OP Malhotra Class 12 Maths Solutions Chapter 11 Applications of Derivatives Ex 11(d)

Effective OP Malhotra Maths Class 12 Solutions Chapter 11 Applications of Derivatives Ex 11(d) can help bridge the gap between theory and application.

S Chand Class 12 ICSE Maths Solutions Chapter 11 Applications of Derivatives Ex 11(d)

Question 1.
Show that the function f(x) = \(\frac { 3 }{ x }\) + 7 is decreasing for x ∈ R (x ≠ 0).
Solution:
Given f(x) = \(\frac { 3 }{ x }\) + 7
∴ f (x) = – \(\frac { 3 }{ x² }\) < 0 ∀ x ∈ R, x ≠ 0 [∵ x² > 0 ∀ x ∈ R]
Thus f(x) is decreasing for all x ∈ R, x ≠ 0

Question 2.
Show that the function x + \(\frac { 1 }{ x }\), x ≥ 1 is increasing.
Solution:
Let f(x) = x + \(\frac { 1 }{ x }\)
∴ f’ (x) = 1 – \(\frac { 1 }{ x² }\)
for x ≥ 1 ⇒ x² > 1
⇒ \(\frac { 1 }{ x² }\) ≤ 1 ⇒ – \(\frac { 1 }{ x² }\) ≥ – 1
⇒ 1 – \(\frac { 1 }{ x² }\) ≥ 1 – 1 = 0
⇒ f'(x) > 0
Thus f(x) is descreasing for all x > 1.

Question 3.
State when a function is said to be increasing function of [a, b]. Test whether the function f (x) = x³ – 8 is increasing on [1, 2].
Solution:
If f'(x) > 0 at each point x∈(a, b)
Thus f is increasing on [a, b]
given f(x) = x³ – 8
∴ f(x) = 3x²
since x ∈ (1, 2)
⇒ 1 < x < 2 ⇒ 1< x² < 4
⇒ 3 < 3x² < 12
⇒ 3 < f'(x) < 12 Thus f(x) > 0 ∀ x ∈ (1, 2)
Therefore f(x) is increasing on [1, 2]

Question 4.
Prove that the function
f(x) = x³ – 6x² + 12x – 18 is increasing on R
Solution:
Let f(x) = x³ – 6x² + 12x – 18
∴ f(x) = 3x² – 12x + 12
= 3(x² – 4x + 4)
= 3(x – 2)² > 0 ∀ x ∈ R
Thus f(x) is increasing on R.

Question 5.
Determine the values of x for which f (x) = \(\frac{x-2}{x+1}, x \neq-1\) is increasing or decreasing.
Solution:
Given f(x) = \(\frac{x-2}{x+1}\) ; x ≠ -1
∴ f'(x) = \(\frac{(x+1) \times 1-(x-2) \times 1}{(x+1)^2}\)
= \(\frac{3}{(x+1)^2}\) > 0 ∀ x ∈ R ; x ≠ -1
Thus f (x) is increasing on R – (-1}

Question 6.
For which values of x, the function x
f(x) = \(\frac{x}{x^2+1}\) is increasing and for which values of x, it is decreasing?
Solution:
Given f (x) = \(\frac{x}{x^2+1}\) …(1)
Diff. both sides w.r.t. x ; we have

Thus, the function f (x) is decreasing in (- ∞, – 1] ∪ [1, ∞).
Let (x₁, y₁) be any point on given curve

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 7.
Show that f(x) = sin x is an increasing function on {-π/2, π/2).
Solution:
Given f (x) = sin x
∴ f’ (x) = cos x
Now for – \(\frac { π }{ 2 }\) < x < \(\frac { π }{ 2 }\) ⇒ cos x > 0 ⇒ f'(x) > 0
Thus f (x) is increasing on (- \(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\)).

