Mahesh

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(h) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(h)

I.

Question 1.
Find the points of local extrema (if any) and local extrema of the following functions each of whose domain is shown against the function.
i) f(x) – x², ∀ x ∈ R.
Solution:
f(x) = x²
f'(x) = 2x ⇒ f”(x) = 2
For maximum or minimum f'(x) = 0
2x = 0
x = 0
Now f”(x) = 2 > 0 ,
∴ f(x) has minimum at x = 0
Point of local minimum x = 0
Local minimum = 0.

ii) f(x) = sin x, [0, 4π)
Solution:
Given f(x) = sinx
⇒ f'(x) = cosx
⇒ f”(x) = -sinx
For max on min,
f'(x) = 0
cos x = 0
⇒ x = \(\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}\)

i) f”(\(\frac{\pi}{2}\)) = -sin\(\frac{\pi}{2}\) = -1 < 0
f(x) = sin \(\frac{\pi}{2}\) = 1
∴ Point of local maximum x = \(\frac{\pi}{2}\)
local maximum – 1

ii) f”(\(\frac{3\pi}{2}\)) = – sin \(\frac{3\pi}{2}\) = -1 > 0
f(x) = sin \(\frac{3\pi}{2}\) = -1
∴ Point of local minimum x = \(\frac{3\pi}{2}\)
local minimum x = -1

iii) f”(\(\frac{5\pi}{2}\)) = – sin \(\frac{5\pi}{2}\) = -1 > 0
f(x) = sin \(\frac{5\pi}{2}\) = 1
∴ Point of local maximum x = \(\frac{5\pi}{2}\)
local maximum = 1

iv) f”(\(\frac{7\pi}{2}\)) = – sin \(\frac{7\pi}{2}\) = -1 > 0
f(x) = sin \(\frac{7\pi}{2}\) = -1
∴ Point of local maximum x = \(\frac{7\pi}{2}\)
local maximum = 1

iii) f (x) = x³ – 6x² + 9x + 15 ∀ x ∈ R.
Solution:
f (x) = 3x² – 12x + 9 and f'(x) = 6x – 12
For maximum or minimum f(x) = 0
⇒ 3x² – 12x + 9 = 0
⇒ x² – 4x + 3 = 0
⇒ (x – 1) (x – 3) = 0
⇒ x = 1 or 3
Now f”(1) = 6(1) – 12 = – 6 < 0
∴ f(x) has maximum value at x = 1
Max. valueis f(1)= 1³ – 6(1)² + 9(1) + 15
= 1 – 6 + 9 + 15 = 19

f”(3) = 6(3) – 12 = 18 – 12 = 6 > 0
∴ f(x) has minimum value at x = 3
Min. value is f(3) = 33 – 6.32 + 9.3 + 15
= 27 – 54 + 27+ 15
= 15

iv) f(x) = \(\sqrt{1-x}\) ∀ x ∈ (0, 1)
Solution:

For max. or min. f'(x) = 0

v) f(x) = 1/x² + 2 ∀ x ∈ R
Solution:

∴ For maximum or minimum f (x) = 0
⇒ \(\frac{-2x}{(x^{2}+2)^{2}}\) = 0 ⇒ x = 0
f”(0) = \(\frac{2(0-2)}{(0+2^{3})}=\frac{-4}{8}=\frac{-1}{2}\) < 0
∴ f(x) has max. value at x = 0
Max. value is f(0) = \(\frac{1}{0+2}=\frac{1}{2}\)

vi) f(x) = x²- 3x ∀ x ∈ R
Solution:
f(x) = 3x² – 3 and f”(x) = 6x
∴ For maximum or minimum f(x) = 0
⇒ 3x² – 3 = 0
⇒ x² – 1 = 0
⇒ x = ± 1
Now f”(1) = 6(1) = 6 > 0
∴ f(x) has minimum at x = 1
Minimum value is f(1) = 1³ – 3(1) = -2
f”(-1) = 6(-1) = -6 < 0
∴ f(x) has maximum value at x = -1
Maximum value is f(-1) = (-1)³ – 3(-1)
= -1 + 3 = 2

vii) f(x) = (x -1) (x + 2)² ∀ x ∈ R
Solution:
f(x) = (x – 1) (x + 2)²
f(x) = (x – 1) 2(x + 2) + (x + 2)²
= 2(x – 1) (x + 2) + (x + 2)²
f”(x) = 2(x – 1) + 2(x + 2) + 2(x + 2)
= 2(3x + 3) = 6(x + 1)
∴ For maximum or minimum f'(x) = 0
2(x – 1) (x + 2) + (x + 2)² = 0
(x + 2) [2(x – 1) + (x + 2)] = 0
⇒ (x + 2) (3x) = 0
⇒ x = 0, x = -2
Now f”(0) = 6(0 + 1) = 6 > 0
∴ f(x) has min. value at x = 0
Min. value is f(0) = (0 – 1) (0 + 2)² = -4
f'(-2) = 6 (-2 + 1) = -6 < 0
∴ f(x) has max. value at x = -2
Max. value is f(-2) = (-2 -1) (-2 + 2)² = 0

viii) f(x) = \(\frac{x}{2}+\frac{2}{x}\) ∀ x ∈ (0, ∞)
Solution:
f'(x) = \(\frac{1}{2}-\frac{2}{x^{2}}\) and f”(x) = \(\frac{4}{x^{3}}\)
∴ For max. or min. f'(x) = 0
⇒ \(\frac{1}{2}-\frac{2}{x^{2}}\) = 0 ⇒ x² – 4 = 0 ⇒ x = ± 2
f”(2) = \(\frac{4}{2^{3}}\) = \(\frac{1}{2}\) > 0 (Since x > 0)
∴ f(x) has min. value at x = 2
Min. value is f(2) = \(\frac{2}{2}+\frac{2}{2}\) = 1 + 1 = 2.

