x = r + \(\frac{r}{3}\) = \(\frac{4 r}{3}\)

Question 40.
Show that the height of a right circular cylinder of greatest volume which can be inscribed in a right circular cone of height h and radius r is one-third of the height of the cone and the greatest volume of the cylinder is \(\frac{4}{9}\) times the volume of the cone.
Solution:
Let PAB be the cone with height OP = h
and OA = OB = r
Let a cylinder of base radius OM’ = ON’ = x
and height = OO’ be inscribed in cone.
Now ∆POB ~ ∆NN’B

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 18

Thus, \(\frac{\mathrm{OP}}{\mathrm{NN}^{\prime}}=\frac{\mathrm{OB}}{\mathrm{N}^{\prime} \mathrm{B}}\)
⇒ \(\frac{h}{\mathrm{NN}^{\prime}}=\frac{r}{r-x}\)
NN’ = \(\frac{h(r-x)}{r}\) = height of cylinder
V = volume of right circular cylinder
= πx² (NN’)
⇒ V = πx² \(\frac{h}{r}\) (r – x)
= \(\frac{\pi h}{r}\) (rx² – x³)
∴ \(\frac{d \mathrm{~V}}{d x}=\frac{\pi h}{r}\) (2rx – 3x²)
and \(\frac{d^2 V}{d x^2}\) = \(\frac{\pi h}{r}\) (2r – 6x)
For maxima / minima, \(\frac{d V}{d x}\) = 0
⇒ \(\frac{\pi h}{r}\) (2rx – 3x²) = 0
⇒ 2rx = 3x²
⇒ x = \(\frac{2 r}{3}\)
and \(\frac{d^2 V}{d x^2}\) = \(\frac{\pi h}{r}\) (2r – 4r)
= – 2πh < 0 [∵ x > 0]
Thus V is maximise when x = \(\frac{2 r}{3}\)
∴ NN’ = height of cylinder
= \(\frac{h}{r}\left[r-\frac{2 r}{3}\right]=\frac{h}{3}\)
Hence volume of right circular cylinder which is inscribed in right circular cone be greatest when height of cylinder = \(\frac{1}{3}\) height
of cone and greatest volume of cylinder
= πx²\(\frac{h}{3}\)
= \(\pi\left(\frac{2 r}{3}\right)^2 \frac{h}{3}\)
= \(\frac{4}{27}\) πr²h
= \(\frac{4}{9}\) (\(\frac{1}{3}\) πr²h)
= \(\frac{4}{9}\) (Volume of cone).

Question 41.
The sum of the surface areas of a rectangular parallelopiped with sides x, 2x and \(\frac{x}{3}\) a sphere is given to be constant, prove that the sum of their volumes is minimum if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.
Solution:
S = sum of areas of || piped and sphere
⇒ S = 2 (x × 2x + 2x . \(\frac{x}{3}\) + x . \(\frac{x}{3}\)) + 4πy²
where y = radius of sphere
⇒ S = 2 (2x² + \(\frac{2}{3}\) x² + \(\frac{x^{2}}{3}\)) + 4πy²
⇒ S = 6x² + 4πy² …………..(1)
Let V = volume of ||piped + Volume of sphere
= x (2x) \(\frac{x}{3}\) + \(\frac{4}{3}\) πy³

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 19

Thus V is minimise when x = 3y
∴ V = \(\frac{4}{3} \pi y^3+\frac{2}{3} x^3\)
= \(\frac{4 \pi}{3}\left(\frac{x}{3}\right)^3+\frac{2}{3} x^3\)
V = \(\frac{4 \pi}{81} x^3+\frac{2}{3} x^3\)
= \(\frac{2}{3} x^3\left(1+\frac{2 \pi}{27}\right)\).

Question 42.
A manufacturer plans to construct a cylindrical can to hold one cubic metre of oil. If the cost of constructing top and bottom of the can is twice the cost of constructing the side, what are the dimensions ofthe most economical can?
Solution:
Let r be the radius of cylindrical can and h be the height of cylindrical can.
Then 1 = πr²h volume of cylindrical can ………..(1)
Let Rs. p be the cost of constructing the side
then cost of constructing top and bottom of can be Rs. 2p each.
Let C = total cost of constructing the cylindrical can
⇒ C = 2πr² × 2p + 2πrh × p
⇒ C = 4pπr² + 2πrp × \(\frac{1}{\pi r^2}\) [using eqn. (1)]
⇒ C = 4pπr² + \(\frac{2 p}{r}\)
On differentiating w.r.t. r, we have
\(\frac{d C}{d r}\) = 8pπr – \(\frac{2 p}{r^{2}}\) ;
\(\frac{d^2 \mathrm{C}}{d r^2}=8 p \pi+\frac{4 p}{r^3}\)
For maxima / minima,
\(\frac{d C}{d r}\) = 0
⇒ 8pπr = \(\frac{2 p}{r^{2}}\)
⇒ r³ = \(\frac{1}{4 \pi}\)
⇒ r = \(\left(\frac{1}{4 \pi}\right)^{1 / 3}\)
at r = \(\left(\frac{1}{4 \pi}\right)^{1 / 3}\) ;
\(\frac{d^2 C}{d r^2}\) = 8πp + 4p × 4π
= 24pπ > 0
Thus C is maximise for r = \(\left(\frac{1}{4 \pi}\right)^{1 / 3}\)
∴ from (1) ; we have
h = \(\frac{1}{\pi r^2}\)
= \(\frac{1}{\pi\left(\frac{1}{4 \pi}\right)^{2 / 3}}=\frac{(4 \pi)^{2 / 3}}{\pi}\)
⇒ h = \(\left(\frac{16}{\pi}\right)^{1 / 3}\)
Hence, the required dimensions of the most economical can be r = \(\left(\frac{1}{4 \pi}\right)^{1 / 3}\) metre and
h = \(\left(\frac{16}{\pi}\right)^{1 / 3}\) metre.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8 16

[Since given radius of sphere = r = 12 cm]

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.7

November 22, 2023

Students can cross-reference their work with ML Aggarwal Class 12 Solutions Chapter 7 Applications of Derivatives Ex 7.7 to ensure accuracy.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.7

Question 1.
Find the turning points of the following functions and distinguish between them. Also find the local maximum and minimum values of the functions :
(i) f(x) = 2x³ – 21x² + 36x – 20
(ii) f(x) = x³ – 3x + 3x. (NCERT)
Solution:
(i) Given f(x) = 2x³ – 21x² + 36x – 20 ………….(1)
Diff. both sides of eqn. (1) w.r.t. x ; we have
f'(x) = 6x² – 42x + 36
= 6 (x² – 7x + 6)
For maxima / minima, we put f'(x) = 0
⇒ 6 (x² – 7x + 6) = 0
⇒ 6 (x – 1) (x – 6) = 0
⇒ x = 1, 6
∴ f”(x) = 12x – 42.

Case – I :
When x = 1
f”(1)= 12 – 42 = – 30 < 0
∴ x = 1 is a point of local maxima. and local maximum value
= 2 – 21 + 36 – 20 = – 3

Case – II :
When x = 6
∴ f”(6) = 72 – 42
= 30 > 0
Thus, x = 6 be a point of local minima, and local minimum value = f(6)
= 432 – 756 + 216 – 20
= 648 – 776 = – 128.

(ii) Given f(x) = x³ – 3x² + 3x
∴ f (x) = 3x² – 6x + 3
∴ f”(x) = 6x – 6
For turning points, we have f'(x) = 0
⇒ 3x² – 6x + 3 = 0
⇒3(x²2 – 2x + 1) = 0
⇒ (x – 1)² = 0
⇒ x = 1
Thus, x = 1 be the only critical point.
f”(1) = 6 – 6 = 0
Now f”‘ (x) = 6 i.e. f” (1) = 6 ≠ 0
∴ x = 1 be the point of inflexion.
i.e. x = 1 be a point of neither maxima nor minima.

Question 1(old).
(ii) f(x) = 4x³ + 19x² – 14x + 3
Solution:
Given f(x) = 4x³ + 19x² – 14x + 3 ……………(1)
Diff. both sides eqn. (1) w.r.t. x ; we have
f'(x)= 12x² + 38x – 14
For maxima/minima, we put f'(x) = 0
⇒ 2 (6x² + 19x – 7) = 0
⇒ 6x² + 21x – 2x – 7 = 0
⇒ 3x (2x + 7) – 1 (2x + 7) = 0
⇒ (2x + 7) (3x – 1) = 0
⇒ x = – \(\frac{7}{2}\), \(\frac{1}{3}\)
∴ f”(x) = 24x + 38

Case – I :
at x = – \(\frac{7}{2}\)
∴ f”(- \(\frac{7}{2}\)) = 24 (- \(\frac{7}{2}\)) + 38
= – 84 + 38
= – 46 < 0
∴ x = – \(\frac{7}{2}\) be a point of local maxima. and max. value
= f(- \(\frac{7}{2}\))
= \(-\frac{343}{2}+\frac{931}{4}\) + 49 + 3
= \(\frac{453}{4}\)

Case – II :
When x = \(\frac{1}{3}\)
∴ f”(\(\frac{1}{3}\)) = 8 + 38
= 46 > 0
Thus, x = \(\frac{1}{3}\) be a point of local minima.
and local minimum value = f(\(\frac{1}{3}\))
= \(\frac{4}{27}+\frac{19}{9}-\frac{14}{3}\) + 3
= \(\frac{4+57-126+81}{27}\)
= \(+\frac{16}{27}\)

Question 2.
Find the coordinates of stationary points on the curve y = x³ – 3x² – 9x + 7, and distinguish between points of local maxima and minima.
Solution:
Given f(x) = x³ – 3x² – 9x + 7
Diff. both sides w.r.t. x, we have
f'(x) = 3x² – 6x – 9
∴ f” (x) = 6x – 6
For critical points, we put f’ (x) = 0
⇒ 3 (x² – 2x – 3) = 0
⇒ (x + 1) (x – 3) = 0
⇒ x = – 1, 3.

Case – I :
When x = – 1
∴ f”(- 1) = – 6 – 6 = – 12 < 0
∴ x = – 1 be a point of local maxima and local maximum value
= f (- 1) = – 1 – 3 + 9 + 7 = 12.

Case-II :
When x = 3
∴ f” (3) = 18 – 6 = 12 > 0
Thus, x = 3 be a point of local minima, and local minimum value = f(3)
= 27 – 27 – 27 + 7 = – 20.

Question 3.
Find the points of local maxima and minima (if any) of each of the following functions. Find also the local maximum and minimum values :
(i) f{x) = x² (NCERT)
(ii) f(x) = x³ – 3x (NCERT)
(iii) f(x) = x³ – 12x
(iv) f(x) = x³ – 6x² + 9x + 15 (NCERT)
(v) f(x) = x³ – 3x + 3 (NCERT)
(vi) f(x) = 2x³ – 6x² + 6x + 5 (NCERT)
(vii) f(x) = \(\frac{x}{2}+\frac{2}{x}\), x > 0 (NCERT)
(viii) f(x) = \(\frac{1}{x^2+2}\) (NCERT)
(ix) f(x) = 2 sin x – x,
(x) f(x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\) (NCERT)
(xi) f(x) = 3x⁴ + 4x³ – 12x² + 12. (NCERT)
Solution:
(i) f (x) = x²
∴ f'(x) = 2x
For local max. or min. f'(x) = 0
⇒ 2x = 0
⇒ x = 0
Now f” (x) = 2
∴ f” (0) = 2 > 0
∴ x = 0 is a pt. of minima and Min. value = f( 0) = 0.

(ii) Given f(x) = x³ – 3x
∴ f (x) = 3x² – 3
= 3 (x – 1) (x + 1)
For maxima/minima, f’ (x) = 0
⇒ 3 (x – 1) (x + 1) = 0
⇒ x = 1, – 1
∴ f”(x) = 6x
f”(1) = 6 > 0
∴ x = 1 be a point of minima and min. value of f(x)
= f(1) = 1 – 3 = – 2
and f”(- 1) = – 6 < 0 x = – 1 be a point maxima
and max. value of f(x) = f(- 1)
= – 1 + 3 = 2

Aliter :
Given f(x) = x³ – 3x
∴ f'(x) = 3x² – 3 = 3(x² – 1)
= 3 (x – 1) (x + 1)
For local ‘maxima/minima, f'(x) = 0
⇒ x² – 1 = 0
⇒ x = ± 1

Case – I:
At x = 1 when x slightly < 1

Case – II:
At x = -1 when x slightly < -1 ⇒ \(\frac{d y}{d x}\) = (- ve) (+ ve) = (- ve) when x slightly > -1
∴ \(\frac{d y}{d x}\) = (+ ve) (+ ve) = (+ ve)
∴ f'(x) changes its sign from – ve to + ve
∴ x = 1 is a point of maxima and min. value = f (1) = 1 – 3 = – 2.

Case II:
At x = – 1
When x slightly < – 1 ∴ \(\frac{d y}{d x}\) = (- ve) (- ve) = (+ ve) when x slightly > – 1
∴ \(\frac{d y}{d x}\) = (- ve) (+ ve) = (- ve)
Thus \(\frac{d y}{d x}\) changes its sign from + ve to – ve
∴ x = – 1 is a point of maxima.
∴ max. value = f (- 1) = – 1 + 3 = 2.

(iii) Given f(x) = x³ – 12x ;
f’ (x) = 3×2 – 12 ;
f” (x) = 6x
For local max/minima, f’ (x) = 0
⇒ 3x² – 12 = 0
⇒ x = ± 2
Now, f” (2) = 12 > 0
∴ x = 2 is a pt. of minima
and min value = 8 – 24 = – 16
also, f” (- 2) = – 12 < 0
∴ x = – 2 is a pt. of maxima
and max. value = – 8 + 24 = 16.

(iv) Given f (x) = x³ – 6x² + 9x + 15 ;
Diff. both sides w.r.t. x, we have
∴ f’ (x) = 3x² – 12x + 9 ;
f” (x) = 6x – 12
For maxima/minima, f’ (x) = 0
⇒ 3 (x² – 4x + 3) = 0
⇒ (x – 1) (x – 3) = 0
⇒ x = 1, 3

Case – I :
When x = 1,
f” (1) = 6 – 12 = – 6 < 0
∴ x = 1 be a point of maxima (local) and local maximum value
f(1) = 1 – 6 + 9 + 15 = 19

Case – II :
When x = 3 f” (3) = 18 – 12 = 6 > 0
∴ x = 3 be a point of local minima
and local min value = f(3)
= 27 – 54 + 27 + 15 = 15

(v) Given f(x) = x³ – 3x + 3
∴ f(x) = 3x² – 3;
f”(x) = 6x
For local maxima / minima,
∴ f(x) = 0
⇒ 3 (x² – 1) = 0
⇒ x = ± 1
Now f” (1) = 6 > 0
∴ x = 1 is a point, of minima
and min value = 1 – 3 + 3 = 1
and f” (- 1) = – 6 < 0
∴ x = – 1 is a point of maxima
and max. value = – 1 + 3 + 3 = 5.

(vi) Given f(x) = 2x³ — 6x² + 6x + 5
Diff, both sides w.r.t. x; we have
∴ f(x) = 6x² – 12x + 6
∴ f”(x) = 12x – 12
For maxima/minima, we put f'(x) = 0
⇒ 6(x² – 2x + 1) = 0
⇒ 6(x – 1)²
⇒ x = 1
Now, f” (1) = 12 – 12 = 0
Now, f”‘ (x) = 12
⇒ f”‘(1) = 12 ≠ 0
∴ x = 1 be a point of inflexion.
Thus, x = 1 be a point of neither maxima nor minima.

(vii) Given f(x) = \(\frac{x}{2}+\frac{2}{x}\)
∴ f'(x) = \(+\frac{1}{2}-\frac{2}{x^2}\)
= [/latex]\frac{x^2-4}{2 x^2}[/latex]
The critical points of f(x) are given by
∴ f(x) = 0
⇒ x² – 4 = 0
⇒ x = ± 2
but x > 0
∴ x = – 2 is rejected.
Hence x = 2 be the only critical point.
∴ f”(x) = \(\frac{4}{x^3}\)
f”(2) = \(\frac{4}{8}=\frac{1}{2}\) > 0
Thus x = 2 is a point of minima and local min. value of f(x) at x = 2.
= f(2)
= \(\frac{2}{2}+\frac{2}{2}\) = 2.

(viii) Given f(x) = \(\frac{1}{x^2+2}\)
∴ f'(x) = – \(\frac{1}{\left(x^2+2\right)^2}\) × 2x
The critical points of f(x) are given by putting f'(x) = 0
i.e. \(\frac{2 x}{\left(x^2+2\right)^2}\) = 0
⇒ x = 0
[∵ (x² + 2)² > 0]

signs of f'(x) for different values of x
Clearly f'(x) changes its sign from negative to positive as x increases through 0.
Thus x = 0 is a point of local minima.
∴ local minimum value f(0) = \(\frac{1}{0+2}=\frac{1}{2}\).

(ix) Given f(x) = 2 sin x – x,
Diff both sides w.r.t. x, we have
∴ f(x) = 2 cos x – 1
For maxima/minima, we have f'(x) = 0
⇒ 2 cos x – 1 = 0
⇒ cos x = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
⇒ x = 2nπ ± \(\frac{\pi}{3}\) ∀ n ∈ Z
But – \(\frac{\pi}{2}\) < x < \(\frac{\pi}{2}\)
∴ x = ± \(\frac{\pi}{3}\)
∴ f”(x) = – 2 sin x
When x = \(\frac{\pi}{3}\);
f”(\(\frac{\pi}{3}\)) = – 2 sin \(\frac{\pi}{3}\)
= – 2 × \(\frac{\sqrt{3}}{2}\) = – √3 < 0
∴ x = \(\frac{\pi}{3}\) be a point of local maxima and local maximum value
= f(\(\frac{\pi}{3}\)) = 3 sin \(\frac{\pi}{3}\) – \(\frac{\pi}{3}\)
= √3 – \(\frac{\pi}{3}\)
When x = – \(\frac{\pi}{3}\); f”(- \(\frac{\pi}{3}\))
= – 2 sin (- \(\frac{\pi}{3}\))
= √3 > 0
∴ x = – \(\frac{\pi}{3}\) be a point of local minima.
and local minimum value = f(- \(\frac{\pi}{3}\))
= 2 sin (- \(\frac{\pi}{3}\)) + \(\frac{\pi}{3}\)
= – √3 + \(\frac{\pi}{3}\)

(x) Given f(x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\)
∴ f'(x) = cos x – sin x
For local maxima or minima, f’ (x) = 0
⇒ cos x – sin x = 0
⇒ tan x = 1 = tan \(\frac{\pi}{4}\)
∴ x = \(\frac{\pi}{4}\) ∈ (0, \(\frac{\pi}{2}\))
[∵ tan x = tan α
⇒ x = nπ + α, n ∈ I]
∴ f”(x) = – sin x – cos x
Now at x = \(\frac{\pi}{4}\),
f”(\(\frac{\pi}{4}\)) = \(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\)
= – √2 < 0
∴ x = \(\frac{\pi}{4}\) is a pt. of maxima
and Max. value = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) = √2

(xi) Given f(x) = 3x⁴ + 4x³ – 12x² + 12
∴ f’ (x) = 12x³ + 12x² – 24x ;
f” (x) = 36x² + 24x – 24
For local maxima/minima, f’ (x) = 0
⇒ 12x (x² + x – 2) = 0
∴ x = 0, 1, – 2
at x = 0 ;
f” (x) = – 24 < 0
x = 0 is a point of local maxima and max. value = 12
at x = 1 ;
f” (1) = 36 + 24 – 24 = 36 > 0
∴ x = 1 is a pt. of local minima and min value
= 3 + 4 – 12 + 12 = 7
at x = – 2 ;
f”(- 2) = 144 – 48 – 24
= 72 > 0
∴ x = – 2 is a pt. of local minima and local minimum value
= 48 – 32 – 48 + 12 = – 20.

Question 4.
For what values of a and b, the function f(x) = x³ + ax² + bx – 3 has local maximum value at x = 0 and local minimum value at x = 1 ?
Solution:
Given f(x) = x³ + ax² + bx – 3
Diff. both sides w.r.t. x, we have
f ‘(x) = 3x² + 2ax + b
Since the function f(x) has local maximum value at x = 0
and local minimum value at x = 1.
f’ (0) = 0 = f’ (1)
Now f’ (0) = 0
⇒ 0 + 2a × 0 + b = 0
⇒ b = 0
and f'(1) = 0
⇒ 3 + 2a + b = 0
⇒ 2a + 3 = 0
⇒ a = – \(\frac{3}{2}\)
Thus a = – \(\frac{3}{2}\) and b = 0.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.6

November 22, 2023

Access to comprehensive Class 12 ISC Maths Solutions Chapter 7 Applications of Derivatives Ex 7.6 encourages independent learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.6

Find the maximum and the minimum values (if any) of the following (1 to 7) functions:

Question 1.
(i) f(x) = x²
(ii) f(x) = (2x – 1)² + 3. (NCERT)
Solution:
(i) Given f(x) = x²
Since x² ≥ 0 ∀ x ∈ R
∴ f(x) ≥ 0 ∀ x ∈ R
Thus, 0 be the minimum value of f(x) which attains at x = 0
and f(x) can be made as large as we please.
Therefore the maximum value of f(x) does not exists which can be observed from the below graph.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.6 1

(ii) f(x) = (2x – 1)² + 3 ≥ 3
[∵ (2x – 1)² ≥ 0 ∀ x ∈ R]
Here, D_f = R and f(x) ≥ 3
Hence the minimum value is 3.
However, maximum value does not exists.
[∵ f(x) can be made as large as possible]

Question 2.
(i) f (x) = – (x – 1)² + 10 (NCERT)
(ii) f(x) = 9x² + 12x + 2 (NCERT)
Solution:
(1) Given f(x) = – (x – 1)² + 10 ≤ 10,
D_f = R
∴ f(x) ≤ 10
[∵ (x – 1)² ≥ 0 ∀ x ∈ R]
∴ the maximum value of f(x) = 10.
However minimum value does not exists.
[∵ f(x) can be made as small as we please]

(ii) Given f(x) = 9x² + 12x + 6
= 9 \(\left(x^2+\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}+\frac{6}{9}\right)\)
= 9 \(\left(x+\frac{2}{3}\right)^2\) + 2 ≥ 2
∴ Minimum value 2 and Max. value does not exists.
[Since f(x) can be made as large as possible].

Question 3.
(i) f(x) = |x|
(ii) f(x) = x³ + 1 (NCERT)
Solution:
(i) f(x) = |x| ≥ 0 ∀ x ∈ R
Thus 0 be the minimum value of f(x) which it attains at x = 0.
Further F (x) can be made as large as we please.
Therefore, the maximum value of f(x) does not exists and clearly it can be observed from given below graph.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.6 2

(ii) Given g(x) = x³ + 1
∴ D_f = R
∴ g(x) = (x + 1) (x² – x + 1)
= (x + 1) \(\left[\left(x^2-x+\frac{1}{4}+\frac{3}{4}\right)\right]\)
= (x + 1) \(\left[\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right]\)
Since \(\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\) > 0 ∀ x ∈ R
∴ g (x) > 0 or g (x) < 0 according as x + 1 > 0 or < 0
∴ g (x) has no maximum and minimum value.

Question 4.
(i) f(x) = | x + 2 | – 1 (NCERT)
(ii) f(x) = 3 + | x | (NCERT)
Solution:
(i) Given f(x) = | x + 2 | – 1
∴ f(x) ≥ – 1
[∵ |x + 2| ≥ 0 ∀ x ∈ R]
∴ min. value = – 1 and Max. value does not exists.
[Since f (x) can be made as small as we please].

(ii) f(x) = 3 + | x |
Since | x | ≥ 0 ∀ x ∈ R
⇒ 3 + | x | ≥ 3
⇒ f{x) ≥ 3 ∀ x ∈ R
Hence 3 be the minimum value of f (x) which is attains at x = 0 and f (x) can be made as large as we please.
Thus the maximum value of/(x) does not exists.