Question 8.
Show that f (x) = cos x is a decreasing function on (0, π).
Solution:
Given f (x) = cos x
f’ (x) = – sin x
For x ∈ (0, π) ⇒ 0 < x < π ⇒ sin x > 0
⇒ – sin x < 0 ⇒ f’ (x) < 0
Hence f(x) is decreasing function on (0, π).
For x ∈ (- π, 0) ⇒ – π < x < 0 ⇒ sin x < 0 ⇒ – sin x > 0
⇒ f’ (x) > 0
Hence f(x) is increasing function on (- π, 0)
Thus f (x) is neither increasing nor decreasing on (- π, π).

Question 9.
Show that f(x) = tan x is an increasing function on {-π/2, π/2).
Solution:
Given, f (x) = tan x
∴ f'(x) = sec²x
when x ∈ (- \(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\)) ⇒ sec² x > 0
⇒ f'(x) > 0
∴ f(x) is strictly increasing on (-\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))

Find the intervals in which the following functions are increasing or decreasing.

Question 10.
f (x) = 2x³ – 9x² + 12x + 15
Solution:
Given f (x) = 2x³ – 9x² + 12x + 15
∴ 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
∴ (x – 1) (x – 2) > 0
⇒ (x – 1) (x – 2) > 0
⇒ x > 2 or x < 1
⇒ x ∈ (- ∞, 1) ∪ (2, ∞)

signs of f'(x) for different values of x Thus, f(x) is increasing on (- ∞, 1) ∪ (2, ∞)
For f (x) to be decreasing we must have f'(x) < 0
6 (x – 1) (x – 2) < 0
⇒ (x – 1) (x – 2) < 0
⇒ 1 < x < 2 ⇒ x ∈ (1, 2) Thus, f(x) is decreasing on (1, 2).

Question 11.
f(x) = 2x³ – 15x² + 36x + 1
Solution:
Given f (x) = 2x³ – 9x² + 12x – 5
∴ 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
⇒ x > 2 or x < 1
⇒ X ∈ (- ∞, 1) u (2, ∞)

signs off’ (x) for different values of x
Thus, f (x) is increasing on (- ∞, 1) ∪ (2, ∞)
For f (x) to be decreasing we must have f’ (x) < 0
6 (x – 1) (x – 2) < 0
⇒ (x – 1) (x – 2) < 0
⇒ 1 < x < 2
⇒ x ∈ (1, 2)
Thus, f(x) is decreasing on (1, 2).

Question 12.
f(x) = 6 + 12x – 3x² – 2x³
Solution:
Given f(x) = 6 + 12x – 3x² – 2x³
f'(x) = 12 – 6x – 6x²
= – 6(x² + x – 2)
= – 6 (x – 1) (x + 2)
For f(x) is to be increasing iff f'(x) > 0

⇒ – 6 (x – 1) (x + 2) > 0
⇒ (x – 1) (x + 2) < 0
⇒ x ∈ (-2, 1)
Thus f(x) is increasing in (-2, 1).
For f(x) is to be decreasing iff f'(x) < 0
⇒ – 6 (x – 1) (x + 2) < 0
⇒ (x – 1) (x + 2) > 0
⇒ x ∈ (- ∞, – 2) ∪ (1, ∞)
Thus, f(x) is decreasing in (- ∞, – 2) ∪ (1, ∞)

Question 13.
f(x) = x⁴ – \(\frac{x^3}{3}\)
Solution:
Given f(x) = x⁴ – \(\frac{x^3}{3}\)
∴ f'(x) = 4x³ – x²
= x²(4x – 1) = 4x²(x – \(\frac { 1 }{ 4 }\))
For f(x) is to be increasing iff f'(x) > 0
i.e., 4x²(x – \(\frac { 1 }{ 4 }\)) > 0
⇒ x – \(\frac { 1 }{ 4 }\) > o (∵ x² ≥ 0 [x ∈ 0)
⇒ x > \(\frac { 1 }{ 4 }\)
Thus f(x) is increasing in (\(\frac { 1 }{ 4 }\), ∞).
For f(x) is to be decreasing iff f(x) < 0
i.e. 4x²(x – \(\frac { 1 }{ 4 }\)) < 0
⇒ x – \(\frac { 1 }{ 4 }\) < 0
⇒ x < \(\frac { 1 }{ 4 }\)
Thus f(x) is decreasing in (- ∞, \(\frac { 1 }{ 4 }\)).