ix) f(x) = – (x – 1)³ (x + 1)² ∀ x ∈ R
Solution:
f(x) = -(x – 1)³ (x + 1)² = (1 – x)³ (x + 1)²
f”(x) = (1 – x)³ 2(x + 1) + 3(1 – x)² (-1) (x + 1)²
= (1 – x)² (x + 1) {2(1 – x) – 3(x + 1)}
= (1 – x)² (x + 1) {2 – 2x – 3x – 3}
= (1 – x)² (x + 1) (-1 – 5x)
f”(x) = (1 – x)² (x + 1) (-5) + (1 – x)² (-1 – 5x) + (x + 1) (-1 -5x) 2(1 – x) (-1)
= -5 (1 -x)² (x+ 1) – (1 + 5x) (1 – x)² + (x + 1) (1 + 5x) 2(1 – x)
∴ For maximum or minimum f(x) = 0
(1 – x)² (x + 1) (-1 – 5x) = 0
⇒ x = ± 1 or -1/5
f”(1) = 0 – 0 + 0 ⇒ critical value at x = 0
f”(1 +1)² (-1) = 0 – (1 – 5) + 0 = 16 > 0
∴ f(x) has min. value at x = -1
Min. value = f(-1) = (1 + 1)³ (-1 + 1)² = 0
f”(- \(\frac{1}{5}\)) < o
⇒ f(x) has max. value at x = – \(\frac{1}{5}\)
Min. value is f (-\(\frac{1}{5}\)) = \(\frac{3456}{3125}\)

x) f(x) = x² e^3x ∀ x ∈ R
Solution:
f'(x) = x² e^3x .3 + e^3x. 2x
For maximum or minimum f'(x) = 0
3x² e^3x + 2e^3x. x = 0
x² e^3x (3x + 2) = 0
x = 0, x = \(\frac{-2}{3}\) and e = 0 is not possible
Now f”(x) = 3(x² e^3x. 3 + e^3x 2x)
+ e^3x 2 + 2x e^3x
f”(x) = 9x²e^3x + 6x
e^3x+ 2 e^3x + 6xe^3x
= 9x²e^3x + 12xe^3x + 2e^3x
f”(0) = 2 > 0
∴ Point of local minimum = 0
local minimum = 0
f”(\(\frac{-2}{3}\)) = \(\frac{-2}{e^{2}}\) < 0
∴ point of local maximum = \(\frac{-2}{3}\)
local maximum = \(\frac{4}{9e^{2}}\)

Question 2.
Prove that the following functions do not have absoute maximum and absolute minimum.
i) e^x in R
Solution:
f(x) = e^x and f”(x) = e^x
∴ For maxima or minima f'(x) = 0 ⇒ e^x = 0
⇒ x value is not defined
Hence it has no maxima or minima.

ii) log x in (0, ∞)
Solution:
f(x) = \(\frac{1}{x}\) and f”(x) = – \(\frac{1}{x^{2}}\)
f (x) = 0 ⇒ x value is not defined
⇒ f(x) has no maxima or minima.

iii) x³ + x² + x + 1 in R
Solution:
f (x) = 3x² + 2x + 1 gives imaginary values.
⇒ It has no maximum or minimum values.

II.

Question 1.
Find the absolute maximum value and absoulte minimum value of the following functions on the domain specified against the function.
Solution:
f(x) = x³ on (-2, 2)
f(x) = 3x² and f'(x) – 6x
the value f(-2) = (-2)³ = – 8
Max Value f(2) = 2³ = 8

ii) f(x) = (x – 1)²+ 3 on [-3, 1]
Solution:
f(x) = 2(x – 1) and f'(x) = 2
Max. value f(-3) = (-3 – 1)² + 3 = 16 + 3 = 19
Min. value f(l) = 0 + 3 = 3

iii) f(x) = 2|x| on [-1, 6]
Solution:
f'(x) = \(\frac{2|x|}{x}\)
For max. or min., f'(x) = 0
\(\frac{2|x|}{x}\) = 0 ⇒ x = 0
f(0) = 0
f(-1) – 2|-1| = 2
f(6) = 2(6) = 12
Absolute minimum = 0
Absolute maximum = 12

iv) f(x) = sin x + cos x on [0, π]
Solution:
f'(x) cos x – sin x which exists at all x ∈ (0, π)
Now f'(x) = 0 ⇒ cos x – sin x = 0
⇒ tan x = 1
⇒ x = \(\frac{\pi}{4}\) ∈ (0, π)
Now f(0) = sin 0 + cos 0 = 1
f(\(\frac{\pi}{4}\)) = sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
f(π) = sin π + cos π = 0 – 1 = -1
∴ Minimum value : -1
Maximum value: √2

v) f(x) = x + sin 2x on [0, π]
Solution:
f(x) = x + sin 2x
f(x) = 1 + 2 cos 2x
f (x) = 0 ⇒ 2 cos 2x + 1 = 0
⇒ cos 2x = –\(\frac{1}{2}\) = cos \(\frac{2 \pi}{3}\)
⇒ 2x = \(\frac{2 \pi}{3}\)
⇒ x = \(\frac{\pi}{3}\) ∈ (0, 2π)
Now f(0) = 0 + sin 2(0) = 0
f(\(\frac{\pi}{3}\)) = \(\frac{\pi}{3}\) + sin 2.\(\frac{\pi}{3}\) = \(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\)
f(2π) = 2π + sin 2. 2π = 2π + 0 = 2π
Minimum value = 0
Maximum value is = 2π