Question 5.
(i) f(x) = – | x + 1 | + 3 (NCERT)
(ii) f (x) = sin 2x
Solution:
(i) Given, f(x) = – | x + 1 | + 3 on R
since |x + 1| ≥ 0 ∀ x ∈ R
⇒ – |x + 1 |≤ 0 ∀ x ∈ R
⇒ – | x + 1| + 3 ≤ 3 ∀ x ∈ R
⇒ f (x) ≤ 3 ∀ x ∈ R
⇒ f (x) ≤ f (- 1) ∀ x ∈ R
Thus, 1 be the maximum value of f(x) and it attains at x = – 1.
Since, f(x) can be made as small as we please thus minimum value of f(x) does not exists.

(ii) Given f(x) = sin 2x
since – 1 ≤ sin 2x ≤ 1 ∀ x ∈ R
⇒ – 1 ≤ f(x) ≤ 1 ∀ x ∈ R
Thus maximum value of f(x) = 1
and minimum value of f(x) = – 1.

Question 6.
(i) f(x) = sin 2x + 5 (NCERT)
(ii) f(x) = | sin 4x + 3 | (NCERT)
Solution:
(i) Given f(x) = sin 2x + 5
Since, -1 ≤ sin 2x ≤ 1 ∀ x ∈ R
⇒ – 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5 ; ∀ x ∈ R
⇒ 4 ≤ sin 2x + 5 ≤ 6,∀ x ∈ R
⇒ 4 ≤ f(x) ≤ 6, ∀ x ∈ R
∴ Minimum value of f(x) = 4
and maximum value of f(x) = 6.

(ii) Given f(x) = | sin 4x + 3 | = sin 4x + 3
[∵ | sin x | ≤ 1]
Since – 1 ≤ sin 4x ≤ 1 ∀ x ∈ R
– 1 + 3 ≤ sin 4x + 3 ≤ 1 + 3, ∀ x ∈ R
⇒ 2 ≤ sin 4x + 3 ≤ 4 ∀ x ∈ R
2 ≤ f(x) ≤ 4 ∀ x ∈ R
∴ Min. value of f(x) = 2
and max. value of f(x) = 4.

Question 8.
(i) Is the minimum value of cos x zero ?
(ii) What are the minimum and maximum values of 2 – 3 cos x ?
Solution:
(i) Since, | cos x | ≤ 1 ∀ x ∈ R
⇒ – 1 ≤ cos x ≤ 1
Thus, minimum value of f(x) = cos x be – 1
∴ minimum value of cos x is not 0.

(ii) Since, | cos x | ≤ 1 ∀ x ∈ R
⇒ – 1 ≤ cos x ≤ 1, ∀ x ∈ R
⇒ – 3 ≤ 3 cos x ≤ 3
⇒ – 3 ≤ – 3 cos x ≤ 3
⇒ 2 – 3 ≤ 2 – 3 cos x <≤ 2 + 3
⇒ – 1 ≤ 2 – 3 cos x ≤ 5 ∀ x ∈ R
⇒ – 1 ≤ f(x) ≤ 5, ∀ x ∈ R
∴ Maximum value of f(x) = 5
Minimum value of f(x) = – 1.

Question 9.
What are the maximum and minimum values of 3 sin x + 4 cos x ?
Solution:
Let f(x) = 3 sin x + 4 cos x
put 3 = r cos α ; 4 = r sin α
On squaring and adding ; we have
r = \(\sqrt{16+9}\) = 5 ;
tan α = \(\frac{4}{3}\)
⇒ α = tan^-1 (\(\frac{4}{3}\) )
∴ f(x) = r sin (x + α) = 5 sin (x + α)
since – 1 ≤ sin (x + α) < 1 – 5 < 5 sin (x + α) < 5
⇒ – 5 ≤ f(x) ≤ 5
Thus maximum value of f(x) = 5
and Minimum value of f(x) = – 5.

Question 10.
Find the maximum and minimum values, if any, of the following functions :
(i) x in (0, 1) (NCERT)
(ii) x in [0, 1].
Solution:
(i) Let f(x) = x, x ∈ (0, 1)
So it is differentiable ∀ x ∈ (0, 1)
Diff. w.r.t. x, we have
f'(x) = 1 ≠ 0
Thus, f'(x) ≠ 0 for any x ∈ (0, 1)
Hence given function f(x) has no turning point.
Thus, the given function has neither maximum nor minimum values.

(ii) Let f(x) = x, x ∈ [0, 1] …………..(1)
Clearly it is differentiable ∀ x ∈ [0, 1]
DifF. eqn. (1) w.r.t. x, we have
f'(x) = 1 ≠ 0 ∀ x ∈ (- 1, 1)
and hence f(x) has no critical / turning points.
Also, f(0) = 0 ;
f(1) = 1
Maximum value of f(x) = 1
and Minimum value = 0
The point of maxima is 1 and point of minima be 0.

Question 11.
Prove that the function f(x) = x³ + x² + x + 1 does not have maxima or minima.
Solution:
Given f(x) = x³ + x² + x + 1
∴ f (x) = 3x² + 2x + 1
Now f'(x) = 0
⇒ 3x² + 2x + 1 = 0 does not gives any real values of x.
∴ f(x) has no point of maxima or minima.

Question 12.
Find the (absolute) maximum and the (absolute) minimum values of the following functions in the indicated intervals. Also find the points of (absolute) maxima and minima :
(i) f(x) = (x – 1)² + 3 in [- 3, 1] (NCERT)
(ii) f(x) = 2x³ – 15x² + 36x + 1 on the interval [1, 5]
(iii) f (x) – 3x⁴ – 8x³ + 12x² – 48x + 25 in [0, 3] (NCERT)
(iv) f(x) = x + sin x in [0, 2π]
(v) f(x) = sin x + √3 cos x in [0, π]
(vi) f(x) = 3 + | x + 1 | in [- 2, 3].
Solution:
(i) Given f(x) = (x – 1)² + 3 in [- 3, 1]
∴ f(x) = 2(x – 1)
∴ f(x) = 0
⇒ 2x – 2 = 0
⇒ x = 1 ∈ [- 3, 1]
Now y]_{x = 1} = 3 ;
y]_{x = – 3} = 16 + 3 = 19
Absolute maximum value = 19
and Absolute minimum value = 3.

(ii) Given f(x) = 2x³ – 15x² + 36x + l in [1, 5]
∴ f'(x) = 6x² – 30x + 36
∴ f (x) = 0
⇒ 6(x² – 5x + 6) = 0
⇒ x = 2, 3
Now x = 2, 3 ∈ [1, 5]
Now y]_{x = 1} = 2 – 15 + 36 + 1 = 24 ;
y]_{x = 2} = 16 – 60 + 72 + 1 = 29;
y]_{x = 3} = 54 – 135 + 108 + 1 = 23;
y]_{x = 5} = 250 – 375 + 180 + 1 = 56
∴ y has absolute maximum value is 56
and absolute min value = 24.
Thus point of maxima be at x = 5
and point of minima be x = 1

(iii) Given f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25
f'(x) = 12x³ – 24x² + 24x – 48
= 12 (x³ – 2x² + 2x – 4)
For local maxima/minima, we have f'(x) = 0
⇒ 12 [x³ – 2x² + 2x – 4] = 0
⇒ x² (x – 2) + 2 (x – 2) = 0
⇒ (x – 2) (x² + 2) = 0
⇒x-2 = 0 or x² + 2 = 0 does not gives real values of x
∴ x = 2
Now let us compute the values of f(x) at stationary points and also at the end points of given interval [0, 3].
f(2) = 3 × 16 – 8 × 8 + 12 × 4 – 96 + 25
= 48 – 64 + 48 – 96 + 25 = – 39
f(0) = 0 – 0 + 0 – 0 + 25 = 25
f(3) = 3 × 81 -8 × 27 + 12 × 9 – 48 × 3 + 25
= 243 – 216 + 108 – 144 + 25 = 16
Out of these values, maximum value be f(0) = 25
and least value be f(2) = – 39
Hence absolute maximum value = 25 at x = 0
and absolute minimum value = – 39 at x = 2.

(iv) Given f(x) = x + sin x in [0, 2π]
Diff. both sides w.r.t. x, we have
f'(x) = 1 + cos x
Now, f'(x) = 0
⇒ 1 + cos x = 0
⇒ cos x = – 1 = cos π
But x ∈ [0, 2π]
∴ x = π
Now, f(π) = π + sin π
= π + 0 = π
f(0) = 0 + sin 0
= 0 + 0 = 0
f(2π) = 2π + sin 2π
= 2π + 0 = 2π
Thus, absolute maximum value = 2π
and point of maxima be 2π.
Also, absolute minimum value of f(x) be 0
and it attains at x = 0.

(v) Given f(x) = sin x + √3 cos x
∴ f'(x) = cos x – √3 sin x
Now critical points are given by putting f'(x) = 0
⇒ cos x – √3 sin x = 0
⇒ cos x = √3 sin x
⇒ tan x = \(\frac{1}{\sqrt{3}}\) = tan \(\frac{\pi}{6}\)
⇒ x = nπ + \(\frac{\pi}{6}\), n ∈ Z
But x ∈ [0, π]
∴ x = \(\frac{\pi}{6}\)
Now we computed f(x0 at critical points and end points
i.e. x = 0, \(\frac{\pi}{6}\), π
∴ f(0) = 0 + √3 × 1 = √3;
f(\(\frac{\pi}{6}\)) = \(\frac{1}{2}+\sqrt{3} \times \frac{\sqrt{3}}{2}\) = 2
f(π) = 0 + √3 cos π = – √3
Out pf these values, maximum value of f(x) be 2 at x = \(\frac{\pi}{6}\)
and minimum value of f(x) = – √3 at x = π
∴ points of maxima and minima are x = \(\frac{\pi}{6}\) and π.

Question 12 (old).
(i) f(x) = x³ in [- 2, 2] (NCERT)
(ii) f(x) = 4x – x² in [- 2, \(\frac{9}{2}\)]
(iii) f(x) = (\(\frac{1}{2}\) – x)² + x³ in [- 2, 2-5]
(iv) f(x) = sin x + cos x in [0, π] (NCERT)
Solution:
(i) Given f(x) = x³ in [- 2, 2]
∴ f'(x) = 3x² i.e. f(x) = 0
⇒ 3x² = 0
⇒ x = 0 ∈ [- 2, 2]
Now y]_{x = 0} = 0;
y]_{x = 2} = 8;
y]_{x = – 2} = – 8
absolute max. value = 8 ;
absolute min value = – 8.

(ii) Given f(x) = 4x – \(\frac{1}{2}\) x² ; x ∈ [- 2, \(\frac{9}{2}\)]
∴ f'(x) = 4 – x
Now f'(x) = 0
⇒ x = 4 ∈ [- 2, \(\frac{9}{2}\)]
∴ y]_{x = – 2} = – 8 – \(\frac{1}{2}\) × 4 = – 10;
y]_{x = \(\frac{9}{2}\)} = 18 – \(\frac{1}{2} \times \frac{81}{4}\)
= \(\frac{144-81}{8}=\frac{63}{8}\)
y]_{x = 4} = 4 × 4 – \(\frac{1}{2}\) (4)² = 8
∴ y is absolute max. at x = 4
and absolute maximum value = 8
and y is absolute minimum at x = -2
and absolute minimum value = -10.

(iii) f(x) = (\(\frac{1}{2}\) – x)² + x³ in [- 2, 2.5]
∴ f'(x) = 2 (\(\frac{1}{2}\) – x) (- 1) + 3x²
∴ f'(x) = 0
⇒ 3x² – 1 + 2x = 0
⇒ (x + 1) (3x – 1) = 0
∴ x = – 1, \(\frac{1}{3}\) ∈ [- 2, 2.5]
Now y]_{x = – 1} = \(\frac{9}{4}\) + (- 1) = \(\frac{5}{4}\) ;
y]_{x = \(\frac{1}{3}\)} = \(\frac{1}{36}+\frac{1}{27}=\frac{7}{108}\)
y]_{x = – 2} = \(\frac{5}{4}\) – 8 = \(\frac{-7}{4}\)
y]_{x = \(\frac{5}{2}\)} = 4 + \(\frac{125}{8}\) = \(\frac{157}{8}\)
∴ absolute max. value = \(\frac{157}{8}\)
and absolute min. value = \(\frac{-7}{4}\)
Thus point of maxima be at x = \(\frac{5}{2}\) and
point of minima be x = – 2.

(iv) Given f(x) = sin x + cos x, [0, π]
∴ f'(x) = cos x – sin x,
f'(x) = 0
⇒ cos x – sin x = 0
⇒ tan x = 1
∴ x = \(\frac{\pi}{4}\) ∈ [0, π]
Now y]_{x = 0} = 1 ;
y]_{x = π} = – 1 ;
y]_{x = \(\frac{\pi}{4}\)} = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) = √2
Absolute max. value = √2 and attains at x = \(\frac{\pi}{4}\)
and Absolute min value = – 1 and attains at x = π

(vi) Given f(x) = 3 + | x + 1 | ∀ x ∈ [- 2, 3]
Clearly f(x) is differentiable ∀ x ∈ [- 2, 3] except at x = – 1.
Diff. both sides w.r.t. x, we have
f'(x) = \(\frac{x+1}{|x+1|}\)
Now f'(x) = 0
⇒ \(\frac{x+1}{|x+1|}\) = 0, x ≠ – 1
which gives no values of x i.e. has no solution.
So f(x) has no critical points.
So, f(- 1) = 3 + |- 1 + 1| = 3
f(- 2) = 3 + |- 2 + 1|
= 3 + |- 1| = 4
and f(3) = 3 + |3 + 1|
= 3+ 4 = 7
Thus absolute max. value of f(x) = 7 and it attains at x = 3.
and Absolute minimum value of f(x) = 3 and it attains at x = – 1.

Question 13.
Find the maximum and minimum values of x + sin 2x on [0, 2π]. Also find points of maxima and minima. (NCERT)
Solution:
Let y = x + sin 2x on [0, 2π]
∴ \(\frac{d y}{d x}\) = 1 + 2 cos 2x,
Now \(\frac{d y}{d x}\) = 0
⇒ 1 + 2 cos 2x = 0
⇒ cos 2x = – \(\frac{1}{2}\)
∴ cos 2x = – cos (\(\frac{\pi}{3}\))
= cos (π – \(\frac{\pi}{3}\))
⇒ cos x = cos \(\frac{2 \pi}{3}\)
⇒ 2x = 2nπ ± \(\frac{2 \pi}{3}\)
[∵ cos θ = cos α
⇒ 0 = 2nπ ± α]
⇒ x = nπ ± \(\frac{\pi}{3}\)
∴ x = \(\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}\)
∴ y]_{x = 0} = 0 ;
y]_{x = 2π} = 2π ;
y]_{x = \(\frac{\pi}{3}\)} = \(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\) ;
y]_{x = \(\frac{2\pi}{3}\)} = \(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\) ;
y]_{x = \(\frac{4 \pi}{3}\)} = \(\frac{4 \pi}{3}+\frac{\sqrt{3}}{2}\) ;
y]_{x = \(\frac{5 \pi}{3}\)} = \(\frac{5 \pi}{3}-\frac{\sqrt{3}}{2}\)
∴ y is absolute maximum at x = 2π
and absolute Max. value = 2π
and y is absolute minimum at x = 0
and absolute minimum value = 0.

Question 14.
Find the absolute maximum and absolute minimum values of the function f(x) = cos² x + sin x, x ∈ [0, π].
Solution:
Given f(x) = cos² x + sin x, x ∈ [0, π]
∴ f’(x) = – 2 cos x sin x + cos x
= cos x (1 – 2 sin x)
The critical points are given by putting f’(x) = 0
⇒ cos x (1 – 2 sin x) = 0
⇒ cos x = 0 or sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
⇒ x = \(\frac{\pi}{2}\) or x = \(\frac{\pi}{6}\), \(\frac{5 \pi}{6}\)
Thus, we compute f(x) at x = 0, \(\frac{\pi}{6}\), \(\frac{\pi}{2}\), \(\frac{5 \pi}{6}\), π
Here, f(0) = 1;
f(\(\frac{\pi}{6}\)) = \(\left(\frac{\sqrt{3}}{2}\right)^2+\frac{1}{2}\)
= \(\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
f(\(\frac{\pi}{2}\)) = 1 ;
f(\(\frac{5 \pi}{6}\)) = \(\left(\frac{\sqrt{3}}{2}\right)^2+\frac{1}{2}=\frac{5}{4}\) ;
f(π) = 1
Thus, maximum value of f(x) = \(\frac{5}{4}\)
and minimum value of f(x) = 1.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5

November 22, 2023

Students appreciate clear and concise ISC Maths Class 12 Solutions Chapter 7 Applications of Derivatives Ex 7.5 that guide them through exercises.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5

Question 1.
Show that the following functions are strictly increasing on R :
(i) f(x) = 3x+ 17 (NCERT)
(ii) f(x) = 7x – 3. (NCERT)
Solution:
(i) Given f(x) = 3x + 17
Diff. both sides w.r.t. x, we have
f’(x) = 3 > 0 ∀ x ∈ R
Thus the functionf(x) is strictly increasing on R.

Aliter:
Given f(x) = 3x + 17 and let x₁, x₂ ∈ R
s.t. x₁ > x₂
⇒ 3x₁ > 3x₂
⇒ 3x₁ + 7 > 3x₂ + 7
⇒ f(x₁) >f(x₂)
∴ f(x) is strictly increasing on R.

(ii) Given f(x) = 7x – 3
∴ f’(x) = 7 > 0 ∀ x ∈ R
Hence, the function f (x) is strictly increasing on R.

Question 2.
(i) Show that the function f given by f(x) = 3 – 7x is strictly decreasing.
(ii) Show that the function f(x) = 1 – \(\frac{1}{x}\) is strictly increasing.
Solution:
(i) Given f(x) = 3 – 7x
∴ f’(x) = – 7 < 0 ∀ x ∈ R Thus, the function f(x) is strictly decreasing on R.

(ii) Given f(x) = 1 – \(\frac{1}{x}\)
∴ f'(x) = \(\frac{1}{x^2}\) > 0 ∀ x ∈ R
[∵ x² ≥ 0 ∀ x ∈ R]
Thus f(x) is strictly increasing on R.

Question 3.
(i) Prove that the function f(x) = 3 + \(\frac{1}{x}\) is strictly decreasing.
(ii) Show that the function f(x) = x – \(\frac{1}{x}\) is strictly increasing.
Solution:
(i) Given f(x) = 3 + \(\frac{1}{x}\)
∴ f'(x) = – \(\frac{1}{x^2}\) < 0
[∵ x² ≥ 0 ∀ x ∈ R
⇒ \(\frac{1}{x^2}\) > 0 ∀ x ∈ R]
Thus, f(x) is strictly decreasing on R.

(ii) Given f(x) = x – \(\frac{1}{x}\)
∴ f'(x) = 1 + \(\frac{1}{x^2}\) ∀ x ∈ R
[∵ x² ≥ 0 ∀ x ∈ R
⇒ \(\frac{1}{x^2}\) > 0
⇒ 1 + \(\frac{1}{x^2}\) > 1 ∀ x ∈ R]
Thus f(x) is strictly increasing on R.

Question 4.
(i) Show that the function f (x) = e^{– 2x} is strictly decreasing on R.
(ii) Show that the function f (x) = 5^x is strictly increasing.
(iii) Show that the function f (x) = log x is strictly increasing. (NCERT)
Solution:
(i) Given f(x) = e^{– 2x}
f (x) = – 2 e^-2x < 0
[∵ e^{– 2x} > 0 ∀ x ∈ R]
Hence f(x) is strictly decreasing on R.

(ii) Given f(x) = 5^x
f'(x) = 5^x log 5 ∀ x ∈ R
[∵ 5^x > 0 ∀ x ∈ R]
Hence f(x) is strictly increasing on R.

(iii) Let x₁, x₂ ∈ (0, ∞) s.t. x₁ < x₂
⇒ log_e x₁ < log_e x₂
⇒ f(x₁) < f(x₂)
Thus x₁ < x₂
⇒ f(x₁) < f(x₂) ∀ x₁, x₂ ∈ (0, ∞)
Hence f(x) is increasing on (0, ∞).

Aliter :
f(x) = log x
∴ f'(x) = \(\frac{1}{x}\) > 0 ∀ x ∈ R (0, ∞)
[∵ as x > 0 ⇒ \(\frac{1}{x}\) > 0]
f(x) is strictly increasing in (0, ∞).

Question 5.
Show that the function f(x) = – x + cos x is decreasing.
Solution:
Given, f(x) = – x + cos x
∴ f'(x) = – 1 – sin x = – (1 + sin x)
Since, – 1 ≤ sin x ≤ 1
⇒ 1 + sin x ≥ 0
⇒ – (1 + sin x) ≤ 0
∴ f'(x) ≤ 0
Hence f(x) is decreasing on R.

Question 6.
For what value of k, the function f(x) = kx³ + 5 is decreasing ?
Solution:
Given f(x) = kx³ + 5 ;
f'(x) = 3kx²
Now f(x) is decreasing if f'(x) ≤ 0
if 3kx² ≤ 0 if 3k < 0
[∵ x² > 0 ∀ x ∈ R] if k< 0.

Question 7.
(i) Prove that the function f(x) = ax + b is strictly decreasing iff a < 0.
(ii) Prove that the function/(x) = [x] – x is strictly decreasing on [0, 1).
Solution:
(i) Given f(x) = ax + b
∴ f'(x) = a
f (x) is strictly decreasing iff f'(x) < 0 iff a < 0.

(ii) Given f(x) = [x] – x, x ∈ [0, 1)
Since x ∈ [0, 1)
∴ 0 < x < 1
⇒ [x] = 0
f(x) = 0 – x = – x, x ∈ [0, 1)
⇒ f'(x) = – 1 < 0
Thus f(x) is strictly decreasing in [0, 1).

Question 8.
Prove that the function f(x) = x³ – 6x² + 12x + 5 is increasing on R. (ISC 2019)
Solution:
Given f(x) = x³ – 6x² + 12x + 5
∴ f'(x) = 3x² – 12x + 12
= 3 (x² – 4x + 4)
= 3 (x – 2)² ≥ 0 ∀ x ∈ R.
Thus f(x) is increasing on R.

Question 8 (old).
Show that the function f given by f(x) = x³ – 3x² + 4x, x ∈ R, is strictly increasing on R. (NCERT)
Solution:
Given f(x) = x³ – 3x² + 4x
∴ f'(x) = 3x² – 6x + 4
= 3 (x² – 2x + \(\frac{4}{3}\))
= 3 [x² – 2x + 1 + \(\frac{1}{3}\)]
= 3 [(x – 1)² + \(\frac{1}{3}\)]
[∵ (x – 1)² ≥ 0
⇒ (x – 1)² + \(\frac{1}{3}\) ≥ \(\frac{1}{3}\) > 0
⇒ 3 [(x – 1)² + \(\frac{1}{3}\)] > 0
Thus, f(x) is strictly increasing on R.

Question 9.
Show that the function f(x) = 4x³ – 18x² + 27x – 7 is strictly increasing on R.
Solution:
Given f(x) = 4x³ – 18x² + 27x – 7
∴ f'(x) = 12x² – 36x + 27
= 3 (4x² – 12x + 9)
= 3 (2x – 3)² ≥ 0 ∀ x ∈ R.

Question 9 (old).
Prove that the function f(x) = x³ – 3x² + 3x – 100 is strictly increasing on R. (NCERT)
Solution:
Given f(x) = x³ – 3x² + 3x – 100
∴ f'(x) = 3x² – 6x + 3
= 3 (x² – 2x + 1)
= 3 (x- 1)² > 0
∴ f(x) is increasing function on R.

Question 10.
Prove that the function f(x) = 5 – 3x + 3x² – x³ is decreasing on R.
Solution:
Given f(x) = 5 – 3x + 3x² – x³ …………(1)
Diff both sides eqn. (1) w.r.t. x, we have
f(x) = – 3 + 6x – 3x²
= – 3 (x² – 2x + 1)
= – 3 (x – 1)²
since, (x – 1)² > 0 ∀ x ∈ R
⇒ – 3 (x – 1)² < 0, ∀ x ∈ R
⇒ f'(x) < 0, ∀ x ∈ R
Thus, f(x) is decreasing on R.

