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) = x<sup>4</sup> – \(\frac{x^3}{3}\)
Solution:
Given f(x) = x<sup>4</sup> – \(\frac{x^3}{3}\)
∴ 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 e2x is strictly increasing on R.
Solution:
Given f (x) = e22x
∴ f (x) = 2e2x > 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, \(\frac { π }{ 6 }\) ]