Question 2.
Use the first derivative test to find local extrema of f(x) = x³ – 12x on R.
Solution:
f(x) = x³ – 12x
f'(x) = 3x² – 12
f”(x) = 6x
For maximum or minimum, f'(x) = 0
3x² – 12 = 0
3x² = 12
x = ± 2
f”(2) = 12 > 0
Point of local minimum at x = 2
Local minimum = – 16
f”(-2) = -12 < 0
Point of local maximum at x = -1
Local maximum =16

Question 3.
Use the first derivative test to find local extrema of f(x) = x² – 6x + 8 on R.
Solution:
f(x) = x² – 6x + 8
f’(x) = 2x -6 ⇒ f”(x) = 2
For maximum or minimum f'(x) = 0
2x – 6 = 0
⇒ x = 3
f”(3) = 2 > 0
∴ Point of local minimum x = 3
Local minimum = -1

Question 4.
Use the second derivative test to find local extrema of the function
f(x) =x³ – 9x² – 48x + 72 on R.
Solution:
f(x) =x³ – 9x² – 48x + 72.
⇒ f'(x) = 3x² – 18x – 48
= 3(x – 8) (x + 2)
Thus the stationary point are – 2 & 8
f”(x) = 6x – 18 = 6(x – 3)
At x= 8, f”(8) = 30 > 0
∴ f (8) = (8)³ – 9(8)² – 48(8) + 72
= 512 – 576 – 384 + 72
= – 376
At x= -2, f”(-2) = – 30 < 0
f(-2) = (-2)³ – 9(-2)² – 48(-2)+72
= -8 – 36 + 96 + 72
= 124
Local minimum = -376
Local maximum = 124

Question 5.
Use the second derivative test to find local extrema of the function. f(x) = -x³ + 12x² – 5 on R.
Solution:
f(x) = -x³ + 12x² – 5
⇒ f'(x) = -3x² + 24x
= -3x(x – 8)
Thus the stationary point are 0, 8
f”(x) = – 6x + 24
At x= 0, f”(0) = 24 > 0.
f(0) = -5
At x= 8, f”(8) = -24 < 0
f(8) = -8³ + 12(8)² – 5
= -512 + 768 – 5 = 251
Local minimum = -5
Local maximum = 251

Question 6.
Find local maximum or local minimum of f(x) = -sin 2x – x defined on [-π/2, π/2].
Solution:
f(x) =-sin 2x – x
f'(x) = -2cos 2x – 1
f”(x) = 4 sin 2x
Thus the starting point are x = \(\frac{\pi}{3},\frac{-\pi}{3}\)

Local minimum = –\(\frac{\sqrt{3}}{2}-\frac{\pi}{3}\)
Local maximum = \(\frac{\sqrt{3}}{2}+\frac{\pi}{3}\)

Question 7.
Find the absolute maximum and absolute minimum of f (x) = 2x³ – 3x² – 36x + 2 on the interval [0, 5].
Solution:
f(x) = 2x³ – 3x² – 36x + 2
f(x) = 6x² – 6x – 36
f(x) = 12x – 6
for maxima or minima, f'(x) = 0
6x² – 6x – 36 = 0
x² – x – 6 = 0
x² -3x + 2x – 6 = 0
x(x – 3) + 2(x – 3) = 0
(x + 2) (x – 3) = 0
x = 3, -2
f”(3) = 30 > 0
f(x) has max/min value at x = 3
f(3) = 2(3)³ – 3(3)² – 36(3) + 2
= 54 – 27 – 108 + 2
= -79
Absolute minimum = – 79
Since 0 ≤ x ≤ 5
∴ f(0) = 0 – 0 – 0 + 2
= 2
∴ Absolute maximum = 2.

Question 8.
Find the absolute extremum of f(x) = 4x – \(\frac{x^{2}}{2}\) on [-2, \(\frac{9}{2}\)]
Solution:
f(x) = 4x – \(\frac{x^{2}}{2}\)
f'(x) =4 – x
f”(x) = – 1
for maxima or minima f'(x) = 0
4 – x = 0
x = 4
f”(4) = -1 < 0
∴ f has maximum value at x = 4
f(4) = 16 – \(\frac{16}{2}\) = 8
Since -2 ≤ x ≤ \(\frac{9}{2}\)
∴ f(-2) = -8 – \(\frac{4}{2}\)
= -8- 2 = -10
∴ Absolute minimum = -10
Absolute maximum = 8

Question 9.
Find the maximum profit that a company can make, if the profit function is given by P(x) = -41 + 72x – 18x²
Solution:
P(x) = – 41 + 72 x – 18x².
\(\frac{dp(x)}{dx}\) = 72 – 36x
for maxima or minima, \(\frac{dp}{dx}\) = 0
72 – 36x = 0
x = 2
\(\frac{d^{2}p}{dx^{2}}\) = -36 < 0
∴ The profit f(x) is maximum for x = 2
The maximum profit will be P(2) =
= -41 + 72(2) – 18(4)
= 31

Question 10.
The profit function P(x) of a company selling x items is given by P(x) = -x³ + 9x² – 15x – 13 where x represents thousands of units. Find the absolute maximum profits if the company can manufacture a maximum of 6000 units.
Solution:
P(x) = -x³ + 9x² – 15x – 13
\(\frac{dp(x)}{dx}\) = -3x² + 18x – 15
for maximum or minimum \(\frac{dp}{dx}\) = 0
-3x² + 18x – 15 = 0
x² – 6x + 5 = 0
x² – 5x – x + 5 = 0
x(x- 5) – 1(x- 5) = 0
(x – 1) (x – 5) = 0
x = 1, 5
P(1) = -1 + 9 – 15 – 13 = -10
P(5) = -125 + 225 – 75 – 13 = 12
∴ Maximum profit = 12.