Question 11.
Show that f(x) = 2x – sin 2x is an increasing function.
Solution:
Given f(x) = 2x – sin 2x
Diff. both sides w.r.t. x, we have
f'(x) = 2 – 2 cos 2x
= 2 (1 – cos 2x)
= 2 × 2 sin² x = 4 sin² x
Since 0 ≤ sin² x ≤ 1 ∀ x ∈ R
⇒ 0 ≤ 4 sin² x ≤ 4
⇒ 0 ≤ f'(x) ≤ 4
Thus, f'(x) ≥ 0 ∀ x ∈ R
Hence, f(x) is increasing on R.

Question 12.
Prove that the function f given by f(x) = log cos x is strictly decreasing on (0, \(\frac{\pi}{2}\)). (NCERT)
Solution:
Given f(x) = log (cos x)
∴ f'(x) = – tan x
when x ∈ (0, \(\frac{\pi}{2}\))
⇒ tan x > 0
⇒ f'(x) < 0
∴ f(x) is strictly decreasing in (0, \(\frac{\pi}{2}\)).

Question 13.
Determine whether the following functions are increasing or decreasing for the stated values of x :
(i) f(x) = \(\frac{1}{x}\), x < 0
(ii) f(x) = \(\frac{1}{1+x^2}\), x ≥ 0
(iii) f(x) = cos² x in [0, \(\frac{\pi}{2}\)]
(iv) f(x) = – \(\frac{x}{2}\) + sin x, – \(\frac{\pi}{3}\) < x < \(\frac{\pi}{3}\)
Solution:
(i) Given f(x) = \(\frac{1}{x}\)
∴ f'(x) = – \(\frac{1}{x^2}\) < 0
[When x < 0
⇒ x² > 0
⇒ \(\frac{1}{x^2}\) > 0
⇒ – \(\frac{1}{x^2}\) < 0]
Thus f(x) is strictly decreasing.

(ii) Given f(x) = \(\frac{1}{1+x^2}\), x ≥ 0
Diff. both sides w.r.t. x, we have
f'(x) = \(\frac{-2 x}{\left(1+x^2\right)^2}\)
Since (1 + x²)² > 0 ∀ x ∈ R
⇒ \(\frac{1}{\left(1+x^2\right)^2}\) ≥ 0
⇒ \(\frac{-2 x}{\left(1+x^2\right)^2}\) ≤ 0 ∀ x ≥ 0
Thus, f'(x) ≤ 0 ∀ x ≥ 0
Hence, f(x) is decreasing for all x ≥ 0.

(iii) f(x) = cos² x
∴ f'(x) = 2 cos x (- sin x) = – sin 2x
Now f'(x) < 0
⇒ – sin 2x < 0 ⇒ sin 2x > 0
⇒ 2x ∈ (0, π)
⇒ x ∈ (0, π/2)
∴ f(x) is strictly decreasing in [0, \(\frac{\pi}{2}\)].

(iv) Given f(x) = – \(\frac{x}{2}\) + sin x
∴ f'(x) = – \(\frac{1}{2}\) + cos x
Now for x ∈ (- \(\frac{\pi}{3}\), \(\frac{\pi}{3}\))
⇒ – \(\frac{\pi}{3}\) < x < \(\frac{\pi}{3}\)
⇒ \(\frac{1}{2}\) < cos x < 1
⇒ – \(\frac{1}{2}\) + cos x > 0
⇒ f'(x) > 0
Hence f'(x) is strictly increasing function on (- \(\frac{\pi}{3}\), \(\frac{\pi}{3}\)).

Question 14.
Which of the following functions are strictly decreasing on ?
(i) cos x
(ii) cos 2x
(iii) sin2 x
(iv) tan x (NCERT)
Solution:
(i) Given f(x) = cos x ;
Diff. both sides w.r.t. x, we have
⇒ f’(x) = – sin x ∀ x ∈ (0, \(\frac{\pi}{2}\)) ;
sin x > 0
⇒ – sin x < 0
⇒ f'(x) < 0
Thus f(x) is strictly decreasing on (0, \(\frac{\pi}{2}\)).

(ii) Given f(x) = cos 2x ;
Diff. both sides w.r.t. x, we get
f'(x) = – 2 sin 2x
When 0 < x < \(\frac{\pi}{2}\) ⇒ 0 < 2x < π ∴ sin 2x > 0
⇒ – 2 sin 2x < 0
⇒ f'(x) < 0
Hence, f(x) is strictly decreasing on (0, \(\frac{\pi}{2}\)).

(iii) Given f(x) = sin² x ;
Diff. both sides w.r.t. x, we have
f'(x) = 2 sin x cos x = sin 2x
when x ∈ (0, \(\frac{\pi}{2}\))
i.e. 0 < x < \(\frac{\pi}{2}\) ⇒ 0 < 2x < π ⇒ sin 2x > 0
⇒ f'(x) > 0
Hence, f(x) is strictly increasing on (0, \(\frac{\pi}{2}\)).

(iv) Given f(x) = tan x;
Diff. both sides w.r.t. x, we have
f'(x) = sec² x > 0 ∀ x ∈ (0, \(\frac{\pi}{2}\))
Thus f(x) is strictly increasing on (0, \(\frac{\pi}{2}\)).

Question 15.
Prove that the function f(x) = sin (2x + \(\frac{\pi}{4}\)) is decreasing for \(\frac{3 \pi}{8}\) ≤ x ≤ \(\frac{5 \pi}{8}\).
Solution:
Given f(x) = sin (2x + \(\frac{\pi}{4}\))
∴ f'(x) = + 2 cos (2x + \(\frac{\pi}{4}\))
Now for x ∈ [\(\frac{3 \pi}{8}\), \(\frac{5 \pi}{8}\)]
⇒ \(\frac{3 \pi}{8}\) ≤ x ≤ \(\frac{5 \pi}{8}\)
⇒ \(\frac{3 \pi}{4}\) ≤ 2x ≤ \(\frac{5 \pi}{4}\)
⇒ \(\frac{3 \pi}{4}+\frac{\pi}{4} \leq 2 x+\frac{\pi}{4} \leq \frac{5 \pi}{4}+\frac{\pi}{4}\)
⇒ π ≤ 2x + \(\frac{\pi}{4}\) ≤ \(\frac{3 \pi}{2}\)
∴ 2x + \(\frac{\pi}{4}\) lies in 3rd quadrant.
∴ cos (2x + \(\frac{\pi}{4}\)) ≤ 0
∴ f'(x) = 2 cos (2x + \(\frac{\pi}{4}\)) ≤ 0
Hence f(x) is decreasing function on \(\left[\frac{3 \pi}{8}, \frac{5 \pi}{8}\right]\).

Question 16.
Prove that the function f(x) = sin x is
(i) strictly increasing in (0, \(\frac{\pi}{2}\))
(ii) strictly decreasing in (\(\frac{\pi}{2}\), π)
(iii) neither increasing nor decreasing in (0, π). (NCERT)
Solution:
(i) Given f(x) = sin x
∴ f’(x) = cos x > 0 ∀ x ∈ (0, \(\frac{\pi}{2}\))
[since in first quadrant, cos x > 0]
Thus f(x) is strictly increasing on (0, \(\frac{\pi}{2}\)).

(ii) Also f’(x) = cos x < 0 ∀ x ∈ (\(\frac{\pi}{2}\), π)
since in 2nd quadrant, cos x is negative.
Thusf(x) is strictly decreasing on (\(\frac{\pi}{2}\), π)

(iii) Hence f(x) is increasing on (0, \(\frac{\pi}{2}\)) and decreasing on (\(\frac{\pi}{2}\), π) Therefore, f (x) is neither increasing nor decreasing on (0, π).

Question 17.
Prove that the function f defined by f(x) = cos x is
(i) strictly incresing in (π, 2π)
(ii) strictly decreasing in (0, 2π)
(iii) neither increasing nor decreasing in (0, 2π). (NCERT)
Solution:
(i) Given f(x) = cos x
∴ f’(x) = – sin x
Now f(x) is strictly increasing if f’(x) > 0
i.e. if – sin x > 0
i.e. sin x < 0
⇒ x ∈ (π, 2π)

(ii) f(x) is strictly decreasing if f’(x) < 0
i.e. – sin x < 0 i.e. sin x > 0
⇒ x ∈ (0, π)

(iii) So f(x) is neither increasing nor decreasing in (0, 2π).
Since f(x) is strictly increasing in (π, 2π) and strictly decreasing in (0, π).

Question 18.
Prove that the following functions are neither increasing nor decreasing :
(i) f(x) = x² – x + 1 on (- 1, 1) (NCERT)
(ii) f(x) = \(\frac{1}{x^2+1}\) on R.
Solution:
(i) Given f(x) = x² – x + 1
∴ f'(x) = 2x – 1
Now f’(x) > 0
⇒ 2x – 1 > 0
⇒ x > \(\frac{1}{2}\)
∴ f(x) is increasing in (\(\frac{1}{2}\), 1).
Now f'(x) < 0
⇒ 2x – 1 < 0
⇒ x < \(\frac{1}{2}\)
∴ f(x) is decreasing on (- 1, \(\frac{1}{2}\)).

(ii) Given f(x) = \(\frac{1}{x^2+1}\) ;
Diff. both sides w.r.t. x ; we have
∴ f'(x) = \(\frac{-2 x}{\left(x^2+1\right)^2}\)
Now f(x) is increasing iff f'(x) ≥ 0
iff \(\frac{-2 x}{\left(x^2+1\right)^2}\) ≥ 0 iff – 2x ≥ 0
[∵ (x² + 1)² ≥ 0, ∀ x ∈ R]
iff x ≤ 0
and f(x) is decreasing iff f'(x) ≤ 0
iff \(\frac{-2 x}{\left(x^2+1\right)^2}\) ≤ 0 iff – 2x ≤ 0
[∵ (x² + 1)² ≥ 0, ∀ x ∈ R]
iff x ≥ 0
Hence, the function f(x) is neither increasing nor decreasing on R.

Question 19.
Prove that the function f(x) = x is
(i) strictly increasing in (0, ∞)
(ii) strictly decreasing in (- ∞, 0).
Solution:
Given f(x) = (x)
= \(\left\{\begin{array}{cc}
x & ; x>0 \\
-x & ; x<0 \\ 0 & ; x=0 \end{array}\right.\) (i) When x > 0
f(x) = x
⇒ f’(x) = 1
∴ f(x) is strictly increasing in (0, ∞)

(ii) When x < 0 f(x) = – x ⇒ f’(x) = – 1
∴ f(x) is strictly decreasing in (∞, 0).

Question 20.
Prove that the function f given by f(x) = \(\frac{4 \sin x}{2+\cos x}\) – x is increasing in [o, \(\frac{\pi}{2}\)].
Solution:
Given f(x) = \(\frac{4 \sin x}{2+\cos x}\) – x
Diff. both sides w.r.t. x, we have
f'(x) = \(\frac{(2+\cos x)(4 \cos x)-4 \sin x(-\sin x)}{(2+\cos x)^2}\) – 1
= \(\frac{8 \cos x+4}{(2+\cos x)^2}\) – 1
= \(\frac{8 \cos x+4-(2+\cos x)^2}{(2+\cos x)^2}\)
= \(\frac{-\cos ^2 x+4 \cos x}{(2+\cos x)^2}\)
= \(\frac{\cos x(4-\cos x)}{(2+\cos x)^2}\)
Since cos x > 0,
4 – cos x > 0 ∀ x ∈ (o, \(\frac{\pi}{2}\))
∴ cos x (4 – cos x) > 0 ∀ x ∈ (o, \(\frac{\pi}{2}\))
Also (2 + cos x)² > 0
∴ f'(x) > 0 ∀ x ∈ (o, \(\frac{\pi}{2}\))
Thus f is strictly increasing in [o, \(\frac{\pi}{2}\)].
[if f'(x) > 0 ∀ x ∈ (a, b) Then f is strictly increasing in [a, b]]

Question 21.
Find the intervals in which the following functions are strictly increasing or strictly decreasing :
(i) f(x) = 2x³ – 3x² – 36x + 7 (NCERT)
(ii) f(x) = 4x³ – 6x² – 72x + 30 (NCERT)
(iii) f(x) = 2x³ – 9x² + 12x + 15
(iv) f(x) = – 2x³ – 9x² – 12x + 1
(v) f(x) = x + \(\frac{1}{x}\), x ≠ 0
(vi) f(x) = \(\frac{4 x^2+1}{x}\), x ≠ 0
(vii) f(x) = x⁴ – 8x³ + 22x² – 24x + 21
(viii) f(x) = \(\frac{1}{4}\) x⁴ – x³ – 5x² + 24x + 12
(ix) f(x) = x⁴ – 4x
(x) f(x) = (x – 1)³ (x – 2)².
Solution:
(i) Given f(x) = 2x³ – 3x² – 36x + 7
∴ f'(x) = 6x² – 6x – 36
= 6 (x² – x – 6)
= 6 (x + 2) (x – 3)
Now f'(x) > 0
⇒ (x + 2) (x – 3) > 0
⇒ x > 3 or x < – 2
∴ f(x) is strictly increasing in (- ∞, – 2] ∪ [3, ∞)
Now f'(x) < 0
⇒ (x + 2) (x – 3) < 0
⇒ – 2 < x < 3
∴ f(x) is strictly decreasing in [- 2, 3].

(ii) Given f(x) = 4x³ – 6x² – 72x + 30
∴ f'(x) = 12x² – 12x – 72
= 12 (x² – x – 6)
= 12 (x + 2) (x – 3)
Now f'(x) > 0
⇒ (x + 2) (x – 3) > 0
⇒ x > 3 or x < – 2
∴ f(x) is stictly increasing in (- ∞, – 2] ∪ [3, ∞)
Now f'(x) < 0
⇒ (x + 2) (x – 3) < 0
⇒ – 2 < x < 3.
∴ f(x) is strictly decreasing in [- 2, 3].

(iii) Given f(x) = 2x³ – 9x² + 12x + 15
∴ f’(x) = 6x² – 18x + 12
= 6 (x² – 3x + 2)
= 6 (x – 1) (x – 2)
Now f'(x) > 0
⇒ 6 (x – 1) (x – 2) > 0
⇒ x > 2 or x < 1
∴ f(x) is stictly increasing in (- ∞, 1] ∪ [2, ∞)
Now f'(x) < 0
⇒ 6 (x – 1) (x – 2) < 0
⇒ (x – 1) (x – 2) < 0
⇒ 1 < x < 2
∴ f(x) is increasing in [1, 2].

(iv) Given f(x) = – 2x³ – 9x² – 12x + 1
∴ f’(x) = – 6x² – 18x – 12
= – 6 (x² + 3x + 2)
= – 6 (x + 1) (x + 2)
For f(x) to be increasing, we must have f'(x) > 0
∴ – 6 (x + 1) (x + 2) > 0
⇒ (x + 1) (x + 2) < 0
⇒ – 2 < x – 1
⇒ x ∈ (- 2, – 1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5 1

signs of f’ (x) for different values of x
Thus, f(x) is decreasing on [- 2, – 1]
For f (x) to be decreasing we must have f’(x) < 0
– 6 (x + 1) (x + 2) < 0 (x + 1) (x + 2) > 0
x < – 2 or x > – 1
i.e. x ∈ (- ∞, – 2) ∪ (- 1, ∞)
Thus,f(x) is decreasing on (- ∞, – 2) ∪ (- 1, ∞).

(v) Given y = x + \(\frac{1}{x}\), x ≠ 0
∴ f'(x) = \(\frac{d y}{d x}\)
= 1 – \(\frac{1}{x^2}\)
= \(\frac{x^2-1}{x^2}\)
Now \(\frac{d y}{d x}\) > 0
⇒ \(\frac{x^2-1}{x^2}\) > 0
⇒ x² > 1
⇒ |x| > 1
⇒ x > 1 or x < – 1
⇒ f(x) is strictly increasing in (- ∞, – 1] ∪ [1, ∞).
Also f'(x) < 0
⇒ \(\frac{x^2-1}{x^2}\) < 0
⇒ x² – 1 < 0
⇒ |x| < 1
⇒ – 1 < x < 1 Thus, f(x) is strictly decreasing in [- 1, 1].

(vi) Given f(x) = \(\frac{4 x^2+1}{x}\)
= 4x + \(\frac{1}{x}\)
∴ f’(x) = 4 – \(\frac{1}{x^2}\)
= \(\frac{4 x^2-1}{x^2}\)
Now f'(x) > 0
⇒ \(\frac{4 x^2-1}{x^2}\) > 0
⇒ 4x² – 1 > 0
⇒ x² > \(\frac{1}{4}\)
⇒ |x| > \(\frac{1}{2}\)
⇒ x > \(\frac{1}{2}\) or x < – \(\frac{1}{2}\)
∴ f(x) is stictly increasing in (- ∞, – \(\frac{1}{2}\)] ∪[\(\frac{1}{2}\), ∞).
Now f'(x) < 0
⇒ \(\frac{4 x^2-1}{x^2}\) < 0
⇒ 4x² – 1 < 0
⇒ |x| < \(\frac{1}{2}\)
⇒ – \(\frac{1}{2}\) < x < \(\frac{1}{2}\)
∴ f(x) is strictly decreasing in [- \(\frac{1}{2}\), \(\frac{1}{2}\)].

(vii) Given f(x) = x⁴ – 8x³ + 22x² – 24x + 21
Diff. both sides w.r.t. x, we have
f'(x) = 4x³ – 24x² + 44x – 24
= 4 (x³ – 6x² + 11x – 6)
= 4 (x – 1) (x – 2) (x – 3)
Now f'(x) > 0
iff 4 (x – 1) (x – 2) (x – 3) > 0
⇒ (x – 1) (x – 2) (x – 3) > 0
⇒ 1 < x < 2 or x > 3
⇒ x ∈ (1, 2) ∪ (3, ∞)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5 2

Thus,f(x) is strictly increasing in [1, 2] ∪ [3, ∞)
Now f’(x) < 0 iff 4 (x – 1) (x – 2)( x – 3) < 0
iff (x- 1) (x – 2) (x – 3) < 0
⇒ x< 1 or 2 < x < 3
⇒ x ∈ (- ∞, 1) ∪ (2, 3)
Thus,f(x) is strictly decreasing in (- ∞, 1] ∪ [2, 3].

(viii) Given f(x) = \(\frac{1}{4}\) x⁴ – x³ – 5x² + 24x + 12
∴ f(x) = x³ – 3x² – 10x + 24
= (x – 2) (x² – x – 12)
= (x – 2) (x – 4) (x + 3)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5 4

Now f'(x) > 0 iff (x – 2) (x – 4) (x + 3) > 0
⇒ x ∈ (4, ∞) ∪ (- 3, 2) [method of intervals)
⇒ x ∈ (- 3, 2) ∪ (4, ∞)
Thus f is strictly increasing in [- 3, 2] ∪ (4, ∞)
⇒ x ∈ (- ∞, – 3) ∪ (2, 4)
Thus f is strictly decreasing in (- ∞, – 3] ∪ [2, 4]

(ix) Given f(x) = x⁴ – 4x
∴ f'(x) = 4x³ – 4
⇒ f'(x) = 4 (x – 1) (x² + x + 1)
= 4 (x – 1) [x² + x + \(\frac{1}{4}\) + \(\frac{3}{4}\)]
= 4 (x – 1) \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\)
For f(x) to be increasing, we must have f'(x) > 0
⇒ 4 (x – 1) \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\) > 0
⇒ (x – 1) > 0
[∵ \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\) > 0]
⇒ x > 1
⇒ x ∈ (1, ∞)
Hence f(x) is increasing on [1, ∞).
For f(x) to be decreasing we must have f'(x) < 0
⇒ 4 (x – 1) \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\) < 0
⇒ (x – 1) < 0 [∵ \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\) > 0]
⇒ x < 1
⇒ x ∈ (- ∞, 1)
Thus f(x) is decreasing on (- ∞, 1].

(x) Given f(x) = (x – 1)³ (x – 2)²
∴f’(x) = (x – 1)³ 2 (x – 2) + (x – 2)² 3 (x – 1)²
= (x – 1)² (x – 2) [2 (x – 1) + 3 (x – 2)]
= (x – 1)² (x – 2) (5x – 8)
Now f’(x) = 0
⇒ (x – 1)² (x – 2) (5x – 8) = 0
⇒ x = 1, 2, \(\frac{8}{5}\)
Now f’(x) > 0
iff (x – 1)² (x – 2) (5x – 8) > 0
iff (x – 2) (5x – 8) > 0
[∵ (x – 1)² > 0 ∀ x ∈ R – {1}]
iff x > 2 or x < \(\frac{8}{5}\) and x ≠ 1
iff x ∈ (2, ∞) or x ∈ (- ∞, 1) (1, \(\frac{8}{5}\))
Thus f(x) is stictly increasing in (- ∞, 1) ∪ (1, \(\frac{8}{5}\)) ∪ [2, ∞)
Now f'(x) < 0
iff (x – 1)² (x – 2) (5x – 8) < 0
iff (x – 2) (5x – 8) < 0
[∵ (x – 1)² > 0 ∀ x ∈ R – {1}]
iff \(\frac{8}{5}\) < x < 2
iff x ∈ (\(\frac{8}{5}\), 2)
Thus f(x) is strictly decreasing in [\(\frac{8}{5}\), 2].

Question 21 (old).
(i) f(x) = x² – 4x + 6 (NCERT)
(ii) f(x) = x² + 2x – 5 (NCERT)
(iii) f(x) = 2x² – 3x (NCERT)
(iv) f(x) = 6 – 9x – x² (NCERT)
(xii) f(x) = \(\frac{1}{4}\) x⁴ + \(\frac{2}{3}\) x³ + \(\frac{5}{2}\) x² – 6x + 9
Solution:
(i) Given f(x) = x² – 4x – 6
⇒ f’(x) = 2x – 4
∴ f(x) is strictly increasing if f’(x) > 0 i.e. if x> 2
∴ f(x) is stictly increasing in [2, ∞)
Now f’(x) < 0 = 2x – 4 < 0 ⇒ x < 2
∴ f(x) is strictly decreasing in (- ∞, 2].

(ii) Given f(x) = x² + 2x – 5
∴ f’(x) = 2x + 2
f(x) > 0
⇒ 2x + 2 > 0
⇒ x > – 1
∴ f (x) is strictly increasing in [- 1, ∞)
Now f'(x) < 0
⇒ 2x + 2 < 0
⇒ x < – 1
∴ f(x) is strictly decreasing in (- ∞. – 1].

(iii) f(x) = 2x² – 3x
∴ f'(x) = 4x – 3
Now f'(x) > 0
⇒ 4x – 3 > 0
⇒ x > \(\frac{3}{4}\)
∴ f(x) is strictly incresing in [\(\frac{3}{4}\), ∞)
Now f'(x) < 0
⇒ 4x – 3 < 0
⇒ x < \(\frac{3}{4}\)
∴ f(x) is strictly incresing in (- ∞, \(\frac{3}{4}\)].

(iv) Given f(x) = 6 – 9x – x²
∴ f(x) = – 9 – 2x
Now f'(x) > 0
⇒ – 9 – 2x > 0
⇒ 2x < – 9
⇒ x < – \(\frac{9}{2}\)
∴ f(x) is strictly incresing in (- ∞, – \(\frac{9}{2}\)].
Now f'(x) < 0
⇒ – 9 – 2x < 0 ⇒ x > – \(\frac{9}{2}\)
∴ f(x) is strictly incresing in [- \(\frac{9}{2}\), ∞).

(xii) Given f(x) = \(\frac{x^4}{4}+\frac{2}{3} x^3-\frac{5}{2} x^2\) – 6x + 9
∴ f(x) = x³ + 2x² – 5x – 6
= (x – 2) (x² + 4x + 3)
= (x – 2) (x + 1) (x + 3)

signs of f(x) for different values of x.
For f(x) to be increasing, we must have
f(x) > 0
⇒ (x – 2) (x + 1) (x + 3) > 0
⇒ x ∈ (- 3, – 1) ∪ (2, ∞).
Thus,f(x) be strictly increasing on [- 3, – 1] ∪ (2, ∞).
For f(x) to be decreasing, we must have f(x) < 0
⇒ (x – 2) (x + 1) (x + 3) < 0
⇒ x ∈ (- ∞, – 3) ∪ (- 1, 2).
Thus, f(x) be strictly decreasing on [- ∞, – 3] ∪ (- 1, 2).