Question 14.
f(x) = 20 – 9x + 6x² – x³
Solution:
Given f(x) = 20 – 9x + 6x² – x³
∴ f'(x) = -9 + 12x – 3x²
= – 3(x² – 4x + 3)
= – 3(x – 1) (x – 3)
Since critical points are given by putting f'(x) = 0 ⇒ x = 1, 3
For f (x) is to be increasing iff f’(x) > 0

i.e. – 3(x – 1) (x – 3) > 0
⇒ (x – 1) (x – 3) < 0
⇒ X ∈ (1, 3)
Thus f(x) is increasing in (1, 3).
∴ f(x) is decreasing iff f'(x) < 0
⇒ – 3(x – 1) (x – 3) < 0
⇒ (x – 1) (x – 3) > 0
⇒ x ∈ ( ∞, 1) ∪ (3, ∞)
Thus f(x) is decreasing in (-∞, 1) ∪ (3, ∞).

Question 15.
f(x) = \(\frac{3}{10} x^4-\frac{4}{5} x^3-3 x^2+\frac{36}{5} x\) + 11.
Solution:
f (x) = \(\frac{3}{10} x^4-\frac{4}{5} x^3-3 x^2+\frac{36}{5} x\) + 11
∴ f'(x) = \(\frac{6}{5} x^3-\frac{12}{5} x^2-6 x+\frac{36}{5}\)
= \(\frac { 6 }{ 5 }\) [x³ – 2x² – 5x + 6]
= \(\frac { 6 }{ 5 }\)(x – 1)(x² – x – 6)
⇒ f'(x) = \(\frac { 6 }{ 5 }\)(x – 1) (x + 2) (x – 3)
For f'(x) = 0 ⇒ x = 1, – 2, 3
Hence we shall discuss four cases:
Case I: When x < – 2
∴ f'(x) = (-ve) (-ve) (-ve) = -ve
∴ f'(x) is strictly decreasing in (-∞, -2).

Case II: when -2 < x < 1
∴ f'(x) = (-ve) (+ve) (-ve) = +ve
∴ f (x) is strictly increasing in (-2, 1).

Case III: When 1 < x < 3
∴ f'(x) = (+ve) (+ve) (-ve) = -ve
∴ f (x) is strictly decreasing in (1, 3).

Case IV: when x > 3
∴ f'(x) = (+ve) (+ve) (+ve) = +ve
∴ f (x) is strictly increasing in (3, ∞).
on combining all four cases, f(x) is strictly
increasing in (-2, 1) ∪ (3, ∞) and strictly
decreasing in (-∞, -2) ∪ (1, 3).

Question 16.
f(x) = sin 3x – cos 3x, 0 < x < π
Solution:
Given f(x) = sin 3x – cos 3x, 0 < x < π

Thus f(x) is decreasing in (\(\frac { π }{ 4 }\), \(\frac { 7π }{ 12 }\)) ∪ (\(\frac { 11π }{ 12 }\), π).

Question 17.
Determine the values of x for which the function f(x) = x² – 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x² – 6x + 9, where the normal is parallel to the line y = x + 5.
Solution:
Given f (x) = x² – 6x + 9
∴ f’ (x) = 2x – 6
Now f(x) is to be increasing so
We must have f’ (x) > 0
∴ 2x – 6 > 0 ⇒ x > 3 ⇒ x ∈ (3, ∞),
Thus f(x) is increasing on (3, ∞).
Now f (x) is to be decreasing so we must have f’ (x) < 0
∴ 2x – 6 < 0 ⇒ x < 3
⇒ x ∈ (- ∞, 3)
Thus, f(x) is decreasing on (- ∞, 3).
given eqn. of curve be y = x² – 6x + 9 …(1)
Let the required point on eqn. (1) be (x₁, y₁)
Then y₁ = x₁² – 6x₁ + 9 …(2)
∴ \(\frac{d y}{d x}=2 x-6 \Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\) = 2x₁ – 6
Since the normal is parallel to given line x – y + 5 = 0
∴ slope of normal at (x₁, y₁) = slope of line x – y + 5 = 0

Thus the coordinates of required point be (\(\frac { 5 }{ 2 }\), \(\frac { 1 }{ 4 }\)).