III.

Question 1.
The profit function P(x) of a company selling x items per day is given by P(x) = (150 – x) x – 1000. Find the number of items that the company should manufacture to get maximum profit. Also find the maximum profit.
Solution:
Given that tbe profit function
P(x) = (150 – x)x -1000
for maximum or minimum \(\frac{dp}{dx}\) = 0
(150 – x(1) -x (-1) = 0
150 – 2x = 0
x = 75
Now \(\frac{d^{2}p}{dx^{2}}\) = -2 < 0
∴ The profit P(x) is maximum for x = 75
The company should sell 75 tems a day the maxma profit will be P (75) = 4625.

Question 2.
Find the absolute maximum and absolute minimum of f(x) = 8x³ + 81x² – 42x – 8 on [-8, 2].
Solution:
f(x) = 8x³ + 81x² – 42x – 8
f'(x) = 24x²+ 162x – 42
For maximum or minimum, f'(x) = 0
24x² + 162x – 42 = 0
4x² + 27x – 7 = 0
4x² + 28x – x – 7 = 0
4x(x + 7) – 1(x + 7) = 0
(x + 7) (4x – 1) = 0
x = – 7 or \(\frac{1}{4}\)

f(- 8) = 8(-8)³ + 81(-8)² – 42(-8) -8
= – 8(512) + 81(64) + 336 – 8
= -4096 + 5184 + 336 – 8
= 5520 – 4104
= 1416

f(2) = 8(2)³ + 81 (2)² – 42(2) – 8
= 64 + 324 – 84 – 8
= 296

f(-7) = 1246
Absolute maximum = 1416
Absolute minimum = \(\frac{-216}{16}\)

Question 3.
Find two positive integers whose sum is 16 and the sum of whose squares is minimum.
Solution:
Suppose x and y are the sum value
x + y =16
⇒ y = 16 – x
f(x) = x² + y² = x² + (16 – x)²
= x² + 256 + x² – 32x
f'(x) = 4x – 32
for maximum or minimum f'(x) = 0
⇒ 4x – 32 = 0
4x = 32 x = 8
f”(x) = 4 > 0
∴ f(x) is minimum when x = 8
y = 16 – x = 16 – 8 = 9
∴ Required number are 8, 8.

Question 4.
Find two positive integers x and y such that x + y = 60 and xy3 is maximum.
Solution:
x + y = 60 ⇒ y = 60 – x ………… (1)
Let p = xy³ = x(60 – x)³
\(\frac{dp}{dx}\) = x³(60 – x)²(-1) + (60-x)³
= -3x (60 – x)² + (60 – x)³
= (60 – x)² [-3x + 60 – x]
= (60 – x)² (60 – 4x) = 4(60 – x)² (15 – x)

\(\frac{d^{2}p}{dx^{2}}\) = 4[(60-x)² (-1) + (15-x) 2(60-x) (-1)]
= 4(60 – x) [-60 + x – 30 + 2x]
= 4(60 – x) (3x – 90)
= 12(60 – x) (x – 30)
For maximum or minimum \(\frac{dp}{dx}\) = 0
⇒ 4(60 – x)2 (15 -x) = 0
⇒ x = 60 or x = 15 ; x cannot be 60
∴ x = 15 ⇒ y = 60 – 15 = 45.
(\(\frac{d^{2}p}{dx^{2}}\))_{x = 15} = 12(60 – 15) (15 – 30) < 0
⇒ p is maximum
∴ Required numbers are 15, 45.

Question 5.
From a rectangular sheet of dimensions 30 cm x 80 cm four equal squares of side x cm are removed at the comers and the sides are then turned up so as to form an open rectangular box. Find the value of x, so that the volume of the box is the greatest?
Solution:
Length of the box = 80 – 2x = l
Breadth of the box = 30 – 2x = b

Height of the box = x = h
Volume = lbh = (80 – 2x) (30 – 2x). x
= x (2400 – 220 x + 4x²)
f(x) = 4x³ – 220 x² + 2400x
f'(x) = 12x² – 440x + 2400
= 4 [3x² – 110 x + 600]
f'(x) = 0
⇒ 3x² – 110 x +600 = 0

f(x) is maximum when x = \(\frac{20}{3}\)
Volume of the box is maximum when x = \(\frac{20}{3}\) cm

Question 6.
A window is in the shape of a rectangle surmounted by a semi-circle. If the peri-meter of the window is 20 feet, find the maximum area.
Solution:
Let length of the rectangle be 2x and breadth be y so that radius of the semi-circle is x.