Question 22.
(i) Find the interval in which the function f(x) = x² e^-x is increasing. (NCERT)
(ii) Find the intervals in which the function given by f (x) = x² e^x is strictly increasing or strictly decreasing.
Solution:
(i) Given f(x) = x² e^-x ;
Diff. both sides w.r.t. x, we have
f'(x) = – x² e^-x + 2x e^-x
= e^-x (2x – x²)
= e^-x (2 – x) x
Now f'(x) ≥ 0
⇒ e^-x (2 – x) ≥ 0
⇒ (2 – x) x ≥ 0
[∵ e^-x > 0]
⇒ – (x – 2) x ≥ 0
⇒ (x – 2) x ≤ 0
⇒ 0 ≤ x ≤ 2

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5 5

Thus f(x) is increasing in [0, 2].

(ii) Given f(x) = x² e^x
Diff. both sides w.r.t. x, we have
f'(x) = x²e^x + 2x e^x
= e^x x (x + 2)
Now f'(x) > 0 iff e^x x (x + 2) > 0
⇒ x (x + 2) > 0
[∵ e^-x > 0 ∀ x ∈ R]
⇒ x < – 2 or x > 0
⇒ x ∈ (- ∞, – 2) ∪ (0, ∞)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5 6

Thus f(x) is strictly increasing in (- ∞, – 2] ∪ [0, ∞)
Now f’(x) < 0 iff e^x x (x+2) < 0
iff x (x + 2) < 0
[∵ e^{– x} > 0 ∀ x ∈ R]
⇒ – 2 < x < 0
⇒ x ∈ (- 2, 0)
Thus,f(x) is strictly decreasing in [- 2, 0].

Question 23.
Find the intervals in which the function f(x) = 2 log (x – 2) – x² + 4x – 5 is strictly increasing or strictly decreasing.
Solution:
Given f(x) = 2 log (x – 2) – x² + 4x – 5,
D_f = (2, ∞)
Diff. both sides w.r.t. x, we have

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5 7

⇒ f'(x) = – 2 (x -1) (x – 3) (x – 2)
[∵ (x – 2)² > 0 ∀ x ∈ R ∵ x ≠ 2]
Now f'(x) > 0
iff – 2 (x – 1) (x – 3) (x – 2) > 0
iff (x – 1) (x – 3) (x – 2) < 0
⇒ 2 < x < 3 or x < 1 ⇒ x ∈ (2, 3) ∪ (- ∞, 1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5 8

But x > 2
Thus,f(x) is strictly increasing function in [2, 3].
Now f’(x) < 0
iff – 2 (x – 1) (x – 2) (x – 3) < 0 iff (x – 1) (x – 2) (x – 3) > 0
⇒ 1 < x < 2 or x > 3
⇒ x ∈ (1, 2) ∪ (3, ∞)
But x > 2
∴ x ∈ (3, ∞)
Hence f(x) is strictly decreasing in [3, ∞).

Question 24.
Find the intervals in which the function f given by f(x) = tan x – 4x, x ∈ (0, \(\frac{\pi}{2}\)) is
(i) strictly increasing
(ii) strictly decreasing
Solution:
(i) Given f(x) = tan x – 4x
∴ f'(x) = sec² – 4
= (sec x – 2) (sec x + 2)
Now f'(x) > 0
⇒ (sec x – 2) (sec x + 2) > 0
⇒ sec x > 2
[∵ x ∈ (0, \(\frac{\pi}{2}\))
⇒ sec x > 0]
⇒ x > \(\frac{\pi}{3}\) but x ∈ (0, \(\frac{\pi}{2}\))
⇒ \(\frac{\pi}{3}\) < x < \(\frac{\pi}{2}\)
∴ f(x) is strictly increasing in [\(\frac{\pi}{3}\), \(\frac{\pi}{2}\))
Now f'(x) < 0
⇒ (sec x – 2) (sec x + 2) < 0
⇒ sec x < 2 [∵ x ∈ (0, \(\frac{\pi}{2}\)) ∴ sec x > 0]
⇒ x < \(\frac{\pi}{3}\) but x ∈ (0, \(\frac{\pi}{2}\))
⇒ 0 < x < \(\frac{\pi}{3}\)
Thus f(x) is strictly decreasing in (0, \(\frac{\pi}{3}\)].

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4

November 21, 2023

Interactive ISC Mathematics Class 12 Solutions Chapter 7 Applications of Derivatives Ex 7.4 engage students in active learning and exploration.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4

Question 1.
Solve for x :
(i) x (x – 2) (x – 5) (x + 3) > 0
(ii) x⁴ – 5x² + 4 ≥ 0.
Solution:
(i) Given, x (x – 2) (x – 5) (x + 3) = 0 ………….(1)
Here, mark the points – 3, 0, 2, 5 on real line.
By method of intervals, the inequality (1) is satisfied when
x < – 3 or 0 < x < 2 or x > 5

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4 1

∴ The required solution set is (- ∞, – 3) ∪ (0, 2) ∪ (5, ∞).

(ii) Given inequality be x⁴ – 5x² + 4 ≥ 0
(x² – 1) (x² – 4) ≥ 0
⇒ (x- 1) (x – 2) (x + 1) (x + 2) ≥ 0 ………….(1)
Here mark the real numbers – 1, – 2, 1, 2 on real line and by using method of intervals, eqn. (1) is satisfied
When x ≤ – 2 or – 1 ≤ x ≤ 1 or x ≥ 2
∴ required solution set be (- ∞, – 2] ∪ [- 1, 1] ∪ [2, ∞)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4 2

Question 2.
Find all real values of x which satisfy
(i) x³ (x – 1) (x – 2) > 0
(ii) x² (x – 1) (x – 2) ≤ 0
Solution:
(i) Given inequality be x³ (x – 1) (x – 2) > 0
⇒ x (x – 1) (x – 2) > 0 ……….(1)
[∵ x² ≥ 0]
Mark the real numbers 0, 1, 2 on real line.
Using method of intervals, eqn. (1) is satisfied when 0 < x < 1 or x > 2
∴ required solution set be given by (0, 1) ∪ (2, ∞).

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4 3

(ii) Given inequality be
x² (x – 1) (x – 2) ≤ 0
(x – 1) (x – 2) ≤ 0 ……….(1)
[∵ x² ≥ 0]
Mark the numbers 1, 2 on real line and by using method of intervals, eqn. (1) is satisfied when 1 ≤ x ≤ 2
also given eqn. is satisfied when x = 0
∴ required solution set = [1, 2] ∪ {0}

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4 4

Question 3.
(i) \(\frac{1}{x-2}\) ≤ 1
(ii) \(\frac{(x+1)(x-3)}{x+2}\) ≥ 0
Solution:
(i) Given inequality be \(\frac{1}{x-2}\) ≤ 1 ……………(1); x ≠ 2
Since (x – 2)² > 0 ∀ x ∈ R, x ≠ 2
⇒ (x – 2) ≤ (x – 2)²
[multiplying by (x – 2)²]
⇒ (x – 2)² – (x – 2) ≥ 0
⇒ (x – 2) (x – 3) ≥ 0
Mark the numbers 2, 3 on real line.
Using method of intervals, eqn. (2) is satisfied when x ≤ 2 or x ≥ 3

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4 5

But x ≠ 2.
Thus, required solution set be (- ∞, 2) ∪ [3, ∞)

(ii) Given inequality \(\frac{(x+1)(x-3)}{x+2}\) ≥ 0 ; x ≠ – 2
∴ (x + 2)² > 0 ∀ x ∈ R, x ≠ – 2
⇒ (x + 1) (x – 3) (x + 2) ≥ 0 ………..(1)
[Multiplying both sides by (x + 2)² > 0]
Mark the numbers – 1, – 2, 3 on real line and by using method of intervals, eqn. (1) is satisfied
When – 2 ≤ x ≤ – 1 or x ≥ 3
But x ≠ – 2

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4 6

∴ required solution is given by (- 2, -1] ∪ [3, ∞).

Question 4.
Find the values of x for which:
(i) \(\frac{x^2+6 x-11}{x+3}\)
(ii) \(\frac{x^2-3 x+24}{x^2-3 x+3}\) < 4
Solution:
(i) Given inequality be \(\frac{x^2+6 x-11}{x+3}\) < – 1
⇒ \(\frac{x^2+6 x-11}{x+3}\) + 1 < 0
⇒ \(\frac{x^2+7 x-8}{x+3}\) < 0
⇒ \(\frac{(x-1)(x+8)}{x+3}\) < 0 ;
Since (x + 3)² > 0 ∀ x ∈ R, x ≠ – 3
(x – 1) (x + 8) (x + 3) < 0
[multiplying by (x + 3)²]
Mark the numbers – 3, – 8 and 1 on real line.
Using methods of intervals, given inequality satisfied when

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4 7

– 3 < x < 1 or x < – 8
∴ required solution set be (- ∞, – 8) ∪ (- 3, + 1).

(ii) Given \(\frac{x^2-3 x+24}{x^2-3 x+3}\) < 4

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4 8

∴ from (1) ;
– 3 (x² – 3x – 4) < 0
⇒ (x – 3x – 4) > 0
⇒ (x + 1) (x – 4) > 0 ………….(2)
Mark the real numbers – 1 and 4 on real line.
Using method of intervals, eqn. (2) is satisfied when x < – 1 or x > 4
∴ required solution set be (- ∞, – 1) ∪ (4, ∞).

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.3

November 20, 2023

Utilizing Understanding ISC Mathematics Class 12 Solutions Chapter 7 Applications of Derivatives Ex 7.3 as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.3

Question 1.
Using differentials, find the approximate values in the following (1 to 5) problems :
(i) \(\sqrt{26}\)
(ii) \(\sqrt{36.6}\) (NCERT)
(iii) \(\sqrt{49.5}\)
(iv) \(\sqrt{401}\) (NCERT)
Solution:
(i) Let us consider the function
y = √x = f(x)
Let x = 25 and x + ∆x = 26
∴ ∆x = 26 – 25 = 1
at x = 25 ;
y = \(\sqrt{25}\)= 5
Let dx = ∆x = 1
Since y = √x
∴ \(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{x}}\)
at x = 25 ;
\(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{25}}\)
= \(\frac{1}{2 \times 5}=\frac{1}{10}\) = 0.1
∴ dy = ∆y
= \(\frac{d y}{d x}\) dx
= 0.1 × 1 = 0.1
Thus, \(\sqrt{26}\) = y + ∆y
= 5 + 0.1 = 5.1

(ii) Let y = √x ……….(1)
Diff. both sides of eqn. (1) w.r.t. x ; we get
\(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
Take x = 36 ;
x + δx = 36.6
∴ δx = dx = 0.6
When x = 36
⇒ y = \(\sqrt{36}\) = 6
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{2 \sqrt{x}}\) δx
= \(\frac{1}{2 \times \sqrt{36}}\) × 0.6
= \(\frac{0.6}{12}=\frac{6}{120}\)
= \(\frac{1}{20}\) = 0.05
⇒ δy = 0.05 [∵ dy ≅ δy]
∴ from (1) ;
y + δy = \(\sqrt{x+\delta x}\) = \(\sqrt{36.6}\)
⇒ \(\sqrt{36.6}\) = 6 + 0.5 = 6.05.

(iii) Let y = f(x) = √x
Take x = 49,
y = \(\sqrt{49}\) = 7,
let dx = Δx = 0.5
Now y = √x
⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=49}=\frac{1}{2 \sqrt{49}}=\frac{1}{14}\)
∴ dy = \(\frac{d y}{d x}\) . dx
= \(\frac{1}{14}\) . (0.5)
= \(\frac{5}{140}=\frac{1}{28}\)
⇒ Δy = \(\frac{1}{28}\)
[∴ Δy ≅ dy]
∴ \(\sqrt{49.5}\) = y + Δy
= 7 + \(\frac{1}{28}\) = 7.036

(iv) Let y = f(x) = √x
Take x = 400, x + Δx = 401
∴ Δx = 1
When x = 400 than y = \(\sqrt{400}\) = 20
Let dx = Δx = 1
Now y = √x
⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=400}=\frac{1}{2 \sqrt{400}}=\frac{1}{40}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{40}\) × 1
= \(\frac{1}{40}\) = Δy
[∴ dy ≅ Δy]
Hence \(\sqrt{401}\) = y + Δy
= 20 + \(\frac{1}{40}\)
= 20 + 0.25 = 20.025.

Question 2.
(i) \(\sqrt{0.26}\)
(ii) \(\sqrt{0.24}\)
(iii) \(\sqrt{0.0037}\)
(iv) \(\sqrt{0.48}\) (NCERT)
Solution:
(i) Let y = f(x) = √x
Take x = 0.25,
x + Δx = 0.26
⇒ Δx = 0.01
When x = 0.25
then y = \(\sqrt{0.25}\) = 0.5
Let dx = Δx = 0.01
Now y = √x
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=0.25}=\frac{1}{2 \sqrt{0 \cdot 25}}\)
= \(\frac{1}{2 \times 0.5}\) = 1
∴ dy = \(\frac{d y}{d x}\) × dx
= 1 × 0.01 = 0.01
∴ \(\sqrt{0.26}\) = y + Δy
= 0.5 + 0.01 = 0.51.

(ii) Let y = f(x) = √x
Take x = 0.25,
x + Δx = 0.24
⇒ Δx = – 0.01
When x = 0.25
then y = \(\sqrt{0.25}\) = 0.5,
Let dx = Δx = – 0.01
Now y = √x
∴ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=0.25}=\frac{1}{2 \sqrt{0.25}}\)
= \(\frac{1}{2 \times 0.5}\) = 1
∴ dy = \(\frac{d y}{d x}\) dx
= – 1 × 0.01
= – 0.01 = Δy
∴ \(\sqrt{0.24}\) = y + Δy
= 0.5 – 0.01 = 0.49.

(iii) Take x = 0.0036,
x + Δx = 0.0037
∴ Δx = 0.0001
When x = 0.0036
then y = \(\sqrt{0.0036}\) = 0.06
Let dx = Δx = 0.0001
Now, y = √x
⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=0.0036}=\frac{1}{2 \times 0.06}=\frac{1}{0.12}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{0.12}\) × 0.0001
⇒ Δy = \(\frac{1}{10000} \times \frac{100}{12}=\frac{1}{1200}\)
∴ \(\sqrt{0.0037}\) = y + Δy
= 0.06 + \(\frac{1}{1200}\) = 0.0608.

(iv) Let y = f(x) = √x
Take x = 0.49,
x + ∆x = 0.48
⇒ ∆x = – 0.01
When x = 0.49
then y = \(\sqrt{0.49}\) = 0.7
Let dx = ∆x = – 0.01
Now y = √x
⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=0.49}=\frac{1}{2 \sqrt{0 \cdot 49}}\)
= \(\frac{1}{2 \times 0.7}=\frac{1}{1.4}=\frac{5}{7}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{5}{7}\) × (- 0.01) = ∆y
[∵ ∆y ≅ dy]
∴ \(\sqrt{0.49}\) = y + ∆y
= 7 – \(\frac{0.05}{7}\) = 0.693.

Question 3.
(i) (66)^1/3
(ii) (26)^1/3 (NCERT)
(iii) (25)^1/3 (NCERT)
(iv) (26.57)^1/3 (NCERT)
Solution:
(i) Let y = f(x) = x^1/3
Take x = 64,
x + ∆x = 66
⇒ ∆x = 2
When x = 64
then y = (64)^1/3 = 4
Let ∆x = dx = 2
Now y = x^1/3
∴ \(\frac{d y}{d x}\) = \(\frac{1}{3}\) x^-2/3
⇒ \(\left.\frac{d y}{d x}\right]_{x=64}=\frac{1}{3}(64)^{-2 / 3}=\frac{1}{48}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{48}\) × 2
= \(\frac{1}{24}\) = ∆y
(∵ ∆y ≅ dy)
∴ (66)^1/3 = y + ∆y
= 4 + \(\frac{1}{24}\)
= \(\frac{97}{24}\) = 4.0416

(ii) Let y = f(x) = x^1/3
Take x = 27,
x + ∆x = 26
⇒ ∆x = – 1
When x = 27
then y = (27)^1/3 = 3,
Let ∆x = dx = – 1
Now y = x^1/3
∴ \(\frac{d y}{d x}\) = \(\frac{1}{3}\) x^-2/3
∴ \(\left.\frac{d y}{d x}\right]_{x=27}=\frac{1}{3}(27)^{-2 / 3}=\frac{1}{27}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{27}\) × (- 1)
= \(-\frac{1}{27}\) = ∆y
[∵ ∆y ≅ dy]
∴ (26)^1/3 = y + ∆y
= 3 – \(\frac{1}{27}\) = \(\frac{80}{27}\)

(iii) Let y = f(x) = x^1/3
Take x = 27,
x + ∆x = 25
⇒ ∆x = – 2
when x = 27
then (27)^1/3 = 3,
let dx = ∆x = – 2
Now y = x^1/3
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{3}\) x^-2/3
⇒ \(\left.\frac{d y}{d x}\right]_{x=27}=\frac{1}{3}(27)^{-2 / 3}=\frac{1}{27}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{27}\) × (- 2)
= \(-\frac{2}{27}\) = ∆y
[∵ ∆y ≅ dy]
∴ (25)^1/3 = y + ∆y
= = 3 – \(\frac{1}{27}\)
= \(\frac{79}{27}\) = 2.926.

(iv) Let y = f(x) = x^1/3
Take x = 27,
x + ∆x = 26.57
⇒ ∆x = 26.57 – 27 = – 0.43
when x = 27
then (27)^1/3 = 3,
Let dx = ∆x = – 0.43
Now y = x^1/3
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{3}\) x^-2/3
∴ \(\left.\frac{d y}{d x}\right]_{x=27}=\frac{1}{3}(27)^{-2 / 3}=\frac{1}{27}\)
∴ dy = \(\frac{d y}{d x}\) × dx
= \(\frac{1}{27}\) × (- 0.43)
= \(-\frac{0.43}{27}\) = ∆y
[∵ ∆y ≅ dy]
∴ \(\sqrt[3]{26 \cdot 57}\) = y + ∆y
= = 3 – \(\frac{0.43}{27}\)
= 2.984.

Question 4.
(i) (82)^1/4 (NCERT)
(ii) (15)^1/4 (NCERT)
(iii) (32.15)^1/5 (NCERT)
(iv) (31.9)^1/5
Solution:
(i) Let y = f(x) = x^1/4
Take x = 81,
x + ∆x = 82
⇒ ∆x = 1
When x = 81,
then y = (81)^1/4 = 3,
Let dx = ∆x = 1
Now y = x^1/4
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{1}\) x^-3/4
∴ \(\left.\frac{d y}{d x}\right]_{x=81}=\frac{1}{4}(81)^{-3 / 4}=\frac{1}{108}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{108}\) × 1
= \(\frac{1}{108}\)
∴ (82)^1/4 = y + ∆y
= 3 + \(\frac{1}{108}\)
= \(\frac{325}{108}\).

(ii) Let y = f(x) = x^1/4
Take x = 16,
x + ∆x = 15
⇒ ∆x = – 1
When x = 16,
then y = (16)^1/4 = 3,
Let dx = ∆x = – 1
Now y = x^1/4
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{1}\) x^-3/4
∴ \(\left.\frac{d y}{d x}\right]_{x=16}=\frac{1}{4}(16)^{-3 / 4}=\frac{1}{32}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{32}\) × (- 1)
= \(-\frac{1}{32}\) = ∆y
[∵ dy ≅ ∆y]
∴ (15)^1/4 = y + ∆y
= 2 – \(\frac{1}{32}\)
= \(\frac{63}{32}\)
= 1.96875

(iii) Let y = f(x) = x^1/5
Take x = 32,
x + ∆x = 32.15
⇒ ∆x = 0.15
When x = 32,
then y = (32)^1/5 = 2,
Let dx = ∆x = 1
Now y = x^1/5
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{5}\) x^-4/5
∴ \(\left.\frac{d y}{d x}\right]_{x=32}=\frac{1}{5}(32)^{-4 / 5}=\frac{1}{80}\)
∴ dy = \(\frac{d y}{d x}\) ∆x
= \(\frac{1}{80} \times \frac{15}{100}=\frac{3}{1600}\)
∴ (32.15)^1/5 = y + δy
= 2 + \(\frac{3}{1600}\)
= 2.00187

(iv) Let y = x^1/5 ………..(1)
Diff. both sides of eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{1}{5} x^{\frac{1}{5}-1}=\frac{1}{5 x^{4 / 5}}\)
Take x = 32 ;
x + δx = 31.9
∴ δx = dx = 31.9 – 32 = – 0.1
When x = 32
∴ y = \((32)^{\frac{1}{5}}\)
= \(\left(2^5\right)^{\frac{1}{5}}\) = 2
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{5 x^{4 / 5}}\) δx
= \(\frac{1}{5 \times(32)^{4 / 5}}\) × (- 0.1)
= – \(\frac{0.1}{80}\)
⇒ δy = dy
= – \(\frac{0.1}{80}=-\frac{1}{800}\)
∴ from (1) ;
y + δy = (x + δx)^1/5
⇒ (31.9)^1/5 = y + δy
= 2 – \(\frac{1}{800}\)
= \(\frac{1599}{800}\)
= 1.99875.

Question 5.
(i) (0.999)^1/10 (NCERT)
(ii) (3.968)^3/2 (NCERT)
Solution:
(i) Let y = f(x) = x^1/10
Take x = 1,
x + ∆x = 0.999
⇒ ∆x = – 0.001
When x = 1
then y = (1)^1/10 = 1,
Let dx = ∆x = – 0.001
Now y = x^1/10
∴ \(\frac{d y}{d x}\) = \(\frac{1}{10}\) x^-9/10
⇒ \(\left.\frac{d y}{d x}\right]_{x=1}=\frac{1}{10}(1)^{-9 / 10}=\frac{1}{10}\)
Therefore dy = \(\frac{d y}{d x}\) ∆x
= \(\frac{1}{10}\) × (- 0.001)
= – 0.0001
= ∆y
[∵ ∆y ≅ dy]
∴ (0.999)^1/10 = y + ∆y
= 1 – 0.0001 = 0.9999.

(ii) Let y = f(x) = x^3/2
When x = 4
then y = (4)^3/2 = 8,
Let dx = ∆x = 0.032
Now y = x^3/2
⇒ \(\frac{d y}{d x}\) = \(\frac{3}{2}\) x^1/2
∴ \(\left.\frac{d y}{d x}\right]_{x=4}=\frac{3}{2}(4)^{1 / 2}=3\)
Therefore dy = \(\frac{d y}{d x}\) ∆x
= – 3 × 0.0032
= – 0.096
= ∆y
[∵ dy ≅ ∆y]
∴ (3.968)^3/2 = y + ∆y
= 8 – 0.096 = 7.904.

Question 6.
Find the approximate value of (1999)⁵. (NCERT Exemplar)
Solution:
Let us consider the function y = x⁵ = f(x)
Let x = 2
and x + ∆x = 1.999
∴ ∆x = 1.999 – 2 = – 0.001
Let dx = ∆x = – 0.001
at x = 2 ;
y = x⁵
= 2⁵ = 32
Since y = x⁵
⇒ \(\frac{d y}{d x}\) = 5x⁴
at x = 2 ;
\(\frac{d y}{d x}\) = 5 × 2⁴ = 80
∴ ∆y = dy
= \(\frac{d y}{d x}\) dx
= 80 × (- 0.001)
= – 0.08
Thus, (1.999)⁵ = y + ∆y
= 32 – 0.08 = 31.92.