Question 18.
Determine the intervals in which the function f (x) = (x – 1) (x + 2)² is increasing or decreasing. At what points are the tangents to the graph of the function parallel to the x-axis ?
Solution:
Given f(x) = (x – 1) (x + 2)²
∴ f'(x) = (x – 1) 2(x + 2) + (x + 2)² x 1
= (x + 2) (2x – 2 + x + 2) = 3x(x + 2)
The critical points are given by f(x) = 0
⇒ x = 0, – 2
For f(x) is to be increasing iff f'(x) > 0

i.e. 3x (x + 2) > 0
⇒ x(x + 2) < 0
⇒ x ∈(-2, 0)
Thus f(x) is decreasing (- 2, 0)
Since the tangent is parallel to x-axis
∴ slope of tangent to curve = 0
⇒ \(\left(\frac{d y}{d x}\right)_{(x, y)}\) = 0
⇒ 3x(x + 2) = 0
⇒ x = 0, – 2
When x = 0
∴ y = f(x) = (0 – 1) (0 + 2)² = – 4
When x = – 2
∴ y = f(x) = (- 2 – 1) (- 2 + 2)² = 0
Thus, required points are (0, – 4) & (- 2, 0).

Question 19.
Find the intervals in which the function f(x) = \(\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\), 0 ≤ x ≤ 2π is
(i) increasing
(ii) decreasing.
Solution:
Given f(x) = \(\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\)

Question 20.
Find the intervals in which the following function is increasing or decreasing :
f(x) = log(1 + x) – \(\frac { x }{ 1 + x }\)
Solution:

Examples

Question 1.
The equation of the tangent to the curve y = x² at (0, 0) is parallel to ………….
Solution:
Given eqn. of curve be
y = x²
∴ \(\frac { dy }{ dx }\) = 2x
Thus, slope of tangent to given curve at
(0, 0) = \(\left(\frac{d y}{d x}\right)_{(0,0)}\) = 0
Thus required eqn. of tangent to given curve at (0, 0) be given by
y – 0 = 0 (x – 0)
⇒ y = 0 i.e. parallel to x-axis.

Question 2.
The equation of the normal to the curve y = tan x at (0, 0) is ……………
Solution:
Given eqn. of curve be y = tan x
∴ \(\frac { dy }{ dx }\) = sec² x
Thus slope of normal to given curve at (0,0)
= \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(0,0)}}=\frac{-1}{\sec ^2 \theta}=\frac{-1}{1}\) = – 1
Hence the eqn. of normal to given curve at (0, 0) be given by y – 0 = – 1 (x – 0)
⇒ x + y = 0

Question 3.
If the side of a cube is increased by 0.1%, then the corresponding increase in the volume of the cube is ……………
Solution:
Let x be the side of cube s.t \(\frac { δx }{ x }\) x 100 = 0.1
Then volume of cube = V = x³
⇒ δV = 3x² δx
⇒ \(\frac{\delta \mathrm{V}}{\mathrm{V}}=\frac{3 x^2 \delta x}{x^3}=\frac{3}{x} \delta x\)
⇒ \(\frac{\delta V}{V} \times 100=3\left(\frac{\delta x}{x} \times 100\right)\)
= 3 x 0.1 = 0.3
∴ Required increase in volume of cube = 0.3%