Perimeter = 2x + 2y + π. x = 20
2y = 20 – 2x – πx.
y = 10 – x – \(\frac{\pi}{2}\). x
Area = 2xy + \(\frac{\pi}{2}\). x²
= 2x(10 – x – \(\frac{\pi}{2}\)) + \(\frac{\pi}{2}\)x²
= 20x – 2x² – πx² + \(\frac{\pi}{2}\) x²
f(x) = 20x – 2x² – \(\frac{\pi}{2}\) x²
f'(x) = 0 ⇒ 20 – 4x – πx = 0
(π + 4)x = 20
x = \(\frac{20}{\pi+4}\)
f”(x) = -4 – π < 0
f(x) is a maximum when x = \(\frac{20}{\pi+4}\)

Question 7.
If the curved surface of right circular a cylinder inscribed in a sphere of radius r is maximum, show that the height of the cylinder is √2r.
Solution:
Suppose r is the radius and h be the height of the cylinder.

From ∆OAB, OA² + AB² = OB²
r² + \(\frac{h^{2}}{4}\) = R² ; r² = R² – \(\frac{h^{2}}{4}\)
Curved surface area = 2πrh

f(h) is greatest when h = √2 R
i.e., Height of the cylinder = √2 R.

Question 8.
A wire of length l is cut into two parts which are bent respectively in the form of a square and a circle. What are the lengths of the pieces of the wire respe-ctively so that the sum of the areas is the least?
Solution:
Suppose x is the side of the square and r is the radius of the circle.

Given 4x + 2πr = l
4x = l – 2πr
r = \(\frac{1-2 \pi r}{4}\)
Sum of the area = x² + πr²

∴ f(r) is least when r = \(\frac{l}{ 2(\pi+4)}\)
Sum of the area is least when the wire is cut into pieces of length \(\frac{\pi l}{\pi+4}\) and \(\frac{4l}{\pi+4}\)

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(g) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(g)

I.

Question 1.
Without using the derivative, show that
i) The function f(x) = 3x + 7 is strictly increasing on R.
Solution:
Let x₁, x₂ ∈ R with x₁ < x₂
Then 3x₁ < 3x₂
Adding 7 on both sides
3x₁ + 7 < 3x₂ + 7
⇒ f(x₁) < f(x₂)
∴ x₁ < x₂ ⇒ f(x₁) < f(x₂) ∀ x₁ x₂ ∈ R
∴ The given function is strictly increasing on R

ii) The function f(x) = (\(\frac{1}{2}\))^x is strictly decreasing on R.
Solution:
f(x) = (\(\frac{1}{2}\))^x
Let x₁, x₂ ∈ R
Such that x₁ < x₂
⇒ (\(\frac{1}{2}\))^x₁ > (\(\frac{1}{2}\))^x₂
⇒ f(x₁) > f(x₂)
∴ f(x) is strictly decreasing on R.

iii) The function f(x) = e^3x is strictly increasing on R.
Solution:
f(x) = e^3x
Let x₁, x₂ ∈ R such that x₁ < x₂
We know that of a > b then e^a > e^b
Then e^3x < e^3x₂
⇒ f(x₁) < f(x₂)
∴ f is strictly increasing on R.

iv) The function f(x) = 5. – 7x is strictly decreasing on R.
Solution:
f(x) = 5 – 7x
Let x₁ x₂ ∈ R
Such that x₁ < x₂
Then 7x₁ < 7x₂
-7x₁ > -7x₂
Adding 5 on bothsides
5 – 7x₁ > 5 – 7x₂
f(x₁) > (x₂)
∴ x₁ < x₂ ⇒ f(x₁) > f(x₂) V x₁ x₂ ∈ R.
The given function f is strictly decreasing on R.

Question 2.
Show that the function f(x) = sin x. Define on R is neither increasing nor decreasing on (0, π).
Solution:
f(x) = sin x
Since 0 < x < n
Consider 0 < x
f(0) < f(x)
sin 0 < sin x
0 < sin x ……….. (1)
Consider x < n
f(x) < f(π)
sin x < sin π 0 > sin x ………….. (2)
From (1) & (2); f(x) is neither increasing nor decreasing.

II.

Question 1.
Find the intervals in which the following functions are strictly increasing or strictly decreasing.
i) x² + 2x – 5
Solution:
Let f(x) = x² + 2x – 5
f'(x) = 2x + 2
f(x) is increasing if f'(x) > 0
⇒ 2x + 2 < 0 ⇒ x+ 1 > 0
x > -1
f(x) is increasing If x ∈ (-1, ∞)
f(x) is decreasing If f'(x) < 0
⇒ 2x + 2 < 0
⇒ x + 1 < 0
⇒ x < -1 f(x) is decreasing if x ∈ (-∞, -1).

ii) 6 – 9x – x².
Solution:
Let f(x) = 6 – 9x – x²
f'(x) = -9 – 2x
f(x) is increasing if f'(x) > 0
⇒ -9 -2x > 0
⇒ 2x + 9 < 0
x < \(\frac{-9}{2}\)
f(x) is increasing if x ∈ (-∞, \(\frac{-9}{2}\))
f(x) is decreasing if f'(x) < 0
⇒ 2x + 9 > 0
⇒ x > \(\frac{-9}{2}\)
f(x) is decreasing of x ∈ (\(\frac{-9}{2}\), ∞)

iii) (x + 1)³ (x – 1)³.
Solution:
Let f(x) = (x + 1)³ (x – 1)³
= (x² – 1)³
x⁶ – 1 – 3x⁴ + 3x ²
f'(x) = 6x⁵ – 12x³ + 6x
= 6(x⁵ – 2x³ + x)
= 6x(x⁴ – 2x² +1)
= 6x(x² – 1)²
f'(x) ≤ 0
⇒ 6x(x² – 1)² < 0
f(x) is decreasing when (-∞, -1) ∪ (-1, 0)
f'(x) > 0
f(x) is increasing when (0, 1) ∪ (1, ∞)

