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

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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.

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.

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 \(\frac{x^2}{4 \pi^2}+\left(\frac{25-x}{4}\right)^2\)

∴ 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.

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;

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)

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:

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.