Question 4.
If the radius of a circular plate is increasing at the rate of 0.01 cm/sec, when the radius is 12 cm, then the rate at which the area increases is …………….
Solution:
Let r be the radius of circular plate
then A = area of circular plate = πr²
∴ \(\frac{d \mathrm{~A}}{d t}=2 \pi r \frac{d r}{d t}\) … (1)
it is given that, \(\frac { dr }{ dt }\) = 0.01 cm/sec
and r = 12 cm
∴ \(\frac { dA }{ dt }\) = (2π x 12 x 0.01)cm²/sec
= 0.24 n cm²/sec

Question 5.
A point on the parallel y² = 19x at which the ordinate increases at twice the rate of the abscissa is ……………..
Solution:
eqn. of given parabola be y² = 18x …(1)
Let (x, y) be any point on eqn. (1).
according to given condition, y = 2x
⇒ \(\frac { dy }{ dt }\) = 2\(\frac { dx }{ dt }\) … (2)
Diff. eqn. (1) both sides w.r.t. t,
2y\(\frac { dy }{ dt }\) = 18\(\frac { dx }{ dt }\) … (3)
From eqn. (2) and (3) ; we have
⇒ 4y\(\frac { dx }{ dt }\) = 18\(\frac { dx }{ dt }\) ⇒ y = \(\frac { 9 }{ 2 }\)
∴ from (1); (\(\frac { 9 }{ 2 }\))² = 18x ⇒ x = \(\frac { 81 }{ 72 }\) = \(\frac { 9 }{ 2 }\)
Thus any point on given curve (1) be (\(\frac { 9 }{ 8 }\), \(\frac { 9 }{ 2 }\)).

Question 6.
An edge of a variable cube is increasing at the rate of 10 cmfs. How fast the volume of the cube will increase when the edge is 5 cm long ? …………..
Solution:
Let x be the length of edge of cube at any time t.
Then V = volume of cube = x³
∴ \(\frac { dV }{ dt }\) = 3x² \(\frac { dx }{ dt }\)
given \(\frac { dx }{ dt }\) = 10 cm/s ; x = 5 cm
Thus, \(\frac { dV }{ dt }\) = 3 x 5² x 10
= 750 cm³/sec dt

Question 7.
Let f : [a, b] → R be a continuous function on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that f’ (c) = ……………
Solution:
f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
[using Lagranges mean value Theorem]

Question 8.
The function x³ – 3x² + 3x – 100 is always ………………. on R.
Solution:
Let f(x) = x³ – 3x² + 3x – 100
∴ f (x) = 3x² – 3x + 3
= 3 (x² – 2x + 1) = 3 (x – 1)² ≥ 0
Thus f (x) is increasing on R.

Question 9.
The interval in which the function
f(x) = – 2x² – 8x is decreasing is ……………
Solution:
Given f(x) = – 2x² – 8x
∴ f’ (x) = – 4x – 8
Now f’ (x) ≥ 0
⇒ – 4x – 8 ≥ 0
⇒ – 4x ≥ 8
⇒ x ≤ – 2
and f’ (x) ≤ 0 ⇒ – 4x – 8 ≤ 0
⇒ – 4x ≤ 8
⇒ x ≥ – 2
Thus, f (x) is decreasing on [- 2, ∞).
[∵ f (x) is decreasing iff f’ (x) < 2]

Question 10.
The value of c is Rolle’s theorem for the function f(x) = x² + 2x – 8, x∈ [- 4, 2] is …………..
Solution:
(i) f (x) is a polynomial function so continuous everywhere.
∴ f(x) is continuous in [- 4, 2],

(ii) f'(x) = 2x + 2 which exists ∀x∈R
Thus f(x) is differentiable in (- 4, 2).

(iii) f(- 4) = (- 4)² + 2 (- 4) – 8 = 0 ;
f (2) = 2² + 2 x 2 – 8 = 0
∴ f(- 4) = f(2)
Thus all the three conditions of rolle’s theorem are satisfied so ∃ atleast one real
number c ∈ (- 4, 2)
s.t f’ (c) = 0 ⇒ 2c + 2 = 0
⇒ c = – 1 ∈ (- 4, 2).