iv) x³(x – 2)²
Solution:
f'(x) = x³. 2(x – 2) + (x – 2)².3x²
= x² (x – 2) [2x + 3 (x- 2)]
= x² (x – 2) (2x + 3x – 6)
= x² (x – 2) (5x – 6) ∀ x ∈ R, x² ≥ 0
For increasing, f'(x) = 0
x²(x – 2) (5x – 6) > 0
x ∈ (-∞, \(\frac{6}{5}\)) ∪ (2, ∞)
For decreasing, f'(x) < 0
x²(x – 2) (5x – 6) < 0
x ∈ (\(\frac{6}{5}\), 2)

v) xe^x
Solution:
f'(x) = x . e^x + e^x. 1 = e^x(x + 1)
e^x is positive for all real values of x
f'(x)>0 ⇒ x + 1 > 6 ⇒ x > – 1
f(x) is increasing when x > – 1
f(x) < 0 ⇒ x + 1 < 0 ⇒ x < -1
f(x) is decreasing when x < – 1

vi) \(\sqrt{(25-4x^{2})}\)
Solution:
f(x) is real only when 25 – 4x² > 0
-(4x² – 25) > 0
-(2x + 5) (2x – 5) > 0
∴ x lies between –\(\frac{5}{2}\) and \(\frac{5}{2}\)
Domain of f = (-\(\frac{5}{2}\), \(\frac{5}{2}\))
f'(x) = \(\frac{1}{2 \sqrt{25-4 x^{2}}}\) (-8x)
= –\(\frac{4x}{\sqrt{25-4 x^{2}}}\)
f(x) is increasing when f'(x) > 0
⇒ \(\frac{-4x}{\sqrt{25-4 x^{2}}}\) > 0
i.e., x < o
f(x) is increasing when (-\(\frac{5}{2}\), 0)
f(x) is decreasing when f'(x) < 0
⇒ –\(\frac{4x}{\sqrt{25-4 x^{2}}}\) < 0
∴ x > 0
f(x) is decreasing when (0, \(\frac{5}{2}\)).

vii) ln (ln(x)); x > 1.
Solution:
f'(x) = –\(\frac{1}{lnx}•\frac{1}{x}\)
f(x) is decreasing when f'(x) > 0
\(\frac{1}{x.ln x}\) >0
⇒ x. In x > 0
ln x is real only when x > 0
∴ ln x < 0 = ln 1
i.e., x > 1
f(x) is increasing when x > 1 i.e., in (1, ∞)
f(x) is decreasing when f'(x) < 0 ⇒ ln x > 0 = ln 1
i.e., x < 1
f(x) is decreasing in (0, 1)

viii) x³ + 3x² – 6x + 12.
Solution:
f(x) = x³ + 3x² – 6x + 12
f(x) = 3x² + 6x – 6
= 3(x² + 2x – 2)
= 3((x + 1)² – 3)
= 3[(x + 1) + √3] [(x + 1) – √3]
= 3(x + (1 + √3) (x + (1 – √3 )
f (x) < 0
⇒ x = -(1+ √3 ) or -(1 – √3 )
x = -1 – √3 or √3 – 1
f(x) is decreasing in (-1 -√3 , √3 -1)
f (x) > 0
f(x) increasing when (-∞, -1 – √3 ) ∪ (√3 – 1, ∞)

Question 2.
Show that f(x) = cos²x is strictly increasing on (0, π/2).
Solution:
f(x) = cos² X
⇒ f(x) = 2 cos x (-sin x)
= -2 sin x cos x
= -sin 2x
Since 0 < x < \(\frac{\pi}{2}\)
⇒ 0 < 2x < π
Since ‘sin x’ is +ve between 0 and π
∴ f(x) is clearly -ve.
∴ f'(x) < 0
∴ f(x) is strictly decreasing.

Question 3.
Show that x + \(\frac{1}{x}\) is increasing on [1, ∞)
Solution:
Let f(x) = x + \(\frac{1}{x}\)
f'(x) = 1 – \(\frac{1}{x^{2}}\) = \(\frac{x^{2}-1}{x^{2}}\)
Since x ∈ [1, ∞) = \(\frac{x^{2}-1}{x^{2}}\) > 0
∴ f'(x) > 0
∴ f(x) is increasing.

Question 4.
Show that \(\frac{x}{1+x}\) < ln (1 + x) < x ∀ x > 0
Solution:
Let f(x) = ln(1 + x)- \(\frac{x}{1+x}\)
= ln(1 + x) – \(\frac{1+x-1}{1+x}\)
= ln(1 + x) – 1 + \(\frac{1}{1+x}\)
f'(x) = \(\frac{1}{1+x}\) – \(\frac{1}{(1+x)^{2}}\)
= \(\frac{1+x-1}{(1+x)^{2}}\)
= \(\frac{x}{(1+x)^{2}}\) > 0 since x > 0
f(x) is increasing when x > 0
∴ f(x) > f(0)
f(0) = ln 1 – \(\frac{0}{1+0}\) = 0 – 0 = 0
Since xe [1, ∞) =
ln (1 + x) – \(\frac{x}{1+x}\) > 0
⇒ ln (1 + x) > \(\frac{x}{1+x}\) ……… (1)
Let g(x) = x – ln (1 + x)
g'(x) = 1 – \(\frac{x}{1+x}=\frac{1+x-1}{1+x}\)
= \(\frac{x}{1+x}\) > 0 since x > 0
g(x) is increasing when x > 0
i.e., g(x) > g(0)
g(0) = 0 – ln (1) = 0 – 0 = 0
∴ x – ln (1 + x) > 0
x > ln(1 + x) ………….. (2)
From (1), (2) we get
\(\frac{x}{1+x}\) < ln (1 + x) < x ∀ x > 0

III.