Question 7.
(i) If f (x) = x⁴ – 10, then find the approximate value of f(2, 1). (NCERT)
(ii) Find the approximate value of f (3.02), where f(x) = 3x² + 5x + 3.
Solution:
(i) Given f(x) = x⁴ – 10 = y
Let x = 2 and x + δx = 2.1
∴ δx = 2 . 1 – 2 . 0
= 0 . 1 = dx
When x = 2
∴ y = f(2)
= 2⁴ – 10
= 16 – 10 = 6
∴ f'(x) = \(\frac{d y}{d x}\) = 4x³
Thus dy = δy
= \(\frac{d y}{d x}\) δx
= 4x³ . δx
= 4 × 2³ × 0.1 = 3.2
∴ f (x + δx) = y + δy
= (x + δx)⁴ – 10
= (2 . 1)⁴ – 10
= f(2 . 1)
⇒ f(2.1) = 6 + 3.2 = 9.2

(ii) Given f(x) = 3x² + 5x + 3
Take x = 3,
x + ∆x = 3.02
⇒ ∆x = 0.02
When x = 3
then y = 3.3² + 15 . 3 + 3
⇒ y = 27 + 15 + 3 = 45,
Let ∆x = dx = 0.02
Now y = 3x² + 5x + 3
∴ \(\frac{d y}{d x}\) = 6x + 5
⇒ \(\left.\frac{d y}{d x}\right]_{x=3}\) = 6 × 3 + 5 = 23
∴ dy = \(\frac{d y}{d x}\) ∆x
= 23 × 0.02
= 0.46 = ∆y
[∵ ∆y ≅ dy]
∴ f(3.02) = y + ∆y
= 45 + 0.46 = ∆y.

Question 7 (old).
(ii) Find the approximate value of f(5.001), where f(x) = x³ – 7x² + 15. (NCERT)
Solution:
Given f(x) = x³ – 7x² + 15
⇒ f'(x) = 3x² – 14x
Take x = 5 and ∆x = 0.001
then f(5.001) = f(x + ∆x)
= (x + ∆x)³ – 7 (x + ∆x)² + 15
Now ∆y = f (x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆y
⇒ f(x + ∆x) = f (x) + f'(x) ∆x
[∵ dy = ∆y = \(\frac{d y}{d x}\) ∆x = f'(x) ∆x]
⇒ f(5.001) = f(5) + f'(5) × 0.001
= [5³ – 7.5² + 15] + [3.5² – 14 × 5] × 0.01
= – 35 + 0.005 = – 34.995

Question 8.
If the radius of a circle increases from 5 cm to 5-1 cm, find the increase in area.
Solution:
Let r be the radius of circle and A be the area of circle
Then A = πr² ………….(1)
Diff. w.r.t. r, we have
\(\frac{d A}{d r}\) = 2πr dr
Since radius r increases from 5 cm to 5.1 cm.
∴ r = 5 and r + δr = 5.1
⇒ δr = dr = 0.1
When r = 5 ; A = π (5)² = 25π
∴ dA = \(\frac{d A}{d r}\) dr
= (2πr) δr
= 2π × 5 × 0.1 = π
⇒ δA = π cm²
[∵ dA ≅ δA]
Thus approximate increase in area = δA = π cm²
and A + δA = 25 π + π = 26 π cm²

Question 9.
If the radius of a spherical balloon shrinks from 10 cm to 9-8 cm, find the approximate decrease in its
(i) volume
(ii) surface area.
Solution:
Let x be the radius of sphere
Then V = volume of sphere = \(\frac{4}{3}\) πx³
Here x = 10 and x + ∆x = 9-8
∴ ∆x = – 0.2 = dx
\(\frac{d V}{d x}\) = \(\frac{4 \pi}{3}\) × 3x²
= 4πx²
∴ dV = \(\frac{d V}{d x}\) dx
= 4πx² × (- 0.2)
= – 0.8 πx²
at x = 10 ;
dV = – 0.8 π × 10² = – 80 π
Thus, dV = ∆V = – 80 π
Thus, approximate decrease in volume = ∆V = 80 π cm³.

(ii) Let S = surface area of spherical balloon = 4πx²
∴ \(\frac{d S}{d x}\) = 8πx
∴ dS = \(\frac{d S}{d x}\) dx
= 8πx dx
= 8π × 10 × (- 0.2)
⇒ dS = – 16π = ∆S
Thus approximate decrease in surface area = ∆S = 16π cm²

Question 10.
(i) Find the approximate change in the volume of a cube of side x metres caused by increasing the side by 1%.
(ii) Find the approximate change in the surface area of a cube of side x metres by decreasing the side by 1%.
Solution:
(i) Given the length of each side of cube be x m.
Then volume of cube = V = x³
∴ \(\frac{d V}{d x}\) = 3x²
Let ∆x be the change in x and the corresponding change in V be ∆V.
∴ ∆V = \(\frac{d V}{d x}\) ∆x
[∵ ∆V = dV ; ∆x = dx]
= 3x² ∆x
also it is given that \(\frac{\Delta x}{x}\) × 100 = 1
⇒ ∆x = \(\frac{x}{100}\) = 0.01 x
∴ ∆V = 3x² × 0.01x
= 0.03 x³ m³

(ii) Let x be the side of cube
then S = surface area of cube = 6x²
Diff. both sides w.r.t. x, we have,
∴ \(\frac{d S}{d x}\) = 12x dx
∴ dS = change in surface area
= \(\frac{d S}{d x}\) . ∆x
∴ dS = 12x . ∆x [∵ dx = ∆x]
= 12x × (- 0.01 x)
[∆x = 1 % decreasing of x = – 0.01 x]
= – 0.12 x² m²
Hence the approximate change in surface area is 0.12 x² m².

Question 11.
(i) If the radius of a sphere is measured as 9 m with an error of 0-03 m, then find the approximate error in calculating its volume. (NCERT)
(ii) Find the percentage error in computing the surface area of a cubical box if an error of 1% is made in measuring the lengths of the edges of the box.
Solution:
(i) Let r be the radius of sphere
then error in r = ∆r = 0 03 m ;
r = 9 cm
Let V = volume of sphere = \(\frac{4 \pi}{3}\) r³
∴ Log V = log \(\frac{4 \pi}{3}\) + 3 log r
∴ \(\frac{1}{V}\) ∆V = 0 + \(\frac{3}{r}\) ∆r
∴ ∆V = \(\frac{3}{r}\) ∆r × \(\frac{4}{3}\) πr³
= 4πr² ∆r
= 4π × 9² × 0.03
= 9.72 πm²

(ii) Let x be the length of the edge of the cube then S = surface area of cubical base = 6x²
Taking logarithm on both sides, we get
log S = log 6 + 2 log x ;
on differentiating
⇒ \(\frac{d S}{S}\) × 100 = 2 (\(\frac{dx}{x}\) × 100) …………….(1)
Given, \(\frac{dx}{x}\) × 100 = 1
∴ from (1) ; we have
\(\frac{d S}{S}\) × 100 = 2 × 1 = 2
∴ error in calculating surface area is 2 %.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.2

November 20, 2023

The availability of step-by-step ML Aggarwal Maths for Class 12 Solutions Chapter 7 Applications of Derivatives Ex 7.2 can make challenging problems more manageable.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.2

Question 1.
What is the slope of the tangent to the following curves :
(i) y = x³ – x at x = 2? (NCERT)
(ii) y = 3x⁴ – 4x at x = 4? (NCERT)
(iii) x² + 3y + y² = 5 at the point (1, 1)?
(iv) y = x³ – 3x + 2 at the point whose x coordinate is 3? (NCERT)
(y) y = 3x² + 4x at the point whose x coordinate is – 2 ?
(vi) y = 2 sin² (3x) at \(\frac{1}{3}\)?
Solution:
(i) Given equation of curve be y = x³ – x
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = 3x² – 1
∴ (\(\frac{d y}{d x}\))_{x = 2} = 3 × 2² – 1
= 12 – 1 = 11
Thus, slope of tangent at x = 2.
= (\(\frac{d y}{d x}\))_{x = 2} = 11

(ii) Given curve is y = 3x⁴ – 4x
∴ \(\frac{d y}{d x}\) = 12x³ – 4
∴ (\(\frac{d y}{d x}\))_{x = 4} = 12 (4)³ – 4 = 764.

(iii) Given eqn. of curve be x³ + 3y + y² – 5 = 0
Diff. both sides w.r.t. x ; we get
2x + 3 \(\frac{d y}{d x}\) + 2y \(\frac{d y}{d x}\) = 0
⇒ (3 + 2y) \(\frac{d y}{d x}\) = – 2x
⇒ \(\frac{d y}{d x}\) = \(\frac{-2 x}{2 y+3}\)
\(\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{-2}{2+3}=\frac{-2}{5}\)
∴ slope of tangent at (1, 1) = (\(\frac{d y}{d x}\))_{(1, 1)}
= \(-\frac{2}{5}\)

(iv) Given curve be, y = x³ – 3x + 2
∴ \(\frac{d y}{d x}\) = 3x² – 3
∴ (\(\frac{d y}{d x}\))_{x = 3} = 3 (3)² – 3 = 24

(v) Given, eqn. of curve be y = 3x² + 4x
Diff. both sides w.r.t. x, we have
∴ Slope of tangent to given curve at x = – 2 = (\(\frac{d y}{d x}\))_{x = – 2}
= 6 × (- 2) + 4
= – 12 + 4 = – 8.

(vi) Given y = sin² (3x)
∴ \(\frac{d y}{d x}\) = 4 sin 3x . cos 3x . 3
= 12 sin 3x cos 3x
∴ slope of tangent to given curve at (x = \(\frac{\pi}{6}\)) = (\(\frac{d y}{d x}\))_{x = \(\frac{\pi}{6}\)}
= 12 × sin \(\frac{\pi}{2}\) cos \(\frac{\pi}{2}\) = 0.

Question 2.
What is the slope of the normal to the following curves :
(i) y = x³ – 5x² – x + 1 at the point(1, – 4)?
(ii) y = 2x² + 3 sin x at x = 0? (NCERT)
Solution:
(i) Given eqn. of curve be,
y = x³ – 5x² – x + 1 …………….(1)
∴ \(\frac{d y}{d x}\) = 4x² – 10x – 1
Thus, the slope of the normal to curve (1) at point (1, – 4) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(1,-4)}}\)
= \(=\frac{-1}{3-10-1}\)
= \(\frac{-1}{-8}=\frac{1}{8}\)

(ii) Given eqn. of curve be,
y = 2x² + 3 sin x
∴ \(\frac{d y}{d x}\) = 4x + 3 cos x
Thus (\(\frac{d y}{d x}\))_{x = 0} = 4 × 0 + 3 × 1 = 3
∴ slope of normal to curve (1) at x = 0 = – \(\frac{1}{\left(\frac{d y}{d x}\right)_{x=0}}\)
= – \(\frac{1}{3}\).

Question 3.
Find the point on the curve y = 3x² – 2x + 1 at which the slope of the tangent is 4.
Solution:
Let P (x₁, y₁) be any point on given curve.
Eqn. of given curve be, y = 3x² – 2x + 1 ………..(1)
∴ \(\frac{d y}{d x}\) = 6x – 2
∴ slope of tangent to the curve (1) at P(x₁, y₁)
= 6x₁ – 2
also, given slope of tangent to given curve = 4
∴ 6x₁ – 2 = 4
⇒ x₁ = 1
Now P (x₁, y₁) also lies on given curve (1)
y₁ = 3x₁² – 2x₁ + 1
y₁ = 3(1)² – 2x₁ + 1
= 4 – 2 = 2
Thus, the required point on curve be (1, 2).

Question 4.
Determine the point on the curve y = 3x² – 1 at which the slope of the tangent is 3.
Solution:
Let P (x₁, y₁) be any point on given curve
and eqn. of given curve be. y = 3x² – 1
∴ \(\frac{d y}{d x}\) = 6x
Thus, slope of tangent to given curve at (x₁, y₁) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 6x₁
also slope of tangent to given curve = 3
∴ 6x₁ = 3
⇒ x₁ = \(\frac{1}{2}\)
Since, the point (x₁ , y₁) be lies on given curve
∴ y₁ = 3x₁² – 1
⇒ y₁ = 3 × \(\frac{1}{4}\) – 1
= \(\frac{1}{4}\)
Thus, the required point on the given curve be \(\left(\frac{1}{2},-\frac{1}{4}\right)\).

Question 5.
At what point on the curve y = x² does the tangent make an angle of 45° with the x-axis?
Solution:
Let the eqn. of given curve be y = x² ………..(1)
Let the required point on the given curve be (x₁, y₁)
∴ y₁ = x₁² …………()
∴ \(\frac{d y}{d x}\) = 2x
⇒ (\(\frac{d y}{d x}\))_{(x₁, y₁)} = 2x₁
It is given that,
slope of tangent at (x₁, y₁) = tan \(\frac{\pi}{4}\)
∴ (\(\frac{d y}{d x}\))_{(x₁, y₁)} = 1
⇒ 2x₁ = 1
⇒ x₁ = \(\frac{1}{2}\)
∴ from (2) ;
y₁ = \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
Hence the required point be \(\left(\frac{1}{2}, \frac{1}{4}\right)\).

Question 6.
Find the point on the curvey = x² – 2x + 3 at which the tangent is parallel to x-axis.
Solution:
Eqn. of given curve y = x² – 2x + 3 ……….(1)
Let P (x, y) be any point on curve (1)
∴ \(\frac{d y}{d x}\) = 2x – 2
Now slope of tangent to given curve (1) is || to x-axis
\(\frac{d y}{d x}\) = 0
⇒ 2x – 2 = 0
⇒ x = 1
∴ from (1) ;
y = 1² – 2 × 1 + 3 = 2
Thus required point on given curve be (1, 2).

Question 6 (old).
Find the slope of the tangent to the curve y = \(\frac{x-1}{x-2}\) at x = 10. (NCERT)
Solution:
Given eqn. of curve be, y = \(\frac{x-1}{x-2}\)
∴ \(\frac{d y}{d x}=\frac{x-2-(x-1)}{(x-2)^2}\)
= \(\frac{-1}{(x-2)^2}\) ; x ≠ 2.
∴ \(\left.\frac{d y}{d x}\right]_{x=10}=\frac{-1}{64}\)
Hence, slope of tangent to given curve at x = 10.
= \(\left(\frac{d y}{d x}\right)_{x=10}=\frac{-1}{64}\).

Question 7.
Find the slope of the normal to the curve x = a cos³ θ, y = a sin³ θ at θ = \(\frac{\pi}{4}\). (NCERT)
Solution:
Given eqn. of curve be x = a cos³ θ ………..(1)
and y = a sin³ θ ……..(2)
Difi. eqn. (1) and (2) w.r.t. θ ; we have
\(\frac{d x}{d \theta}\) = 3a cos² θ (- sin θ)
\(\frac{d y}{d \theta}\) = 3a sin² θ cos θ
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\)
= \(\frac{3 a \sin ^2 \theta \cos \theta}{-3 a \cos ^2 \theta \sin \theta}\)
= – tan θ
Thus slope of normal to given curve at θ = \(\frac{\pi}{4}\)
= \(\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{4}}\)
= \( \frac{-1}{-\tan \frac{\pi}{4}}\)
= 1.

Question 8.
Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos² θ at θ = \(\frac{\pi}{2}\).
Solution:
Given eqn. of curve be
x = 1 – a sin θ …………(1)
and y = b cos² θ …………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. θ ; we have
\(\frac{d x}{d \theta}\) = – a cos θ ………..(3)
\(\frac{d y}{d \theta}\) = 2b cos θ (- sin θ)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\)
= \(\frac{-2 b \cos \theta \sin \theta}{-a \cos \theta}\)
= \(\frac{2 b}{a}\) sin θ
Thus slope of normal to given curve at θ = \(\frac{\pi}{2}\) is given by = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{2}}}\)
= \(\frac{-1}{\frac{2 b}{a} \sin \frac{\pi}{2}}\)
= \(-\frac{a}{2 b}\)

Question 9.
Write the equation of the normal to the curve y² = 8x at the origin.
Solution:
given eqn. of curve be, y² = 8x
∴ 2y \(\frac{d y}{d x}\) = 8
⇒ \(\frac{d y}{d x}\) = \(\frac{4}{y}\)
∴ slope of normal to curve at (0, 0) = – \(\frac{1}{\left(\frac{d y}{d x}\right)_{(0,0)}}\)
= \(-\frac{1}{\infty}\) = 0
Thus, eqn. of normal to given curve at (0, 0) be given by
y – 0 = – \(\frac{1}{\left(\frac{d y}{d x}\right)}\) (x – 0)
⇒ y – 0 = 0
⇒ y = 0.

Question 10.
Find the slopes of the tangents at the three points where the curve y = x (x² – 3x – 4) cuts the x-axis.
Solution:
Given eqn. of curve be
y = x (x² – 3x – 4) …………….(1)
eqn. (1) meets x-axis i.e. y = 0
Thus, 0 = x (x² – 3x – 4)
⇒ 0 = x (x + 1) (x – 4)
⇒ x = 0, – 1, 4
Hence, the curve meets x-axis at (0, 0) ; (- 1, 0) and (4, 0).
Diff. eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = 3x² – 6x – 4
∴ Slope of tangent to given curve at (0, 0) = \(\left(\frac{d y}{d x}\right)_{(0,0)}\)
= 0 – 0 – 4 = – 4
slope of tangent to given curve at (- 1, 0) = \(\left(\frac{d y}{d x}\right)_{(-1,0)}\)
= 3 (- 1)² – 6 (- 1) – 4 = 5
and slope of tangent to given curve at (4, 0) = \(\left(\frac{d y}{d x}\right)_{(4,0)}\)
= 3 (4)² – 6 × 4 – 4
= 48 – 24 – 4 = 20.

Question 11.
Show that the tangents to the curve y = 7x³ + 11 at the points where x = 2 and x = – 2 are yy parallel. (NCERT)
Solution:
Given curve is y = 7x³ + 11
∴ \(\frac{d y}{d x}\) = 21 x²
∴ slope of tangent at x = 2 = \(\left.\frac{d y}{d x}\right]_{x=2}\) = 84
slope of tangent at x = 2 = \(\left.\frac{d y}{d x}\right]_{x=-2}\) = 84
Since both slopes are equal.
∴ tangents to the curve at x = 2 and – 2 are parallel.

Question 12.
If the tangent to tile curve y = x³ + ax + b at (1, – 6) is parallel to the line 2x – 2y + 7 = 0, find a and b.
Solution:
Given eqn. of given curve be y = x³ + ax + b ………(1)
Diff. both sides of eqn. (1) w.r.t. x;
\(\frac{d y}{d x}\) = 3x² + a
∴ \(\left(\frac{d y}{d x}\right)_{(1,-6)}\) = 3 + a
it is given that slope of tangent to curve (1) at (1, – 6) slope of line x – y + 5 = 0
∴ \(\left(\frac{d y}{d x}\right)_{(1,-6)}=\frac{-1}{-1}\)
⇒ 3 + a = 1
⇒ a = – 2
Also the point (1, – 6) lies on curve (1),
∴ – 6 = 1³ + a + b
⇒ a + b = – 7
⇒ – 2 + b = – 7
⇒ b = – 5.

Question 13.
Determine the point of the curve y = 3x² – 5 at which the tangent is perpendicular to a line whose slope is – \(\frac{1}{3}\).
Solution:
Let P (x₁, y₁) be any point on given curve
y = 3x² – 5 ………..(1)
∴ y₁ = 3x₁² – 5 ………..(2)
Diff. eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = 6x
∴ slope of tangent to given curve at (x₁, y₁) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= m₁ = 6x₁
Also tangent ⊥to line whose slope is = – \(\frac{1}{3}\)
∴ m₂ = – \(\frac{1}{3}\)
Thus, m₁m₂ = – 1
⇒ 6x₁ (- \(\frac{1}{3}\)) = – 1
⇒ – 2x₁ = – 1
⇒ x₁ = \(\frac{1}{2}\)
∴ from (2) ; we have
y₁ = 3 (\(\frac{1}{2}\))² – 5
= \(\frac{3}{4}\) – 5
= – \(\frac{17}{4}\)
Thus, the required point on given curve be \(\left(\frac{1}{2},-\frac{17}{4}\right)\).

Question 14.
Determine the points on the curve y = x³ – 3x² – 9x + 7 at which the tangents are Parallel to x-axis. (NCERT)
Solution:
Given eqn. of curve be
y = x³ – 3x² – 9x + 7 ………….(1)
Let P (x₁, y₁) be any point on given curve (1).
∴ y = x₁³ – 3x₁² – 9x₁ + 7 ………..(2)
Diff. (1) both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = 3x² – 6x – 9
Thus, the slope of the tangent to given curve at (x₁, y₁) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 3x₁² – 6x₁ – 9
Since the tangent are parallel to x-axis.
∴ slope of tangent is 0.
Thus \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\) = 0.
⇒ 3 (x₁² – 2x₁ – 3) = 0
⇒ (x₁ + 1) (x₁ – 3) = 0
⇒ x₁ = – 1, 3
When x₁ = – 1
∴ from (2) ; we have
y₁ = – 1 – 3 + 9 + 7 = 12
when x₁ = 3
∴ from (2) ; we have
y₁ = 27 – 27 – 27 + 7 = – 20
Thus, the required points on given curve be (- 1, 12) and (3, – 20).

Question 15.
Find the points on the curve x² + y² – 2x – 3 = 0 at which the tangents are parallel to x-axis.
Solution:
Let the eqn. of given curve be
x² + y² – 2x – 3 = 0 ……….(1)
Let the required point on given curve (1) be (x₁, y₁).
∴ x₁² + y₁² – 2x₁ – 3 = 0 ………..(2)
Diff. eqn. (1) w.r.t. x ; we have
2x + 2y \(\frac{d y}{d x}\) – 2 = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{1-x}{y}\)
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{1-x_1}{y_1}\)
Since it is given that tangent is parallel to x-axis.
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\) = 0
⇒ \(\frac{1-x_1}{y_1}\) = 0
⇒ x₁ = 1
∴ from (2) ;
1 + y₁² – 2 – 3 = 0
⇒ y₁ = ± 2
Hence the required points are (1, ± 2).

Question 16.
Find the points on the curve \(\frac{x^2}{4}+\frac{y^2}{25}\) = 1 at which the tangents are
(a) parallel to x-axis
(b) parallel to y-axis. (NCERT)
Solution:
(a) Given curve is,
\(\frac{x^2}{4}+\frac{y^2}{25}\) = 1 …………..(1)
Differentiate both sides w.r.t. x, we get
∴ \(\frac{2 x}{4}+\frac{2 y}{25} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{25 x}{4 y}\)
Since tangent is parallel to x-axis.
∴ \(\frac{d y}{d x}\) = 0
⇒ \(\frac{- 25 x}{4 y}\) = 0
⇒ x = 0
∴ From (1) ;
y = ± 5
Here required points are (0, ± 5).

(b) Since tangent is || to y-axis
∴ \(\frac{d y}{d x}\) = ∞
⇒ \(\frac{d y}{d x}\) = 0
∴ y = 0
∴ From (1) ; x = ± 2.
Hence required points are (± 2, 0).

Question 17.
Find the points on the curve x² + y² = 13, the tangent at each one of which is parallel to the line 2x + 3y = 7.
Solution:
Eqn. of given curve be x² + y² = 13 ………….(1)
Let the required point be on given curve (1) be (x₁, y₁).
∴ x₁² + y₁² = 13 ………..(2)
Since the tangent is parallel to given line 2x + 3y = 7
∴ slope of tangent at (x₁, y₁) = slope of line 2x + 3y = 7
⇒ \(\left(\frac{d y}{d x}\right)_{x_1 y_1}=-\frac{2}{3}\) …………(3)
Diff. eqn. (1) w.r.t. x; we get
2x + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}-\frac{x}{y}\)
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1 y_1\right)}=-\frac{x_1}{y_1}\)
∴ From (3) ; we have
⇒ \(-\frac{x_1}{y_1}=-\frac{2}{3}\)
⇒ 3x₁ = 2y₁
∴ from (2) ;
x₁² + \(\frac{9}{4}\) x₁² = 13
⇒ \(\frac{13}{4}\) x₁² = 13
⇒ x₁ = ± 2
When x₁ = 2
∴ from (4) ;
y₁ = 3
When x₁ = – 2
∴ from (4) ;
y₁ = – 3
Hence the required points are (2, 3) and (- 2, – 3).