Question 11.
The equation of the tangent to the curve y – x³ – 6x + 5 at (2,1) is
(a) 6x – y – 11 = 0
(b) 6x – y – 13 = 0
(c) 6x + y + 11 = 0
(d) 6x – y + 11 = 0
(e) x – 6y – 11 = 0
Solution:
Given eqn. of curve be
y = x³ – 6x + 5 … (1)
∴ \(\frac { dy }{ dx }\) = 3x² – 6
Thus, slope of tangent to curve at (2, 1)
= \(\left(\frac{d y}{d x}\right)_{(2,1)}\) = 3 x 2² – 6 = 6
∴ required eqn. of tangent to curve (1) at point (2, 1) be given by
y – 1 = 6(x – 2)
⇒ 6x – y – 11 = 0

Question 12.
Slope of the normal to the curve
y = x² – \(\frac { 1 }{ x² }\) at (- 1, 0) is
(a) – \(\frac { 1 }{ 4 }\)
(b) – 4
(c) \(\frac { 1 }{ 4 }\)
(d) 4
Solution:
Given eqn. of curve be,

Question 13.
The rate of change of volume of a sphere with respect to its surface area when the radius is 4 cm is
(a) 2 cm³/cm²
(b) 4 cm³/cm²
(c) 8 cm³/cm²
(d) 6 cm³/cm²
Solution:
Let V = volume of sphere = \(\frac { 4 }{ 3 }\)πr³
∴ \(\frac { dV }{ dr }\) = 4πr²
and S = surface area of sphere = 4 πr²

Question 14.
A stone is dropped into a quiet lake and waves move is circles at the speed of 5 cm/s. At that instant, when the radius of circular wave is 8 cm, how fast is the enclosed area increasing?
(a) 6n cm²/s
(b) 8π cm²/s
(c) \(\frac { 8 }{ 3 }\) cm²/s
(d) 80π cm²/s
Solution:
Let r be the radius of circular wave.
Then area of circular wave A = πr²
⇒ \(\frac { dA }{ dt }\) = 2πr\(\frac { dr }{ dt }\)
Given \(\frac { dr }{ dt }\) = 5 cm/s ; r = 8 cm
∴ \(\frac { dr }{ dt }\) = 2π x 8 x 5 = 80π cm²/sec

Question 15.
The function f(x) = x² + 2x – 5 is strictly increasing in the interval.
(a) [- 1, ∞)
(b) (- ∞, – 1)
(c) (- ∞, – 1]
(d) (- 1, ∞)
Solution:
Given f(x) = x² + 2x – 5
∴ f'(x) = 2x + 2
Now f'(x) > 0 ⇒ 2x + 2 > 0 ⇒ x > – 1
Thus, f(x) is strictly increasing in (- 1, ∞)

Question 16.
If y = 8x³ – 60x² + 144x + 27 is strictly decreasing function in the interval.
(a) (- 5, 6)
(b) (- oo, 2)
(c) (5, 6)
(d) (- 1, ∞)
Solution:
Given y = 8x³ – 60x² + 144x + 27
∴ \(\frac { dy }{ dx }\) = 24x² – 120x +144 dx
= 24 (x² – 5x + 6)
Now y is decreasing iff \(\frac { dy }{ dx }\) < 0
iff x² -5x + 6 < 0
iff (x – 2) (x – 3) < 0
iff 2 < x < 3
Therefore, f(x) i.e. y is decreasing in (2, 3)

Question 17.
The approximate value of \(\sqrt[5]{33}\) correct to 4 decimal places is
(a) 2.0000
(b) 2.1001
(c) 2.0125
(d) 2.0500
Solution:

Question 18.
The rate of change of area with respect to its radius at r = 2 cm is
(a) 4
(b) 2%
(c) 2
(d) 4 n
Solution:
Let A = area of circle = πr²
∴ \(\frac { dA }{ dr }\) = 2πr at r = 2; \(\frac { dA }{ dr }\) = 4π