Question 1.
Show that \(\frac{x}{1+x^{2}}\) < tan^-1 x < x when x > 0.
Solution:
Let f(x) = tan^-1 x – \(\frac{x}{1+x^{2}}\)

f(x) is increasing when x > 0
f(x) > f(0)
But f(0) = tan^-1 0 – 0 = 0 – 0 = 0
i.e., f(x) > 0

g(x) is increasing when x > 0
g(x) > g(0)
g(0) = 0 – tan^-1 0 = 0 – 0 = 0
∴ x – tan^-1 x > 0
⇒ x > tan^-1 x ………. (2)
From (1), (2) we get
\(\frac{x}{1+x^{2}}\) <tan^-1 x< x for x > 0

Question 2.
Show that tan x > x for all (0, \(\frac{\pi}{2}\))
Solution:
Let f(x) = tan x – x
f'(x) = sec² x – 1 > 0 for every
x ∈ (0, \(\frac{\pi}{2}\))
f(x) is increasing for every x ∈ (0, \(\frac{\pi}{2}\))
i.e., f(x) > f(0)
f(0) = tan 0 – 0 = 0 – 0 = 0
∴ tan x – x > 0
⇒ tan x > x for every x ∈ (0, \(\frac{\pi}{2}\))

Question 3.
If x ∈ (0, \(\frac{\pi}{2}\)) then show that \(\frac{2x}{\pi}\) < sin x < x.
Solution:
Let f(x) = x – sin x
f(x) = 1- cos x > 0 for every x ∈ (0, \(\frac{\pi}{2}\))
f(x) is increasing for every x ∈ (0, \(\frac{\pi}{2}\))
⇒ f(x) > f(0)
f(0) = 0 – sin 0 = 0 – 0 = 0
∴ x – sin x > 0
⇒ x > sin x ………….. (1)
Let g(x) = sin x – \(\frac{2x}{\pi}\)
g'(x) = cos x – \(\frac{2}{\pi}\) > 0 for every x ∈ (0, \(\frac{\pi}{2}\))
g(x) is increasing in (0, \(\frac{\pi}{2}\))
g(x) > g(0)
g(0) = sin 0 – 0 = 0 – 0 = 0
∴ sin x – \(\frac{2x}{\pi}\) > 0
⇒ sin x > \(\frac{2x}{\pi}\) ………… (2)
From (1), (2) we get
\(\frac{2x}{\pi}\) < sin x < x for every x ∈ (0, \(\frac{\pi}{2}\))

Question 4.
If x e (0,1) then show that 2x < ln [latex]\frac{(1+x)}{(x-1)}[/latex] < 2x [1 + \(\frac{x^{2}}{2(1+x^{2})}\)] Solution:
Let f(x) = ln \(\frac{(1+x)}{1-x}\) – 2x
= ln (1 + x) – ln (1 – x) – 2x

f(x) is increasing ih (0, 1)
i.e., x > 0 ⇒ f(x) > f(0)
f(0) = ln 1 – 0 = 0 – 0 = 0


g(x) is increasing when x > 0
g(x) > g(0)
g(0) = 0 – ln 1 = 0 – 0 = 0

for x ∈ (0,1)

Question 5.
At what point the slopes of the tangents y = \(\frac{x^{3}}{6}-\frac{3x^{3}}{2}+\frac{11x}{2}\) + 12 increases?
Solution:
Equation of the curve is
y = \(\frac{x^{3}}{6}-\frac{3}{2}x^{2}+\frac{11x}{2}\) + 12

Slope = m = \(\frac{x^{2}}{c}-3x+\frac{11}{2}\)
\(\frac{dm}{dx}\) = \(\frac{2x}{2}\) -3 = x – 3
Slope increases ⇒ m > 0
x – 3 > 0
x > 3
The slope increases in (3, ∝)

Question 6.
Show that the functions ln \(\frac{(1+x)}{x}\) and \(\frac{x}{(1+x)ln(1+x)}\) are decrasing on (0, ∞).
Solution:
i)


∴ f(x) is decreasing for x ∈ (0, ∝)

ii) let f(x) = \(\frac{x}{(1+x)ln(1+x)}\)

∴ f(x) is decreasing for x ∈ (0, ∝)

Question 7.
Find the intervals in which the function f (x) = x3 – 3×2 + 4 is strictly increasing all x e R.
Solution:
f(x) = x³ – 3x² + 4
f'(x) = 3x² – 6x
f(x) is increasing if f'(x) > 0
3x² – 6x > 0
3x(x – 2) > 0
(x – 0)(x – 2) > 0
f(x) is increasing if x £ (-∞, 0) u (0, ∞)
f(x) is decreasing if f'(x) < 0
(x – 0) (x – 2) < 0
x ∈ (0, 2)

Question 8.
Find the intervals in which the function f(x) = sin⁴x + cos⁴x ∀ x ∈ [0, \(\frac{\pi}{2}\)] is increasing and decreasing.
Solution:
f(x) = sin⁴x + cos⁴x
f(x) = (sin²x)² + (cos²x)²
= (sin²x + cos²x)² – 2sin²x cos²x
= 1 – \(\frac{1}{2}\) sin² 2x
f'(x) = \(\frac{-1}{2}\) 2sin 2x. cos 2x(2)
= -2 sin 2x. cos 2x
= -sin 4x
Let 0 < x < \(\frac{\pi}{4}\)
∴ f(x) is decreasing if f'(x) < 0
⇒ -sinx < 0 ⇒ sinx > 0
∴ x ∈ (0, \(\frac{\pi}{4}\))
f(x) is increasing if f'(x) > 0
⇒ – sinx > 0
⇒ sinx < 0
∴ x ∈ (\(\frac{\pi}{4}\), \(\frac{\pi}{2}\))

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f)

I.