Question 18.
(Find the points on the curve y = x³ – 3x² + 2x at which the tangent lines are parallel to the line y – 2x + 3 = 0.
Solution:
Given eqn. of curve be y = x³ – 3x² + 2x …………(1)
Let P (x₁, y₁) be any point on given curve (1).
∴ y₁ = x₁³ – 3x₁² + 2x₁ ………….(2)
Diff. both sides of eqn. (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = 3x² – 6x + 2
∴ slope of tangent to given curve (1) at (x₁, y₁) = 3x₁² – 6x₁ + 2
slope of given Line y – 2x + 3 = 0
= – \(\frac{(-2)}{1}\) = 2
Since the tangent to given curve is parallel to given line
∴ 3x₁² – 6x₁ + 2 = 2
⇒ 3x₁ (x₁ – 2) = 0
⇒ x₁ = 0, 2
When x₁ = 0
∴ from (2) ; we have
y₁ = 0 – 0 + 0 = 0
When x₁ = 2
∴ from (2); we have
y₁ = 8 – 12 + 4 = 0
Thus the required points on given curve be (0, 0) and (2, 0).

Question 19.
Find the equations of the tangent and the normal to each of the following curves at the given point :
(i) y = x³ at (1,1)
(ii) y = x⁴ – 6x³ + 13x² – 10x + 5 at (1, 3) (NCERT)
(iii) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5) (NCERT)
(iv) y = x² at (0, 0) (NCERT)
(v) y = sin² x at x = \(\frac{\pi}{2}\)
(vi) \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 at (x₀, y₀). (NCERT)
Solution:
(i) Given eqn. of curve be y = x³ ………..(1)
∴ \(\frac{d y}{d x}\) = 3x²
∴ slope of tangent to curve (1) at given point (1, 1) = \(\left(\frac{d y}{d x}\right)_{(1,1)}\)
= 3 × 1² = 3
Thus, eqn. of tangent to given curve at (1, 1) is given by
y – 1 = 3 (x – 1)
⇒ 3x – y – 2 = 0
Slope of normal to given curve at (1, 1) = \(-\frac{1}{\left(\frac{d y}{d x}\right)_{(1,1)}}=-\frac{1}{3}\)
∴ required eqn. of normal to given curve (1) at(1, 1) is given by
y – 1 = – \(\frac{1}{3}\) (x – 1)
⇒ x + 3y —4 = 0.

(ii) Given,
y = x⁴ – 6x³ + 13x² – 10x + 5
∴ \(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10
when x = 1
∴ from (1) we have
y = 1 – 6 + 13 – 10 + 5 = 3
∴ slope of tangent at (1, 3) = \(\left(\frac{d y}{d x}\right)_{(1,3)}\)
= 4— 18 + 26 – 10 = 2
and slope of Normal at (1, 3) = – \(\frac{1}{2}\)
∴ eqn. of tangent at (1, 3) is given by
y – 3 = 2 (x – 1)
⇒ 2x – y + 1 = 0
and eqn. of Normal at (1, 3) is given by y – 3 = – \(\frac{1}{2}\) (x – 1)
⇒ x + 2y – 7 = 0.

(iii) Given curve be y = x⁴ – 6x³ + 13x² – 10x + 5
∴ \(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10
Thus slope of tangent at (0, 5) = \(\left(\frac{d y}{d x}\right)_{(0,5)}\) = – 10
∴ slope of normal at (0, 5) = \(-\frac{1}{\left(\frac{d y}{d x}\right)_{(0,5)}}\)
= \(-\frac{1}{-10}=\frac{1}{10}\)
Thus eqn. of tangent at (0, 5) is given by y – 5 = – 10 (x – 0)
⇒ y + 10x – 5 = 0
and eqn. of Normal to curve at (0, 5) is given by y – 5 = \(\frac{1}{10}\) (x – 0)
⇒ 10y – 50 = x
⇒ x – 10y + 50 = 0.

(iv) Given curve be, y = x²
∴ \(\frac{d y}{d x}\) = 2x
∴ slope of tangent at (0, 0) = \(\left(\frac{d y}{d x}\right)_{(0,0)}\)
= 2 × 0 = 0
and slope of normal at (0, 0) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(0,0)}}\) = ∞
∴ eqn. of tangent at (0, 0) is y – 0 = \(\frac{-1}{\left(\frac{d y}{d x}\right)}\) (x – 0) = 0
and eqn. of normal at (0, 0) is y – 0 = \(\frac{-1}{\left(\frac{d y}{d x}\right)}\) (x – 0)
⇒ x = 0.

(v) Given curve be y = sin2x
∴ \(\frac{d y}{d x}\) = 2 sin x cos x = sin 2x
∴ slope of tangent at x = \(\frac{\pi}{2}\)
= \(\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}}\) = 0
and slope of normal at x = \(\frac{\pi}{2}\) = ∞
When x = \(\frac{\pi}{2}\)
⇒ y = sin² \(\frac{\pi}{2}\)
= (1)² = 1
∴ eqn. of tangent to given curve at given point (\(\frac{\pi}{2}\), 1) is
y – 1 = 0 (x – \(\frac{\pi}{2}\))
⇒ y – 1 = 0
and eqn. of normal at (\(\frac{\pi}{2}\), 1) is
y – 1 = \(\frac{-1}{\left(\frac{d y}{d x}\right)}\) (x – \(\frac{\pi}{2}\))
⇒ x – \(\frac{\pi}{2}\) = 0.

Question 19 (old).
Find the equations of all lines having slope 2 and being tangents to the curve y + \(\frac{2}{x-3}\) = 0. (NCERT)
Solution:
Given eqn. of curve be y + \(\frac{2}{x-3}\) = 0
⇒ y = – \(\frac{2}{x-3}\) …………….(1)
Let the point (x₁, y₁) be on curve at which slope of tangent be equal to 2.
∴ \(\frac{d y}{d x}=\frac{2}{(x-3)^2}\)
Thus slope of tangent to given curve (1) at (x₁, y₁) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= \(\frac{2}{\left(x_1-3\right)^2}\)
\(\frac{2}{\left(x_1-3\right)^2}\) = 2
⇒ (x₁ – 3)² = 1
⇒ x₁ – 3 = ± 1
⇒ x₁ = 3 ± 1
⇒ x₁ = 4, 2
Also the point (x₁, y₁) lies on eqn. (1)
∴ y₁ = – \(\frac{2}{x_1-3}\) …………..(2)
When x₁ = 4
∴ from (2) ; y₁ = \(\frac{-2}{4-3}\) = – 2
When x₁ = 2
∴ from (2) ; y₁ = \(\frac{-2}{2-3}\) = 2
Hence the points on given curve be (4, – 2) and (2, 2).
∴ eqn. of line having slope 2 and passes through (4, – 2) is given by
y + 2 = 2 (x – 4)
⇒ 2x – y – 10 = 0
and the eqn. of line through (2, 2) and having slope 2 is given by
y – 2 = 2 (x – 2)
⇒ 2x – y – 2 = 0

Question 20.
Find the equation of the normal to the curve ay² = x³ at the point (am², am³)
Solution:
The given curve be, ay² = x³
∴ 2ay \(\frac{d y}{d x}\) = 3x²
⇒ \(\frac{d y}{d x}\) = \(\frac{3 x^2}{2 a y}\)
⇒ \(\left.\frac{d y}{d x}\right]_{\left(a m^2, a m^3\right)}=\frac{3\left(a m^2\right)^2}{2 a\left(a m^3\right)}\)
= \(\frac{3 a^2 m^4}{2 a^2 m^3}=\frac{3}{2} m\)
∴ slope of tangent at (am², am³) = \(\frac{3 m}{2}\)
∴ slope of normal at (am², am³) = \(\frac{-1}{\frac{3 m}{2}}=\frac{-2}{3 m}\)
∴ eqn. of Normal at (am², am³) is given by y – am³ = \(\frac{-2}{3 m}\) (x – am²)
⇒ 3my – 3am⁴ = – 2x + 2am²
⇒ 2x + 3my = 3am⁴ + 2am².

Question 21.
(i) Find the equation of the tangent to the curve √x + √y = a at the point \(\left(\frac{a^2}{4}, \frac{a^2}{4}\right)\).
(ii) Find the equation of the tamgent and the normal to the curve x^2/3 + y^2/3 = 2 at (1, 1) (NCERT)
Solution:
(i) Equation of the curve is, √x + √y = a …………(1)
Differentiate eqn. (1) w.r.t. x, we have
\(\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{\sqrt{y}}{\sqrt{x}}\)
slope of tangent at \(\left.\left(\frac{a^2}{4}, \frac{a^2}{4}\right)=\frac{d y}{d x}\right]\left(\frac{a^2}{4}, \frac{a^2}{4}\right)\)
= \(-\frac{a / 2}{a / 2}\) = – 1
Thus, eqn. of tangent at \(\left(\frac{a^2}{4}, \frac{a^2}{4}\right)\) is given by
y – \(\frac{a^2}{4}\) = – 1 (x – \(\frac{a^2}{4}\))
⇒ x + y – \(\frac{a^2}{2}\) = 0.

(ii) Given, eqn. of the curve x^2/3 + y^2/3 = 2 ……….(1)
Diff. both sides of eqn. (1) w.r.t. x ; we have
\(\frac{2}{3} x^{-\frac{1}{3}}+\frac{2}{3} y^{-\frac{1}{3}} \frac{d y}{d x}\) = 0
⇒ \(\frac{1}{y^{1 / 3}} \frac{d y}{d x}=-\frac{1}{x^{1 / 3}}\)
⇒ \(\frac{d y}{d x}=-\frac{y^{1 / 3}}{x^{1 / 3}}\)
\(\left(\frac{d y}{d x}\right)_{(1,1)}=-\frac{1}{1}\) = – 1
The required eqn. of tangent at (1, 1) is given by
y – 1 = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) (x – 1)
⇒ y – 1 = – 1 (x – 1)
⇒ y – 1 = – x + 1
⇒ x + y = 2
The required eqn. of Normal at (1, 1) is given by y – 1 = – \(\frac{1}{\left(\frac{d y}{d x}\right)_{(1,1)}}\) (x – 1)
⇒ y – 1 = 1 (x – 1)
⇒ y – x = 0.

Question 21 (old).
Find the equations of the tangents to the curve y = x³ + 2x – 4 which are perpendicular to the line x + 14y +3 = 0. (NCERT)
Solution:
Given eqn. of curve be y = x³ + 2x – 4 ………..(1)
Let (x₁, y₁) be any point on the curve (1) at which tangents are to be drawn.
∴\(\frac{d y}{d x}\) = 3x² + 2
Thus slope of tangent to curve (1) at (x₁, y₁) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 3x₁² + 2
Also it is given that. tangent is ⊥ to the given line. x + 14y + 3 = 0, whose slope is – \(\frac{1}{14}\).
Thus, (3x₁² + 2) (- \(\frac{1}{14}\)) = – 1
⇒ 3₁² + 2 = 14
⇒ 3₁² = 12
⇒ x₁² = 4
⇒ x₁ = ± 2
Since the point (x₁, y₁) lies on given curve (1).
∴ y₁ = x₁ + 2x₁ – 4 ……….(2)
When x₁= 2
∴ from (2) ; we have
y₁ = 8 + 4 – 4 = 8
When x₁ = – 2
∴ from (2) ; we have
y₁ = – 8 – 4 – 4 = – 16
∴ required points on given curve are (2, 8) and (- 2, – 16).
Thus eqn. of tangent at (2, 8) is given by y – 8 = 14 (x – 2)
⇒ 14x – y – 20 = 0
and eqn. of tangent at (- 2,- 16) is given by
y + 16 = 14 (x + 2)
⇒ 14x – y + 12 = 0.

Question 22.
(i) Find the equation of tangent to the curve given by x = a sin³ t, y = b cos³ t at a point where t = \(\frac{\pi}{2}\). (NCERT)
(ii) Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at t = \(\frac{\pi}{4}\).
Solution:
(i) Given eqn. of curve be,
x = a sin³ t …………..(1)
and y = b cos³ t …………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. t, we have
\(\frac{d x}{d t}\) = 3a sin² t cos t
\(\frac{d y}{d t}\) = 3b cos² t (- sin t)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{3 b \cos ^2 t(-\sin t)}{3 a \sin ^2 t \cos t}\)
= – \(\frac{b}{a}\) cot t
Thus slope of tangent to given curve at t = \(\frac{\pi}{2}\)
= \(\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{2}}\)
= – \(\frac{b}{a}\) cot \(\frac{\pi}{2}\) = 0
Therefore, the required eqn. of tangent at x = a sin³ \(\frac{\pi}{2}\) = a
and y = b cos³ \(\frac{\pi}{2}\) = 0
i.e. (a, 0) is given by y – 0 = 0 (x – a)
⇒ y = 0.

(ii) Given curve be x = sin 3t,
y = cos 2t ………..(1)
∴ \(\frac{d x}{d t}\) = 3 cos 3t
and \(\frac{d y}{d t}\) = – 2 sin 2t
Thus, \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)
= – \(\frac{2 \sin 2 t}{3 \cos 3 t}\)
∴ slope of tangent at t = \(\left.\frac{\pi}{4} \frac{d y}{d x}\right]\)
= \(-\frac{2}{3} \frac{\sin \frac{\pi}{2}}{\cos \frac{3 \pi}{4}}\)
= \(-\frac{2}{3} \cdot \frac{1}{-\frac{1}{\sqrt{2}}}\)
= \(\frac{2 \sqrt{2}}{3}\)
Also from (1) ;
when t = \(\frac{\pi}{4}\)
⇒ x = \(\frac{1}{\sqrt{2}}\) and y = 0
eqn. of tangent at \(\left(\frac{1}{\sqrt{2}}, 0\right)\), is given by y – 0 = \(\frac{2 \sqrt{2}}{3}\left(x-\frac{1}{\sqrt{2}}\right)\)
⇒ 2√2x – 3y – 2 = 0.

Question 23.
(i) Find the equations of all lines having slope 2 and that are tangents to the curve y = \(\frac{1}{x-3}\). (NCERT)
(ii) Find the equation of the tangent to the curve y = x + \(\frac{4}{x^2}\) which is parallel to x-axis.
Solution:
(i) Given curve is y = \(\frac{1}{x-3}\), x ≠ 3
∴ \(\frac{d y}{d x}=\frac{2}{(x-3)^2}\)
According to given condition, we have \(\frac{2}{(x-3)^2}\) = 2 which is not possible
L.H.S. is -ve and R.H.S. is +ve.
∴ There is no such tangent is possible whose slope is 2.

(ii) Given equation of curve be
y = x + \(\frac{4}{x^2}\) ………….(1)
∴ Slope of tangent to given curve (1) at any point (x, y) = \(\frac{d y}{d x}\)
= 1 – \(\frac{8}{x^3}\)
Since tangent is parallel to x-axis
∴ \(\frac{d y}{d x}\) = 0
⇒ 1 – \(\frac{8}{x^3}\) = 0
⇒ x = 2
⇒ from (1) ;
y = 2 + \(\frac{4}{2^2}\) = 3
Hence any point on curve (1) be (2, 3).
Thus, the required equation of tangent to given curve (1) at point (2, 3) and parallel to x-axis is given by
y – 3 = 0 (x – 2)
⇒ y = 3.

Question 24.
Find the equation of the tangent to the curve y = x² – 2x + 7 which is
(i) parallel to the line 2x – y + 9 = 0
(ii) perpendicular to the line 5y – 15x = 13. (NCERT)
Solution:
(i) eqn. of given curve be
y = x² – 2x + 7 ……………(1)
∴ \(\frac{d y}{d x}\) = 2x – 2
Thus slope of tangent = \(\frac{d y}{d x}\) = 2x – 2
But tangent is || to the line 2x – y + 9 = 0, whose slope is 2.
∴ 2x – 2 = 2 [∵ m₁ = m₂]
∴ from (1) ;
y = 4 – 4 + 7 = 7
Hence point of contact is (2, 7).
∴ eqn. of tangent at (2, 7) is given by
y – 7 = 2(x – 2)
⇒ 2x – y + 3 = 0.

(ii) But the tangent is ⊥ to line 5y – 15x – 13 = 0 whose slope is 3.
∴ product of slopes = – 1
⇒ (2x – 2) (3) = – 1
⇒ 6x = 5
⇒ x = \(\frac{5}{6}\)
∴ from (1) ;
y = \(\frac{25}{36}-\frac{10}{6}\) + 7
⇒ y = \(\frac{25-60+252}{36}=\frac{217}{36}\)
∴ required point of contact is \(\left(\frac{5}{6}, \frac{217}{36}\right)\).
Thus eqn. of tangent at \(\left(\frac{5}{6}, \frac{217}{36}\right)\) is given by
y – \(\frac{217}{26}\) = – \(\frac{1}{3}\left(x-\frac{5}{6}\right)\)
⇒ 36 y – 217 = – 12x + 10
⇒ 12x + 36y = 227.

Question 25.
(i) Find the equations of tangents to the curve y = 4x³ – 3x + 5 which are perpendicular to the line 9y + x + 3 =0.
(ii) Find the equations of the tangents to the curve 3x² – y² = 2 which are perpendicular to the line x + 3y = 2.
Solution:
(i) Given, eqn. of curve be, y = 4x³ – 3x + 5 …………(1)
∴ \(\frac{d y}{d x}\) = 12x² – 3
∴ slope of tangent to given curve = \(\left(\frac{d y}{d x}\right)_{(x, y)}\)
= 12x² – 3
∴ eqn. of given line be 9y + x + 3 = 0 ………..(2)
∴ slope of eqn. (1) = – \(\frac{1}{9}\)
Since tangent to curve (2) is ⊥ to line (2).
∴ (12x² – 3) (- \(\frac{1}{9}\)) = – 1
⇒ 12x² – 3 = 9
⇒ x² = 1
⇒ x = ± 1
∴ from (1) ;
y = 4 – 3 + 5 = 6
and y = – 4 + 3 + 5 = 4
Thus point on given curves are (1, 6) and (- 1, 4)
Hence slope of tangent to given curve (1) = \(\left(\frac{d y}{d x}\right)_{x= \pm 1}\)
= 12 – 3 = 9.
Thus eqn. of tangent to curve (1) at point (1, 6) be given by
y – 6 = 9 (x – 1)
⇒ 9x – y – 3 = 0
and eqn. of tangent to curve (1) at point (- 1, 4) be given by
y – 4 = 9 (x + 1)
⇒ 9x – y +1 2 = 0

(ii) Given, eqn. of curve be 3x² – y² = 2 ……….(1)
Let P (x₁, y₁) be point on the curve, at which tangent to the curve are drawn.
Difff. eqn. (1) w.r.t. x, we have
6x – 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{6 x}{2 y}=\frac{3 x}{y}\)
Thus slope of tangent to curve (1) at (x₁, y₁) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= \(\frac{3 x_1}{y_1}\)
Also slope of given line x + 3y – 2 = 0 be – \(\frac{1}{3}\).
Since the tangent to given curve is ⊥ to given line
∴ \(\left(\frac{3 x_1}{y_1}\right)\left(-\frac{1}{3}\right)\) = – 1
[∵ m₁ m₂ = – 1]
⇒ x₁ = y₁ ………..(2)
Also the point (x₁, y₁) lies on given curve (1).
∴ 3x₁² – y₁² = 2
3x₁² – x₁² = 2 [using (2)]
⇒ 2x₁² = 2
⇒ x₁² = ± 1
∴ from(2) ;
y₁ = ± 1
Thus, the points on given curve be (1, 1) and (- 1, – 1) and slope of tangent to curve be equal to 3.
Therefore, the eqn. of tangent to given curve at (1, 1) is given by
y – 1 = 3 (x – 1)
⇒ 3x – y – 2 = 0
and eqn. of tangent to curve at (- 1, – 1) is given by
y + 1 = 3 (x + 1)
⇒ 3x – y + 2 = 0.

Question 26.
Find the equations of the normals to the une 3x² – y² = 8 which are parallel to the line x + 3y = 4. (NCERT Exemplar)
Solution:
Given eqn. of curve be 3x² – y² = 8 ……….(1)
Diff. eqn. (1) w.r.t. x, we have
6x – 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=\frac{3 x}{y}\)
Hence slope of tangent = \(\frac{d y}{d x}=\frac{3 x}{y}\)
∴ slope of normal = – \(\frac{y}{3 x}\)
But normal is || to line x + 3y = 4, whose slope is – \(\frac{1}{3}\).
∴ \(\frac{-y}{3 x}=-\frac{1}{3}\) [∵ m₁ = m₂]
⇒ y = x
∴ from (1);
3x² – x² = 8
⇒ x² = 4
⇒ x = ± 2
When x = 2, from (1) ;
y² = 4
⇒ y = ± 2
Hence points on curve are (2, 2) and (2, – 2)
When x = – 2;
from(1);
y² = 4
y = ± 2
and pts on curve are (- 2, 2) and (- 2, – 2)
Thus eqn. of normal at (2, 2) is
y – 2 = – (x – 2)
⇒ x + 3y – 8 = 0
eqn. of normal at (2, – 2) is
y + 2 = – (x – 2)
⇒ x + 3y + 4 = 0
eqn. of normal at (- 2, 2) is
y – 2 = – \(\frac{1}{3}\) (x + 2)
⇒ x + 3y – 4 = 0
and eqn. of normal at (- 2, – 2) is
y + 2 = – \(\frac{1}{3}\) (x + 2)
⇒ x+ 3y+ 8 = 0.

Question 27.
Find the equation of be normal to the curve y = x + \(\frac{1}{x}\), x > 0, perpendicular to the line 3x – 4y = 7.
Solution:
Given eqn. of curve be y = x + \(\frac{1}{x}\) ………….(1)
∴ \(\frac{d y}{d x}\) = 1 – \(\frac{1}{x^2}\)
eqn. of given line be,
3x – 4y – 7 = 0 …………(2)
∴ slope of line (2) = \(\frac{-3}{-4}=\frac{3}{4}\)
∴ slope of normal to given curve (1) = – \(\frac{1}{\left(\frac{d y}{d x}\right)_{(x, y)}}\)
Since normal to curve (1) is ⊥ to line (2)
– \(\left(\frac{1}{1-\frac{1}{x^2}}\right)\left(\frac{3}{4}\right)\) = – 1
⇒ 1 – \(\frac{1}{x^2}\) = \(\frac{3}{4}\)
⇒ \(\frac{1}{x^2}=\frac{1}{4}\)
⇒ x = ± 2
but x > 0
∴ x = 2
∴ from (1) ;
y = 2 + \(\frac{1}{2}\) = \(\frac{5}{2}\)
Thus any point on given curve (1) be (2, \(\frac{5}{2}\))
and slope of normal = – \(\frac{1}{\left(\frac{d y}{d x}\right)_{\left(2, \frac{5}{2}\right)}}\)
= \(\frac{-1}{1-\frac{1}{4}}=-\frac{4}{3}\)
Hence eqn. of normal to given curve at (2, \(\frac{5}{2}\)) be given by
y – \(\frac{5}{2}\) = – \(\frac{4}{3}\) (x – 2)
⇒ 6y – 15 = – 8x + 16
⇒ 8x + 6y – 31 = 0.

Question 28.
Find the equation of the normal to the curve x2 = 4y which passes through the point (- 1, 4).
Solution:
eqn. of given curve be x² = 4y …………(1)
Let the normal at P (x₁, y₁) on curve (1) pass through the point (- 1, 4).
Diff. (1) w.r.t. x; we have
2x = 4 \(\frac{d y}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{x}{2}\)
∴ slope of tangent to curve at P = (\(\frac{d y}{d x}\))_P
= \(\frac{x_1}{2}\)
Also point P lines on curve (1)
∴ x₁² = 4y₁ …………..(2)
Thus eqn. of normal to given curve (1) at P(x₁, y₁) be given by
y – y₁ = – \(\frac{1}{\left(\frac{d y}{d x}\right)_P}\) (x – x₁)
⇒ y – y₁ = – \(\frac{2}{x_1}\) (x – x₁) ………..(3)
eqn. (3) passes through point (- 1, 4)
∴ 4 – y₁ = – \(\frac{2}{x_1}\) (- 1 – x₁)
⇒ 4 – y₁ = \(\frac{2}{x_1}\) + 2
⇒ y₁ = 2 – \(\frac{2}{x_1}\)
∴ from (2) ;
x₁² = 4 (2 – latex]\frac{2}{x_1}[/latex])
⇒ x₁³ = 8x₁ – 8
⇒ x₁³– 8x₁ + 8 = 0
⇒ x₁ = 2
∴ from (2) ;
y₁ = 1
Thus eqn. of normal to given curve at point (2, 1) be given by
y – 1 = – \(\frac{1}{1}\) (x – 2)
⇒ x + y – 3 = 0.