Question 19.
The point on the curve 6y = x³ + 2 at which ordinate is changing 8 times as fast as abscissa is
(a) (4, 11)
(b) (4, – 11)
(c) (- 4,11)
(d) (- 4, – 1)
Solution:
(i) Given eqn. of curve be,
6y = x³ + 2 … (1)
Let P (x, y) be any point on curve (1). according to given condition ; we have
\(\frac { dy }{ dt }\) = 8\(\frac { dx }{ dt }\) … (2)
diff. eqn. (1) w.r.t. x ; we have
6\(\frac { dy }{ dt }\) = 3x² \(\frac { dx }{ dt }\)
⇒ 48 \(\frac { dx }{ dt }\) = 3x²\(\frac { dx }{ dt }\) [using (2)]
⇒ x² = 16
⇒ x = ± 4
When x = 4 ∴ from (1); 6y = 64 + 2 ⇒ y = 11
When x = – 4 ∴ from (1); 6y = – 64 + 2 ⇒ y = – \(\frac { 31 }{ 3 }\)
∴ required point on given curve be (4, 11)

Question 20.
The volume of a sphere is increasing at the rate of 1200 cu. cmfsec. The rate of increase in its surface area when the radius is 10 cm is
(a) 120 sq cm/sec
(b) 240 sq cm/sec
(c) 200 sq cm/sec
(d) 100 sq cm/sec
Solution:
Let r cm be the radius of sphere at any instant t

Question 21.
An edge of a variable cube is increasing at the rate of 10 cm/sec. How fast the volume of the cube will increase when the edge is 5 cm long ?
(a) 750 cm³/sec
(b) 75 cm³/sec
(c) 300 cm³/sec
(d) 150 cm³/sec
Solution:
Let x cm be the length of edge of cube
Then V = volume of cube = x³
∴ \(\frac { dV }{ dt }\) = 3x²\(\frac { dx }{ dt }\) … (1)
given \(\frac { dx }{ dt }\) = 10 cm/sec ; x = 5 cm
∴ from (1); \(\frac { dV }{ dt }\) = 3 x 10 x 5²
= 750 cm³/sec

Question 22.
A ladder 5 m long is learning against a wall The bottom of the ladder is dragged from the wall along the ground at the rate of 2 m/sec. How fast is the height of the wall decreasing when the foot of the ladder is 4 m away from the wall ?
(a) \(\frac { 3 }{ 8 }\)m/s
(b) \(\frac { 8 }{ 3 }\)m/s
(c) \(\frac { 5 }{ 3 }\)m/s
(d) \(\frac { 2 }{ 3 }\)m/s
Solution:

Question 23.
If the distance s covered by a particle in time t is proportional to the cube root of its velocity, then the acceleration is
(a) a constant
(b) ∝ s³
(c) ∝ \(\frac { 1 }{ s³ }\)
(d) ∝ s⁵
Solution:
According to given condition ;

Question 24.
The function f (x) = cos² x is strictly decreasing on
(a) [0, \(\frac { π }{ 2 }\)]
(b) [0, \(\frac { π }{ 2 }\))
(c) (0, \(\frac { π }{ 2 }\))
(d) (0, \(\frac { π }{ 2 }\)]
Solution:
Given f (x) = cos² x
∴ f’ (x) = – 2 cos x sin x = – sin 2x
f(x) is strictly decreasing if f’ (x) < 0
iff – sin 2x < 0 ; iff sin 2x > 0
iff 2x ∈ (0, π) ; iff 2x ∈ (0, \(\frac { π }{ 2 }\))

Question 25.
For the function f(x) = x + \(\frac { 1 }{ x }\), x ∈[1, 3] the value of c for mean value theorem is
(A) 1
(B) \(\sqrt{3}\)
(C) 2
(D) none of these
Solution:
Given f(x) = x + \(\frac { 1 }{ x }\), x ∈ [1, 3]
since f(x) has a unique value for each x ∈ [1, 3]
Thus f is continuous in [1, 3].
Now f'(x) = 1 – \(\frac { 1 }{ x² }\) exists ∀ x ∈ (-1, 3)
Thus f be diferentiable in (1, 3).
Then by L.M.V theorem ∃ atleast one real number c ∈ (1, 3)