Question 29.
Find the angle of intersection between the following curves :
(i) y = 6 and x²y = 12
(ii) y=4 – x²and y = x² (NCERT Exemplar)
(iii) y = x and x² = y. (NCERT Exemplar)
Solution:
(i) The eqn’s of given curves are xy = 6 ………..(1)
and x²y = 12 ………..(2)
For point of intersection of eqn. (1) and eqn. (2),
we have to solve eqn. (1) and (2) simultaneously, we get
From (2) ;
x . (xy) = 12
⇒ 6x = 12 [using (1)]
⇒ x = 2
∴ from (1) ;
2y = 6
⇒ y = 3
Thus the point of intersection of two given curves be (2, 3).
Diff. eqn (1) and eqn. (2) w.r.t. x ; we have
x + \(\frac{d y}{d x}\) y = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{y}{x}\)
∴ slope of tangent to curve (1) at (2, 3) = m₁
= \(\left(\frac{d y}{d x}\right)_{(2,3)}=\frac{-3}{2}\)
Also, x² \(\frac{d y}{d x}\) + 2xy = 0
⇒ \(\frac{d y}{d x}=-\frac{2 x y}{x^2}=-\frac{2 y}{x}\)
Thus slope of tangent to curve (2) at (2, 3) = m₂
= \(\left(\frac{d y}{d x}\right)_{(2,3)}\)
= – \(\frac{2 \times 3}{2}\) = – 3
Let θ be the acute angle between two given curves
∴ tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{-\frac{3}{2}+3}{1-\frac{3}{2} \times(-3)}\right|\)
⇒ tan θ = \(\left|\frac{\frac{3}{2}}{\frac{11}{2}}\right|\)
⇒ tan θ = \(\frac{3}{11}\)
⇒ θ = tan^-1 (\(\frac{3}{11}\))

(ii) Given eqns. of curve be y = 4 – x² ……….(1)
and y = x² …………(2)
From (1) and (2) ; we get
x² = 2
⇒ x = ± √2
From (1) ;
y = 4 – 2 = 2
Hence the point of intersection of two curves be (± √2, 2).
Diff. (1) w.r.t. x ; we get
\(\frac{d y}{d x}\) = – 2x …………….(3)
Diff. (2) w.r.t. x ; we get
\(\frac{d y}{d x}\) = 2x …………..(4)

at (√2, 2):
using (3) ;
\(\left(\frac{d y}{d x}\right)_{(\sqrt{2}, 2)}\) = m₁ = – 2√2
using (4) ;
\(\left(\frac{d y}{d x}\right)_{(\sqrt{2}, 2)}\) = m₂ = 2√2
Let θ be the angle of intersection of two curves
Then tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{-2 \sqrt{2}-2 \sqrt{2}}{1-8}\right|\)
= \(\frac{4 \sqrt{2}}{7}\)
∴ θ = tan^-1 \(\frac{4 \sqrt{2}}{7}\)

at (- √2, 2):
using (3) ;
m₁ = \(\left(\frac{d y}{d x}\right)_{(-\sqrt{2}, 2)}\) = 2√2
using (4) ;
m₂ = \(\left(\frac{d y}{d x}\right)_{(-\sqrt{2}, 2)}\) = – 2√2
∴ tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{4 \sqrt{2}}{1-8}\right|\)
= \(\frac{4 \sqrt{2}}{7}\)
∴ θ = tan^-1 (\(\frac{4 \sqrt{2}}{7}\))

(iii) given eqns. of curve be y² = x ……….(1)
and x² = y ………(2)
From (1) and (2) ; we have
x⁴ = x
⇒ x (x³ – 1) = 0
⇒ x = 0 and x³ – 1 = 0
i.e. (x – 1) (x² + x + 1) = 0
⇒ x = 1 or x² + x + 1 = 0
∴ only real values of x are 0 and 1.
When x = 0
∴ from (2) ; y = 0
When x = 1
∴ from (2) ; y = 1
Thus both curves intersects at (0, 0) and (1, 1)
Diff. both sides w.r.t. x, we get
2y \(\frac{d y}{d x}\) = 1
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2 y}\) ……….(3)
Diff. (2) both sides w.r.t. x ; we get
2x = \(\frac{d y}{d x}\) ……….(4)

at (0, 0):
∴ m₁ = \(\left(\frac{d y}{d x}\right)_{(0,0)}\) = ∞
Thus tangent to curve (1) at (0, 0) is parallel to y-axis
and m₂ = \(\left(\frac{d y}{d x}\right)_{(0,0)}\)
= 2 × 0 = 0
∴ tangent to curve (2) at (0, 0) is parallel to x axis.
Hence the angie between the tangents to given two curves at (0, 0) is right angle.
Hence given two curves intersects at right angle at (0, 0).

At (1, 1):
m₁ = \(\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{1}{2}\) [using (3)]
and m₂ = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = 2 [using (4)]
Let θ be the angle of intersection of two curves.
Then tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{\frac{1}{2}-2}{1+\frac{1}{2} \times 2}\right|=\left|\frac{3}{4}\right|\)
∴ θ = tan^-1 \(\left(\frac{3}{4}\right)\).

Question 30.
Find the angle of intersection of the curves x² + y = 4 and (x – 2)² + y² = 4 at the point ¡n tbe first quadrant.
Solution:
Given eqns. of curves are x² + y² = 4 ………….(1)
and (x – 2)² + y² = 4 …………..(2)
For point of intersection of both curves,
we solve eqn. (1) and eqn. (2) simultaneously.
eqn. (2) – eqn. (1) ; we have
(x – 2)² – x² = 0
⇒ x² – 4x + 4 – x² = 0
⇒ – 4x + 4 = 0
⇒ x = 1
∴ from (1) ;
y² = 3
⇒ y = ± √3
So point of intersection of both curves are (1, ± √3) but we want to find the angle of intersection at the point which lies in first quadrant.
∴ point of intersection of both curves be P(1, √3).
Diff. eqn. (1) both sides w.r.t. x;
we have 2x + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{x}{y}\)
∴ slope of tangent to given curve (1) at P(1, √3) = m₁
= \(\left(\frac{d y}{d x}\right)_{(1, \sqrt{3})}=-\frac{1}{\sqrt{3}}\)
Diff. eqn. (2) both sides w.r.t. x ; we have
2 (x – 2) + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{(x-2)}{y}\)
∴ slope of tangent to given curve (2) at P (1, √3) = m₂
= \(\left(\frac{d y}{d x}\right)_{(1, \sqrt{3})}\)
= \(-\frac{(1-2)}{\sqrt{3}}=\frac{1}{\sqrt{3}}\)
Let θ be the acute angle between given curves
Then tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{-\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}}\right|\)
= \(\left|\frac{-\frac{2}{\sqrt{3}}}{\frac{2}{3}}\right|\) = √3
⇒ θ = \(\frac{\pi}{3}\)

Question 31.
Prove that the curves y = x² – 3x + 1 and x (y + 3) = 4 intersect at right angles at the point (2, — 1).
Solution:
Eqn’s of given curve are;
y = x² – 3x + 1 ………..(1)
and x (y + 3) = 4 ………..(2)
Diff. eqn. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = 2x – 3
∴ slope of tangent to given curve (1) at (2, – 1) = m₁
= \(\left(\frac{d y}{d x}\right)_{(2,-1)}\)
= 2 2 – 3 = 1
Diff. eqn. (2) w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{4}{x^2}\)
[∵ y = \(\frac{4}{x}\) – 3]
∴ slope of tangent to given curve (2) at (2, – 1) = m₂
= \(\left(\frac{d y}{d x}\right)_{(2,-1)}\)
= – \(\frac{4}{2^2}\) = – 1
Here, m₁m₂ = 1 × (- 1) = – 1
Thus both curve sintersect at right angle at the point (2, – 1).

Question 32.
Find the condition that the curves 2x = y² and 2xy = k intersect orthogonally. (NCERT Exemplar)
Solution:
Given curves are 2x = y² ………..(1)
and 2xy = k ………….(2)
From (1) and (2) ;
y³ = k
⇒ y = k^1/3
∴ from (1) ; we have
x = \(\frac{k^{2 / 3}}{2}\)
Thus both curves intersects at P (\(\frac{k^{2 / 3}}{2}\), k^1/3)
Differentiate (1) ; w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{y}\)
∴ m₁ = \(\left.\frac{d y}{d x}\right]_{a t \mathrm{P}}=\frac{1}{k^{1 / 3}}\)
Differentiate (2) ; w.r.t. x, we get
x \(\frac{d y}{d x}\) + y = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{y}{x}\)
∴ m₂ = \(\left.\frac{d y}{d x}\right]_{\mathrm{at} P}\)
= \(\frac{-2 k^{1 / 3}}{k^{2 / 3}}=\frac{-2}{k^{1 / 3}}\)
Since the given curves cut right angles at P.
∴ m₁ m₂ = – 1
⇒ \(\left(\frac{1}{k^{1 / 3}}\right)\left(\frac{-2}{k^{1 / 3}}\right)\) = – 1
⇒ 2 = k^2/3
⇒ k² = 8.

Question 33.
Prove that the curves xy = 4 and x² + y² = 8 touch each other. (NCERT Exampler)
Solution:
Given eqns. of curve be xy = 4 ………..(1)
and x² + y² = 8 ………..(2)
From (2) ;
x² + y² – 2 × 4 = 0
⇒ x² + y² – 2xy = 0 [using (1)]
⇒ (x – y)² = 0
⇒ x = y
∴ From (1) ;
x² = 4
⇒ x = ± 2
∴ From (3) ;
⇒ y = ± 2
Hence the points of intersection of two curves be (± 2, ± 2).
Diff. both sides of eqn. (1) w.r.t. x ; we get
x \(\frac{d y}{d x}\) + y = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{y}{x}\) ………..(3)
Diff, both sides eqn. (2) w.r.t. x; we get
2x + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{x}{y}\)

at P (2, 2):
\(\left(\frac{d y}{d x}\right)_{c_1}=-\frac{2}{2}\) = – 1 and
\(\left(\frac{d y}{d x}\right)_{c_2}\) = – 1
∴ \(\left(\frac{d y}{d x}\right)_{c_1}=\left(\frac{d y}{d x}\right)_{c_2}\)
So two curves touch each other at p (2, 2).
Similarly both curves touch each other at other points (- 2, – 2); (- 2, 2) and (2, – 2).

Question 34.
Find the angle of Intersection between the curves y² = 4x and x² = 4y.
Solution:
Given equations of curves are
y² = 4x …………..(1)
and x² = 4y …………….(2)
from (2) ;
y = \(\frac{x^2}{4}\)
∴ from (1) ;
\(\left(\frac{x^2}{4}\right)^2\) = 4x
⇒ x⁴ = 64x
⇒ x (x³ – 64) = 0
⇒ x = 0, 4
∴ y = 0, 4
Thus points of intersection of two curves are Q (0, 0) and P(4, 4).
Diff. (1) and (2) w.r.t. x both sides ; we have
2y \(\frac{d y}{d x}\) = 4
⇒ \(\frac{d y}{d x}\) = \(\frac{2}{y}\)
and 2x = 4 \(\frac{d y}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{x}{2}\)
∴ m₁ = slope of tangent to curve (1) at P
= \(\left(\frac{d y}{d x}\right)_{P(4,4)}\)
= \(\frac{2}{4}=\frac{1}{2}\)
m₂ = slope of tangent to curve (2) at P (4, 4)
= \(\left(\frac{d y}{d x}\right)_{P(4,4)}\)
= \(\frac{4}{2}\) = 2
Let θ be the angle between the tangents to given curves then
tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{\frac{1}{2}-2}{1+\frac{1}{2} \times 2}\right|=\frac{3}{4}\)
⇒ θ = tan^-1 \(\frac{3}{4}\)
and at point Q (0, 0) m₁ → ∞ m₂ → 0
(i.e. one tangent is y-axis and other tangent is x-axis)
Thus both tangents to given curves cuts orthogonally.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1

November 18, 2023

Students often turn to ML Aggarwal Class 12 ISC Solutions Chapter 7 Applications of Derivatives Ex 7.1 to clarify doubts and improve problem-solving skills.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1

Question 1.
Find the rate of change of the area of a circle with respect to its radius when the radius is 6 cm. (NCERT)
Solution:
Let r be the radius of circle
Then area of circle = A = πr²
∴ \(\frac{d A}{d r}\) = 2πr
When r = 6 cm ;
\(\frac{d A}{d r}\) = 2 π × 6 cm
= 12 π cm
Thus, the area of the circle changing at the rate of 12 π cm²/cm.

Question 2.
If the radius of a circle is increasing at the rate of 0.5 cm/sec, at what rate is its circumference increasing?
Solution:
Let r be the radius of circle at any time t
and let S be the circumference of circle.
Then S = 2πr
∴ \(\frac{d S}{d t}\) = 2π \(\frac{d r}{d t}\)
Given \(\frac{d r}{d t}\) = 0.5 cm/sec
Thus, \(\frac{d S}{d t}\) = 2π × 0.5 = π cm/sec
Hence, the rate at which circumference of circle is increasing be π cm/sec.

Question 2 (old).
If the radius of a circle is increasing at the rate of 0.7 cm/sec, at what rate is its circumference increasing ?(NCERT)
Solution:
Let r be the radius of circle then
\(\frac{d r}{d t}\) = 0.7 cm/sec.
Let c = circumference of circle = 2πr
∴ \(\frac{d c}{d t}\) = 2π × \(\frac{d r}{d t}\)
= 2π × 0.7
= 1.4π cm/sec.

Question 3.
If the radius of a circle is increasing at the rate of 3 cm/sec, at what rate is its area increasing when its radius is 2 cm?
Solution:
Let r be the radius of circle at any time t
Let A = area of circle = πr²
Given \(\frac{d r}{d t}\) = 3cm/sec;
r = 2 cm
∴ \(\frac{d A}{d t}\) = 2πr = 2π × 2 × 3
= 12π cm²/sec.

Question 3(old).
If the radius of a circle is increasing at the rate of 3 cm/sec, at what rate is its area increasing when its radius is 10 cm ? (NCERT)
Solution:
Let A be the area of circle with radius r
Then A = πr²
On differentiating w.r.t. t, we have
\(\frac{d A}{d t}\) = 2πr \(\frac{d r}{d t}\) ………….(1)
Given, \(\frac{d r}{d t}\) = 3 cm/sec
and r = 10 cm
∴ from (1) ;
\(\frac{d A}{d t}\) = (2π × 10 × 3) cm² / sec
Thus, the required area of circle is increasing at the rate of 60π cm²/sec.

Question 4.
If the sides of a square are decreasing at the rate of 1.5 cm/sec, at what rate is its perimeter decreasing ?
Solution:
Let P be the perimeter of square with side x cm.
Then P = 4x
Diff. both sides w.r.t. t, we have
\(\frac{d p}{d t}\) = 4 \(\frac{d x}{d t}\) …………(1)
Given, \(\frac{d r}{d t}\) = – 1.5 cm/sec.
∴ from (1) ;
\(\frac{d p}{d t}\) = – (4 × 1.5) cm/sec = – 6 cm/sec.
Hence the perimeter of square is decreasing at the rate of 6 cm/sec.

Question 5.
If an edge of a variable cube is increasing at the rate of 10 cmlsec, at what rate is its volume increasing when its edge is 5 cm?
Solution:
Let x be the edge of variable cube at any time t
Let V = volume of cube = x³
Given, \(\frac{d x}{d t}\) = 10 cm / sec ;
x = 5 cm
∴ \(\frac{d V}{d t}\) = 3x² \(\frac{d x}{d t}\)
= 3 × 5² × 10 cm³ / sec
= 750 cm³ / sec

Question 5 (old).
If an edge of a variable cube is increasing at the rate of 3 cm/sec, at what rate is its volume increasing when its edge is 10 cm ? (NCERT)
Solution:
Let x be the length of edge of cube s.t. \(\frac{d x}{d t}\) = 3 cm/sec.
Then V = volume of cube = x³
∴ \(\frac{d V}{d t}\) = 3x² \(\frac{d x}{d t}\)
Given, x = 10 cm
∴ \(\frac{d V}{d t}\) = 3 × (10)² × 3 = 900 cm³/sec.

Question 6.
A balloon which always remains spherical has a variable radius. Find the rate at which its volume is increasing with respect to radius when the radius is 10 cm. (NCERT)
Solution:
Let V be the volume of spherical balloon with radius r.
Then V = volume of spherical baloon = \(\frac{4}{3}\) πr³
Differentiating both sides w.r.t. r. we have
\(\frac{d V}{d r}\) = \(\frac{4 \pi}{3}\) × 3r² = 4πr²
When r = 10 cm ;
\(\frac{d V}{d r}\) = 4π × 10² cm² / cm
= 400 π cm³ / cm
Hence, the volume of spherical balloon is increasing at the rate of 400 cm³/cm.

Question 7.
If the radius of a balloon which always remains spherical is increasing at the rate of 1.5 cm/sec, at what rate is ¡t surface area increasing when its radius is 12 cm?
Solution:
Let r be the radius of spherical balloon at any time t
and let S be the surface area of spherical balloon.
Then S = 4πr²
⇒ \(\frac{d S}{d t}\) = 8πr \(\frac{d r}{d t}\) …………….(1)
Given r = 12 cm;
\(\frac{d r}{d t}\) = 1.5 cm/sec
from (1);
\(\frac{d S}{d t}\) = 8π × 12 × 15 cm² /sec
= 144 π cm²/sec.
Thus the surface area of spherical balloon is increasing at the rate 144 π cm²/sec.

Question 8.
If the radius of a soap bubble is increasing at the rate of \(\frac{1}{2}\) cm/sec, at what rate is its volume increasing when the radius is 1 cm ? (NCERT)
Solution:
Let r be the radius of bubble and V be the volume of air bubble
Then V = \(\frac{4}{3}\) πr³
and \(\frac{d r}{d t}\) = 0.5 cm/sec.
Diff. both sides w.r.t. t; we have
∴ \(\frac{d V}{d t}\) = \(\frac{4 \pi}{3} \times 3 r^2 \frac{d r}{d t}\)
= 4πr² \(\frac{d r}{d t}\)
When r = 1 cm
∴ \(\frac{d V}{d t}\) = (4π × 1² × 0.5) cm³/sec
= 2π cm³/sec.

Question 9.
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Find the rate at which the depth of the wheat is increasing, take π = 3.14. (NCERT)
Solution:
Let V be the volume of cylindrical tank with radius r.
Then V = πr²h ……….(1)
where h be the height of cylindrical tank.
On differentiating eqn. (1) w.r.t. t, we get
\(\frac{d \mathrm{~V}}{d t}=\pi r^2 \frac{d h}{d t}\) …………..(2)
Given \(\frac{d V}{d t}\) = 314 m³/hr ;
r = 10 cm
∴ from (2) ;
314 = 3.14 × (10)³
⇒ \(\frac{d h}{d t}\) = \(\frac{314}{314}\) = 1 m/hr
Thus, the required rate at which the depth of wheat is increasing be 1 m/hr.

Question 10.
(i) The total revenue in rupees received from the sale of x units of a product is given by
R (x) = 3x² + 36x + 5.
Find the marginal revenue when x = 5. (NCERT)
(ii) The total cost C (x) associated with the production of x units of an item is given by C (x) = 0.005x³ – 0.02x³ + 30x + 5000. Find the marginal cost when 3 units are produced.
Solution:
(i) Given R (x) = 3x² + 3 6x + 5
We know that marginal revenue is the rate of change of total revenue w.r.t. to number of units sold.
∴ \(\frac{d R}{d t}\) = 6x + 36
When x = 5 then,
\(\frac{d R}{d t}\) = 30 + 36 = Rs. 66

(ii) The total cost C (x) associated with the production of x unitsof an item is given by
C (x) = 0.005 x³ – 0.02 x² + 30x + 5000
∴ Marginal cost = \(\frac{d}{d x}\) C(x)
= 0.005 × 3x² – 0.02 × 2x + 30
= 0 015 x² – 0.04x + 30
∴ required marginal cost when 3 units are produced = [C’ (x)]_{x = 3}
= 0.015 × 3² – 0.04 × 3 + 30
= 0.135 – 0.12 + 30
= Rs. 30.015.

Question 11.
Find the rate of change of area of a circle of radius r with regard to its radius. How V\ fast is the area changing with respect to the radius when the radius is 5 cm ? (NCERT)
Solution:
Let A be the area of circle of radius r
Then A = π r² ;
differentiating both sides w.r.t. r, we have
\(\frac{d A}{d r}\) = 2πr
where r = 5 cm
∴ \(\frac{d A}{d r}\) = (2 π × 5) cm²/cm
= 10π cm²/cm

Question 12.
Find the rate of change of the whole surface of a closed circular cylinder of radius r and height h with respect to change in radius.
Solution:
Let S be the total surface area of closed cylinder with radius r and height h.
Then S = 2πr² + 2πrh
= 2π (2r + h) unit² / unit
On differentiating eqn. (1) w.r.t. r, we have
\(\frac{d S}{d r}\) = 4πr + 2πh
= 2π (2 r + h) unit² /unit.
Hence, the required rate of change of surface area of cylinder w.r.t. radius be 2π (2r + h) units² /units.

Question 13.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/sec. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing ? (NCERT)
Solution:
Let r be the radius of circular wave, then area of circular wave = A = πr²
given \(\frac{d r}{d t}\) = 5 cm/sec
and r = 8 cm
∴ \(\frac{d A}{d t}\) = 2π × r × \(\frac{d r}{d t}\)
= 2π × 8 × 5
= 80 π cm²/sec

Question 14.
A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec. At the instant when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing ? (NCERT)
Solution:
Let r the radius of circular wave, then area of circular wave = A = πr²
given \(\frac{d r}{d t}\) = 5 cm/sec
and r = 8 cm
∴ \(\frac{d A}{d t}\) = 2π × r × \(\frac{d r}{d t}\)
= 2π × 8 × 5
= 80π cm²/sec

Question 15.
The length x of a rectangle is decreasing at the rateof 5 cm/minute and the width y is increasing at the rate of 4 cm/ minute. When x = 8 cm and y = 6 cm, find the rate of change of
(i) the perimeter
(ii) the area of the rectangle.
Solution:
(i) Perimeter of rectangle = 2 (x + y) = p
∴ \(\frac{d p}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)\)
Given \(\frac{d x}{d t}\) =- 5 cm/min
\(\frac{d y}{d t}\) = 4 cm/min
∴ \(\frac{d p}{d t}\) = 2 (- 5 + 4) = – 2 cm/min.

(ii) Let A = area of rectangle = xy
Diff. both sides w.r.t. t;
\(\frac{d p}{d t}=\frac{d y}{d t}+y \frac{d x}{d t}\)
= 8 × 4 + 6 × (- 5)
= 32 – 30
= 2 cm²/min

Question 16.
The volume of a cube is increasing at the rate of 9 cubic centimeters per \ 1o second. How fast is the surface area increasing when the length of the edge is 10 centimeters ? (NCERT)
Solution:
Let x be the length of edge of cube
Then V = volume of cube = x³ ………….(1)
and S = surface area of cube = 6x² …….(2)
From (1) ;
\(\frac{d V}{d t}\) = 3x² \(\frac{d x}{d t}\) ;
Also \(\frac{d V}{d t}\) = 9 cm³/sec.
9 = 3x² \(\frac{d x}{d t}\); also x = 10 cm
⇒ \(\frac{9}{3 \times 10^2}=\frac{d x}{d t}\)
⇒ \(\frac{d x}{d t}=\frac{3}{100}\) cm / sec
From (2) ;
\(\frac{d S}{d t}\) = 12x \(\frac{d x}{d t}\)
∴ \(\frac{d S}{d t}\) = 12 × 10 × \(\frac{3}{100}\)
= \(\frac{360}{100}\) cm² /sec.
⇒ \(\frac{d S}{d t}\) = 3.6 cm²/sec
Hence the surface area of the cube is increasing at the rate of 3.6 cm²/sec.

Question 16 (old).
The top of a ladder 6 metres long is resting against a vertical wall. Suddenly, c the ladder begins to slide outwards. At the instant when the foot of the ladder is 4 metres from the wall, it is sliding away at the rate of 0.5 m/sec. How fast is the top sliding downwards at this moment ? how far is the foot from the wall when the foot and the top are moving at the same rate ?
Solution:
Let PQ be the wall and AB be the ladder such that
AB = 6 m,
let AQ = y m and
QB = x m
Here x = 4 m and \(\frac{d x}{d t}\) = 0.5 m/sec

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1 1

In right angled ΔAQB, we have
x² + y² = 36 ………..(1)
∴ 2x \(\frac{d x}{d t}\) + 2y \(\frac{d y}{d t}\) = 0
⇒ \(\frac{d y}{d t}\) = – \(\frac{x}{y}\) \(\frac{d x}{d t}\)
at x = 4 ;
\(\frac{d y}{d t}\) = – \(\frac{4}{y} \times \frac{1}{2}\)
= – \(\frac{2}{y}\) m / sec ……….(2)
From (1) ;
y² = 36 – x²
= 36 – 4² = 20
⇒ y = 2√5 m
∴ From (2) ; we have
\(\frac{d y}{d t}=\frac{-2}{2 \sqrt{5}}\) m/sec dt
= – \(\frac{1}{\sqrt{5}}\) m/sec
So the ladder top is sliding at the rate of \(\frac{1}{\sqrt{5}}\) m/sec.
Now we want to find the value of x such that
\(\frac{d x}{d t}=-\frac{d y}{d t}\)
∴ \(\frac{d x}{d t}=\frac{x}{y} \frac{d x}{d t}\)
⇒ x = y
from (1) ;
x² + x² = 36
⇒ x² = 18
⇒ x = 3√2 m
When foot and top are moving at the same rate and foot of wall is 3√2 m away from the wall.

Question 17.
A particle moves along the curve y = \(\frac{2}{3}\) x³ + 1. Find the points on the curve at which the y-coordinate is changing twice as fast as the x- coordinate.
Solution:
Let P (x, y) be any point on given curve
y = \(\frac{2}{3}\) x³ ……………(1)
also it is given that,
\(\frac{d y}{d t}\) = 2 \(\frac{d x}{d t}\) …………(2)
From (1);
\(\frac{d y}{d t}\) = \(\frac{2}{3}\) × 3x² \(\frac{d x}{d t}\)
= 2x² \(\frac{d x}{d t}\)
∴ From (2) ; we have
2 \(\frac{d x}{d t}\) = 2x² \(\frac{d x}{d t}\)
⇒ x² = ± 1
When x = 1
∴ from (1); we have
y = \(\frac{2}{3}\) + 1 = \(\frac{5}{3}\)
When x = – 1
∴ from (2); we have
y = – \(\frac{2}{3}\) + 1 = \(\frac{1}{3}\)
Hence the required points on given curve be (1, \(\frac{5}{3}\)) and (- 1, \(\frac{1}{3}\)).

Question 18.
A man of height 2 meters walks at a uniform speed of 5 km/h away from a lamp post which is 6 metres high. Find the rate at which the length of shadow increases.
Solution:
Let PQ be the lamp post and MN be the position of man at any time t and x metres be the distance of man from lamp post and y meters be the length of its shadow.
Then PQ = 6 m ;
MN = 2 m;

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1 2

In similar Δ’s MLN and PLQ
\(\frac{\mathrm{MN}}{\mathrm{PQ}}=\frac{\mathrm{LN}}{\mathrm{LQ}}\)
⇒ \(\frac{2}{6}=\frac{y}{y+x}\)
⇒ 3y = y + x
⇒ y = \(\frac{x}{2}\)
∴ \(\frac{d y}{d t}=\frac{1}{2} \frac{d x}{d t}\)
but \(\frac{d x}{d t}\) = 5 km/hr
⇒ \(\frac{d y}{d t}=\frac{5}{2}\) km/hr.

Question 19.
A girl of height 1.6 m walks at the rate of 50 metres per minute away from a lamp which is 4 m above the ground. mHow fast is the girl’s shadow lengthening?
Solution:
Let AB be the lamp post where A be the position of the lamp then AB 4 cm.
Let CD be the position of girl at any time t and x be the distance of girl from lamp post.
Let y be the length of the shadow of girl.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1 3

Thus AECD and AEBA are similar A’s.
In similar A’s ECD and EBA, we have
\(\frac{E C}{E B}=\frac{C D}{B A}\)
⇒ \(\frac{y}{x+y}=\frac{16}{4}\)
⇒ \(\frac{y}{x+y}=\frac{16}{40}=\frac{2}{5}\)
⇒ 5y = 2x + 2y
⇒ 3y = 2x …………………(1)
Diff. eqn. (1) both sides w.r.t. t, we have
3 \(\frac{d y}{d t}\) = 2 \(\frac{d x}{d t}\)
But given, \(\frac{d x}{d t}\) \(\frac{d x}{d t}\) = 50 m/min
∴ \(\frac{d y}{d t}\) = (2 × \(\frac{50}{3}\)) m/min
= \(\frac{100}{3}\) m/min
Thus the length of girl’s shaddow is increasing at the rate of 33\(\frac{1}{3}\) m/min.

Question 20.
A kite is 120 in high and 130 m of string c is out. If the kite is moving horizontally y at the rate of 5.2 m/sec, find the rate at which the string is being paid out at that instant.
Solution:
Let at any time t, Q be the position of kite.
Then in right angled ΔQPR we have,
y² = x² + (120)² ……………(1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1 4

∴ 2y \(\frac{d y}{d t}\) = 2x \(\frac{d x}{d t}\) …………(2)
given,
\(\frac{d x}{d t}\) = 5.2 m/sec
and y = 130
∴ from (1) ;
(130)² – (120)² = x²
⇒ x² = 16900 – 14400
⇒ x² = 2500
⇒ x = 50
∴ from (2) ;
\(\frac{d y}{d t}=\frac{x}{y} \frac{d x}{d t}\)
= \(\frac{50}{130}\) × 5.2
= \(\frac{260}{130}\)
= 2 m/sec
Thus the required rate at which the dtring is being paid out be 2 m/sec.

Question 21.
A circular cone, with semi-vertical angle 45°, is fixed with its axis vertical and its vertex downwards. Water is poured into the cone at the rate of 2 cm3 per second. Find the rate at which the depth of the water is increasing when the depth is 4 cm.
Solution:
Let V be the volume of right circular cone with radius r and height h.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1 5

Also tan α = tan 45° = \(\frac{r}{h}\)
⇒ 1 = \(\frac{r}{h}\)
⇒ h = r
Then V = \(\frac{1}{3}\) πr² h
= \(\frac{\pi}{3}\)h³
On differentiating both sides w.r.t. t, we have
\(\frac{d \mathrm{~V}}{d t}=\frac{\pi}{3} \times 3 h^2 \frac{d h}{d t}\)
= πh² \(\frac{d h}{d t}\) …………(1)
Given \(\frac{d V}{d t}\) = 2 cm³ / sec ; h = 4 cm
∴ from (1) ;
2 = π × 4² \(\frac{d h}{d t}\)
⇒ \(\frac{d h}{d t}\) = \(\frac{1}{\pi}\) cm / sec
Thus the required rate at which the depth of water is increasing be \(\frac{1}{8 \pi}\) cm/sec.

Question 22.
Sand is pouring from a pipe at the rate of 12 cm³/sec. The falling sand forms a ,, cone on the ground in such a way that the height of the cone is always one- sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm ?
Solution:
Let r be the radius and h be the height of cone such that h = \(\frac{r}{6}\)
⇒ r = 6 h cm
Let V be the volume of circular cone
Then, V = \(\frac{\pi}{3}\)r²h
On differentiating both sides w.r.t. t, we have
⇒ \(\frac{d V}{d t}\) = 36πh² × \(\frac{d h}{d t}\) …………..(1)
Given, \(\frac{d V}{d t}\) = 12cm³ / sec ;
h = 4 cm
∴ from (1) ; we have
12 = 36π × 4² × \(\frac{d h}{d t}\)
⇒ \(\frac{d h}{d t}\) = \(\frac{12}{36 \pi \times 16}\) cm / sec
⇒ \(\frac{d h}{d t}=\frac{1}{48 \pi}\) cm / sec
Thus, the height of sand cone is increasing at the rate of \(\frac{1}{48 \pi}\) cm/sec.

Question 23.
A conical vessel whose height is 4 metres and of base radius 2 metres is being filled with water at the rate of 0.75 cubic metres per minute. Find the rate at which the level of the water is rising when the depth of water is 1.5 metres.
Solution:
Here 2 m is the radius and 4 m is the height of inverted cone.
If V be the volume at any time t
∴ \(\frac{d V}{d t}\) = \(\frac{3}{4}\) m³/mt. ………………(1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1 6

At that instant, let the water form a cone of height h m and radius r m.
∴ Δ’s OBE and ODC are similar
i.e. \(\frac{E B}{C D}=\frac{E O}{C O}\)
⇒ \(\frac{2}{r}=\frac{4}{h}\)
⇒ h = 2r
⇒ r = \(\frac{h}{2}\)
Now volume of inverted cone V = \(\frac{1}{3}\) πr²h
⇒ V = \(\frac{1}{3} \pi\left(\frac{h}{2}\right)^2\) × h
V = \(\frac{1}{12}\) πh³
∴ \(\frac{d \mathrm{~V}}{d t}=\frac{1}{4} \pi h^2 \frac{d h}{d t}\)
⇒ \(\frac{3}{4}=\frac{\pi}{4} h^2 \frac{d h}{d t}\) [using (1)]
∴ \(\frac{d h}{d t}=\frac{3}{\pi h^2}\)
When the water level rising about the base = h = 1.5 m
Thus \(\frac{d h}{d t}=\frac{3}{\pi} \times \frac{4}{9}\) m/min
= \(\frac{4}{3 \pi}\) m/min.

Question 24.
Water is dripping out at a steady rate of 1 cu cm/sec through a tiny hole at the vertex of the conical vessel, whose axis is vertical. When the slant height of water in the vessel is 4 cm, find the rate of decrease of slant height, where the semi-vertical angle of the cone is \(\frac{\pi}{6}\). (NCERT Exampler)
Solution:
Let l be the slant height of conical vessel at any time t.
Then
V = \(\frac{\pi}{3}\) r²h

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1 7

where r = radius of cone
h = height of cone
Here h = l cos \(\frac{\pi}{6}\) and
r = l sin \(\frac{\pi}{6}\)
∴ from (1) ;
V = \(\frac{\pi}{3}\) (l sin \(\frac{\pi}{6}\))² × l cos \(\frac{\pi}{6}\)
⇒ V = \(\frac{\pi}{3} l^3 \times\left(\frac{1}{2}\right)^2 \times \frac{\sqrt{3}}{2}\)
= \(\frac{\pi l^3 \sqrt{3}}{24}\)
Diff. both sides w.r.t. t, we have
\(\frac{d \mathrm{~V}}{d t}=\frac{\pi \sqrt{3}}{24} \times 3 l^2 \frac{d l}{d t}\)
= \(\frac{\sqrt{3} \pi}{8} l^2 \frac{d l}{d t}\) ………….(2)
Given \(\frac{d}{d x}\) = – 1 cu cm/ sec
and l = 4 cm
∴ from (2) ; we have
– 1 = \(\frac{\sqrt{3} \pi}{8} \times 4^2 \frac{d l}{d t}\)
⇒ \(\frac{d l}{d t}=-\frac{1}{2 \sqrt{3} \pi}\) cm / sec
Hence slant height of tne cone be decreasing at the rate of \(\frac{1}{2 \sqrt{3} \pi}\) cm/sec.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms MCQs

November 17, 2023

Accessing ML Aggarwal Class 12 ISC Solutions Chapter 6 Indeterminate Forms MCQs can be a valuable tool for students seeking extra practice.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms MCQs

Choose the correct answer from the given four options in questions (1 to 6) :

Question 1.
\(\ {Lt}_{x \rightarrow 0} \frac{\sin ^{-1} x}{x}\) is equal to
(a) 0
(b) 1
(c) – 1
(d) does not exist
Solution:
(b) 1

\(\ {Lt}_{x \rightarrow 0} \frac{\sin ^{-1} x}{x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{\frac{1}{\sqrt{1-x^2}}}{1}\)
[using L’Hopital’s rule]
= \(\frac{1}{\sqrt{1-0^2}}\) = 1

Question 2.
\(\ {Lt}_{x \rightarrow 0} \frac{\tan 3 x}{\sin 2 x}\) is eqaul to
(a) 1
(b) \(\frac{2}{3}\)
(c) \(\frac{3}{2}\)
(d) does not exist
Solution:
(c) \(\frac{3}{2}\)

\(\ {Lt}_{x \rightarrow 0} \frac{\tan 3 x}{\sin 2 x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{3 \sec ^2 3 x}{2 \cos 2 x}\)
[using L’Hopital’s rule]
= \(\frac{3 \times 1}{2 \times 1}=\frac{3}{2}\)

Question 3.
\(\ {Lt}_{x \rightarrow 1} \frac{x^3+3 x-4}{2 x^2+x-3}\) is equals to
(a) \(\frac{3}{2}\)
(b) \(\frac{6}{5}\)
(c) \(\frac{5}{6}\)
(d) none of these
Solution:
(b) \(\frac{6}{5}\)

\(\ {Lt}_{x \rightarrow 1} \frac{x^3+3 x-4}{2 x^2+x-3}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 1} \frac{3 x^2+3}{4 x+1}\) [usingh L’Hopital’s ruel]
= \(\frac{3 \times 1^2+3}{4 \times 1+1}=\frac{6}{5}\).

Question 4.
\(\ {Lt}_{x \rightarrow 0} \frac{(1+x)^n-n x-1}{x^2}\), n > 1, is equal to
(a) none
(b) n (n – 1)
(c) \(\frac{n(n-1)}{2}\)
(d) does not exist
Solution:
(c) \(\frac{n(n-1)}{2}\)

\(\ {Lt}_{x \rightarrow 0} \frac{(1+x)^n-n x-1}{x^2}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{n(1+x)^{n-1}-n}{2 x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{n(n-1)(1+x)^{n-2}}{2}\)
= \(\frac{n(n-1)}{2}\) (1 + 0)^n-2
= \(\frac{n(n-1)}{2}\)

Question 5.
\(\ {Lt}_{x \rightarrow \infty} \frac{\sin \left(\frac{1}{x}\right)}{\tan ^{-1} \frac{1}{x}}\) is equal to
(a) 1
(b) 2
(c) \(\frac{1}{2}\)
(d) does not exist
Solution:
(a) 1

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms MCQs 1

= 1 × (0 + 1) = 1.

Question 6.
\(\underset{x \rightarrow \infty}{\mathbf{L t}}\) x sin^-1 \(\frac{3}{x}\) is equal to
(a) \(\frac{1}{3}\)
(b) 3
(c) 1
(d) does not exist
Solution:
(b) 3

\(\underset{x \rightarrow \infty}{\mathbf{L t}}\) x sin^-1 \(\frac{3}{x}\) [0 . ∞ form]
= \(\ {Lt}_{x \rightarrow \infty} \frac{\sin ^{-1} \frac{3}{x}}{\frac{1}{x}}\)
= \(\ {Lt}_{x \rightarrow \infty} \frac{\frac{1}{\sqrt{1-\left(\frac{3}{x}\right)^2}} \cdot\left(-\frac{3}{x^2}\right)}{-\frac{1}{x^2}}\)
= \(\ {Lt}_{x \rightarrow \infty} \frac{3 x}{\sqrt{x^2-9}}\)
= \(\ {Lt}_{x \rightarrow \infty} \frac{3 x}{x \sqrt{1-\frac{9}{x^2}}}\)
= \(\frac{3}{\sqrt{1-0}}\) = 3.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Chapter Test

November 17, 2023

Practicing ML Aggarwal Class 12 Solutions ISC Chapter 6 Indeterminate Forms Chapter Test is the ultimate need for students who intend to score good marks in examinations.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Chapter Test

Question 1.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 1} \frac{1-x^2}{\sin \pi x}\)
(ii) \(\ {Lt}_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{x-\cos \left(\sin ^{-1} x\right)}{1-\tan \left(\sin ^{-1} x\right)}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 1} \frac{1-x^2}{\sin \pi x}\)
[\(\frac{0}{0}\) form, using L’Hopital’s rule]
= \(\ {Lt}_{x \rightarrow 1} \frac{-2 x}{\pi \cos \pi x}\)
= \(\frac{-2}{\pi \cos \pi}\)
= \(\frac{-2}{\pi(-1)}=\frac{2}{\pi}\).

(ii) \(\ {Lt}_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{x-\cos \left(\sin ^{-1} x\right)}{1-\tan \left(\sin ^{-1} x\right)}\)
[\(\frac{0}{0}\) form, using L’Hopital’s rule]
put sin^-1 x = t
⇒ x = sin t
as x → \(\frac{1}{\sqrt{2}}\)
⇒ t → \(\frac{\pi}{4}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{t \rightarrow \frac{\pi}{4}} \frac{\sin t-\cos t}{1-\tan t}\)
= \(\ {Lt}_{t \rightarrow \frac{\pi}{4}} \frac{\cos t+\sin t}{-\sec ^2 t}\)
= \(\frac{\cos \frac{\pi}{4}+\sin \frac{\pi}{4}}{-\sec ^2 \frac{\pi}{4}}\)
= \(\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{-(\sqrt{2})^2}\)
= \(\frac{-\sqrt{2}}{2}=-\frac{1}{\sqrt{2}}\).

Question 2.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^2+x \log (1-x)}\)
(ii) \(\ {Lt}_{x \rightarrow 1} \frac{x^x-x}{1-x+\log x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^2+x \log (1-x)}\)
[\(\frac{0}{0}\) form, using L’Hopital’s rule]

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Chapter Test 1

(ii) \(\ {Lt}_{x \rightarrow 1} \frac{x^x-x}{1-x+\log x}\)
[\(\frac{0}{0}\) form, using L’Hopital’s rule]
= \(\ {Lt}_{x \rightarrow 1} \frac{x^x(1+\log x)-1}{-1+\frac{1}{x}}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 1} \frac{x^x \cdot \frac{1}{x}+(1+\log x)^2 x^x}{-\frac{1}{x^2}}\)
= \(\frac{1 \times 1+(1 \times 0)^2 \cdot 1}{-1}\)
= – 2.

Question 3.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}-x^2-2}{\sin ^2 x-x^2}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{x \cos x-\sin x}{x^2 \sin x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}-x^2-2}{\sin ^2 x-x^2}\)
[\(\frac{0}{0}\) form, using L’Hopital’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}-2 x}{\sin 2 x-2 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{2 \cos 2 x-2}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}}{-4 \sin 2 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}}{-8 \cos 2 x}\)
= \(\frac{1+1}{-8}=-\frac{1}{4}\).

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{x \cos x-\sin x}{x^2 \sin x}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Chapter Test 2

Question 4.
Evaluate the following limits:
(i) \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) log (1 – x) cot \(\frac{\pi}{2}\) x
(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (cosec² x – \(\frac{1}{x^2}\))
Solution:
(i) \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) log (1 – x) cot \(\frac{\pi}{2}\) x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\log (1-x)}{\tan \frac{\pi}{2} x}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\frac{1}{1-x}(-1)}{\frac{\pi}{2} \sec ^2 \frac{\pi x}{2}}\)
= \(-\frac{2}{\pi} \ {Lt}_{x \rightarrow 1^{-}} \frac{\cos ^2 \frac{\pi x}{2}}{1-x}\)
= \(\frac{2}{\pi} \ {Lt}_{x \rightarrow 1^{-}} \frac{2 \cos \frac{\pi x}{2}\left(-\sin \frac{\pi x}{2}\right) \frac{\pi}{2}}{-1}\)
= – \(\frac{2}{\pi}\) × 2 × 0 × 1 × \(\frac{\pi}{2}\) = 0.

(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (cosec² x – \(\frac{1}{x^2}\))

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Chapter Test 3

Question 5.
Evaluate the following limits:
(i) \(\underset{x \rightarrow 0}{\ {Lt}}\left(\frac{4}{x^2}-\frac{2}{1-\cos x}\right)\)
(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (x – sin^-1 x) cosec³ x
Solution:
(i) \(\underset{x \rightarrow 0}{\ {Lt}}\left(\frac{4}{x^2}-\frac{2}{1-\cos x}\right)\) (∞ – ∞ form)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Chapter Test 4

(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (x – sin^-1 x) cosec³ x (0 . ∞ form)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Chapter Test 5

Question 6.
Evaluate the following limits:
(i) \(\underset{x \rightarrow \frac{\pi}{2}}{\ {Lt}} \frac{\ {cosec} 6 x}{\ {cosec} 2 x}\)
(ii) \(\underset{x \rightarrow-\infty}{\mathbf{L t}}\) x² e^x
Solution:
(i) \(\underset{x \rightarrow \frac{\pi}{2}}{\ {Lt}} \frac{\ {cosec} 6 x}{\ {cosec} 2 x}\)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}} \frac{\sin 2 x}{\sin 6 x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}} \frac{2 \cos 2 x}{6 \cos 6 x}\)
= \(\frac{1}{3} \frac{\cos \pi}{\cos 3 \pi}=\frac{1}{3}\)

(ii) \(\underset{x \rightarrow-\infty}{\mathbf{L t}}\) x² e^x
= \(\ {Lt}_{x \rightarrow-\infty} \frac{x^2}{e^{-x}}\left(\frac{\infty}{\infty} \text { form }\right)\)
using L’Hopital’s rule
= \(\ {Lt}_{x \rightarrow-\infty} \frac{2 x}{-e^{-x}} \quad\left(\frac{\infty}{\infty} \text { form }\right)\)
= \(\ {Lt}_{x \rightarrow-\infty} \frac{2}{+e^{-x}}=\frac{2}{\infty} \rightarrow 0\)

Question 7.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 1}(2-x)^{\tan \frac{\pi x}{2}}\)
(ii) \(\ {Lt}_{x \rightarrow 1} x^{\frac{1}{x-1}}\)
Solution:
(i) Let y = \(\ {Lt}_{x \rightarrow 1}(2-x)^{\tan \frac{\pi x}{2}}\)
⇒ log y = tan \(\frac{\pi x}{2}\) log (2 – x)
∴ \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) log y = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) tan \(\frac{\pi x}{2}\) log (2 – x) [0 . ∞ form]
= \(\ {Lt}_{x \rightarrow 1} \frac{\log (2-x)}{\cot \frac{\pi x}{2}}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 1} \frac{2 \sin ^2 \frac{\pi x}{2}}{\pi(2-x)}\)
= \(\frac{2}{\pi} \frac{(1)^2}{2-1}=\frac{2}{\pi}\)
log (\(\underset{x \rightarrow 1}{\mathrm{Lt}}\) y) = \(\frac{2}{\pi}\)
\(\underset{x \rightarrow 1}{\mathrm{Lt}}\) y = e^2/π
∴ \((2-x)^{\tan \frac{\pi x}{2}}\) = e^2/π

(ii) Let y = x^{\(\frac{1}{x-1}\)}
⇒ log y = \(\frac{\log x}{x-1}\)
∴ \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) log y = \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) \(\frac{\log x}{x-1}\)
(\(\frac{0}{0}\) form)
using L’Hopital’s rule
= \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) \(\frac{\frac{1}{x}}{1}=\frac{1}{1}\) = 1
∴ log (\(\underset{x \rightarrow 1}{\mathrm{Lt}}\) y) = 1
⇒ \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) y = e’ = e
Thus, \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) x^{\(\frac{1}{x-1}\)} = e.

Question 8.
If f(x) = \(\left\{\begin{array}{cc}
\frac{\log (3+x)-\log (3-x)}{x} & , x \neq 0 \\
k & , x=0
\end{array}\right.\), find the value of k so that the function f may be continuous at x = 0?
Solution:
\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\log (3+x)-\log (3-x)}{x}\)
(\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{\frac{1}{3+x}-\frac{1}{3-x}(-1)}{1}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{3-x+3+x}{(3+x)(3-x)}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{6}{9-x^2}=\frac{6}{9}=\frac{2}{3}\)
Also f(0) = k
Since the function f be continuous at x = 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
⇒ \(\frac{2}{3}\) = k.

Category: ML Aggarwal Solutions

ISC Mathematics Class 12 Solutions – ML Aggarwal Class 12 ISC Solutions

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Chapter Test

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.7

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.6

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.3

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.2

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms MCQs

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Chapter Test