Question 26.
Find the slope of the tangent to the curve y = 2 sin² 3x at x = \(\frac { π }{ 6 }\).
Solution:
Eqn. of given curve be, y = 2 sin² 3x

Question 27.
Find the slope of the normal to the curve y = x³ – x at the point (2, 6).
Solution:
Given eqn. of curve be,

Question 28.
Find the rate of change of the area of a circle with respect to its radius r, where r = 3 cm.
Solution:
Let r cm be the radius of circle.
Then A = area of circle = πr²
∴ \(\frac { dA }{ dr }\) = 2πr at r = 3 ; \(\frac { dA }{ dr }\) = 6π cm²

Question 29.
Find the equation of the normal to the curve y = tan x at (0, 0)?
Solution:
Given eqn. of curve be y = tan x

Hence the eqn. of normal to given curve at (0, 0) be given by y – 0 = – 1 (x – 0)
⇒ x + y = 0

Question 30.
Find the value of k, if the tangent to the curve y² + 3x – 7 = 0 at the point (h, k) is parallel to the line x – y = 4.
Solution:
eqn. of given curve be, y² + 2x – 7 = 0
diff. both sides w.r.t. x ; we have

Since it is given that, slope of tangent to given curve is parallel to given line.

Question 31.
Find the curve y = 5x – 2x³, find the rate at which the slope changes at x = 3 if x changes at the rate of 2 units/sec.
Solution:
Given eqn. of curve be, y = 5x – 2x³

Question 32.
The radius r of a right circular cylinder is increasing at the rate of 5 cmfmin and its height h is decreasing at the rate of 4 cmfmin, when r = 8 cm and h – 6 cm. Find the rate of change of volume of the cylinder.
Solution:
Let r be the radius of right circular cylinder at any time t.
Then V = volume of cylinder = πr²h
∴ \(\frac{d \mathrm{~V}}{d t}=\pi\left[r^2 \frac{d h}{d t}+2 r h \frac{d r}{d t}\right]\) … (1)
Given \(\frac {dr}{dt}\) = 5 cm/min
\(\frac {dh}{dt}\) = – 4 cm/min ; r = 8 cm ; h = 6 cm
∴ from (1) ; we have
\(\frac {dV}{dt}\) = π [8² (- 4) + 2 x 8 x 6 x 5]
= π[ – 256 + 480]
= 224 π cm³/sec

Question 33.
A balloon which always remain spherical has a variable diameter \(\frac { 2 }{ 3 }\) (3x + 1). Find the rate of change of its volume with respect to x.
Solution:
Let r be the radius of spherical balloon at any instant.

Question 34.
Prove that the function e^2x is strictly increasing on R.
Solution:
Given f (x) = e2^2x
∴ f (x) = 2e^2x > 0 ∀ x ∈ R
Thus, f (x) is strictly increasing on R.

Question 35.
The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of increase of its circumference.
Solution:
Let r be the radius of circle at any instant t.
Then circumference of circle = C = 2 π r
∴ \(\frac { dC }{ dt }\) = 2π\(\frac { dr }{ dt }\) = 2 π x 0.7 = 1.4 cm/sec

Question 36.
Find the intervals in which the function
sin 3x, 0 ≤ x \(\frac { π }{ 2 }\) is increasing.
Solution:
Given f(x) = sin 3x
∴ f’ (x) = 3 cos 3x
Now f (x) is increasing iff f’ (x) ≥ 0
iff 3 cos 3x ≥ 0
iff 3x ∈ [0, \(\frac { π }{ 2 }\) ] [given 0 ≤ x ≤ \(\frac { π }{ 2 }\) ]
iff x ∈ [0, \(\frac { π }{ 6 }\) ]