Question 1.
Verify Rolle’s theorem for the following functions.
i) x² – 1 on [-1, 1]
Solution:
Let f(x) = x² – 1
f is continuous on [-1,1]
since f(-1) = f(1) = 0 and
f is differentiable on [-1, 1]
∴ By Rolle’s theorem ∃ c ∈ (-1, 1) such that f'(c) = 0
f'(x) = 2x = 0
∴ = f'(c) = 0
2c = 0
c = 0
The point c = 0 ∈ (-1, 1)
Then Rolle’s theorem is verified.

ii) sin x – sin 2x on [0, π].
Solution:
Let f(x) = sin x – sin 2x
f is continuous on [0, π]
since f(0) = f(π)= 0 and
f is differentiable on [0, π]
By Rolle’s theorem ∃ c ∈ (0, π)
such that f'(c) = 0
f'(x) = cos x – 2 cos 2x
f'(c) = 0 ⇒ cosc – 2 cos 2c = 0
⇒ cosc – 2(2cos²c – 1):
cosc – 4 cos²c + 2=0 2
4 cos² c – cosc – 2 = 0

iii) log (x² + 2) – log 3 on [-1, 1]
Solution:
Let f(x) = log (x² + 2) – log 3
f is continuous on [-1, 1]
Since f(-1) = f(1) = 0 and f is
Differentable on [-1, 1]
By Rolle’s theorem ∃ c ∈ (-1, 1)
Such that f'(c) = 0
f'(x) = \(\frac{1}{x^{2}+2}\) = (2x)
f'(x) = \(\frac{2c}{c^{2}+2}\) = 0
2c = 0
c = 0
c = 0 ∈ (-1, 1).

Question 2.
It is given that Rolle’s theorem holds for the function f(x) = x³ + bx² + ax on [1, 3] with c = 2t + \(\frac{1}{\sqrt{3}}\). Find the values of a and b.
Solution:
Given f(x) = x³ + bx² + ax
f'(x) = 3x² + 2bx + a
∴ f'(c) = 0 6 ⇔ 3c² + 2bc + a = 0

⇔ b = 6 and b² – 3a = 3
36 – 3 = 3a
33 = 3a
a = 11
Hence a = 11 and b = -6.

Question 3.
Show that there is no real number k, for which the equation x² – 3x + k = 0 has two distinct roots in [0, 1].
Solution:
Clearly f(0) = f(c)
0 – 0 + k = 1 – 3 + k
0 = -2
Which is not possible
∴ There is no real number K.

Question 4.
Find a point on the graph of the curve y = (x – 3)², where the tangent is parallel to the chord joining (3, 0) and (4, 1).
Solution:
Given points (3, 0) and (4, 1)
The slope of chord = \(\frac{1-0}{4-3}\) = 1
Given y = (x- 3)²
\(\frac{dy}{dx}\) = 2(x – 3)
⇒ Slope = 2(x – 3)
1 = 2(x – 3)
\(\frac{1}{2}\) = x – 3
x = \(\frac{1}{2}\) + 3 = \(\frac{7}{2}\)
y = (x – 3)² = (\(\frac{7}{2}\) – 3)² = \(\frac{1}{4}\)
∴ The point on the curve is (\(\frac{7}{2}\), \(\frac{1}{4}\))

Question 5.
Find a point on the graph of the curve y = x³, where the tangent is parallel to the chord joining (1, 1) and (3, 27).
Solution:
Given points (1, 1) and (3, 27)
Slope of chord = \(\frac{27-1}{3-1}\) = 13
Given y = x³
\(\frac{dy}{dx}\) = 3x²
⇒ Slope = 3x²
13 = 3x²

∴ The point on the curve is (\(\frac{\sqrt{39}}{3}\), \(\frac{13\sqrt{39}}{9}\))

Question 6.
Find ‘c’, so that f'(c) = \(\frac{f(b)-f(a)}{b-a}\) in the following cases.
i) f(x) = x² – 3x – 1, a = \(\frac{-11}{7}\), b = \(\frac{13}{7}\).
Solution:

f'(x) = 2x – 3
f'(c) = 2c – 3
Given f'(c)= \(\frac{f(b)-f(a)}{b-a}\)

ii) f(x) = e^x ; a = 0, b = 1
Solution:
f(b) = f(1) = e’ = e
f(a) = f(0) = e° = 1
Given f(x) = e^x
f'(x) = e^x
Given condition f'(c) = \(\frac{f(b)-f(a)}{b-a}\)

Question 7.
Verify the Rolle’s theorem for the function (x² – 1) (x – 2) on [-1, 2]. Find the point in the interval where the derivate vanishes.
Solution:
Let f(x) = (x² – 1) (x- 2) = x³ – 2x² – x + 2
f is continous on [-1, 2]
since f(-1) = f(2) = 0 and f is
Differentiable on [-1, 2]
By Rolle’s theorem ∃ C ∈ (-1, 2)
Let f'(c) = 0
f'(x) = 3x² – 4x – 1
3c² – 4c – 1 = 0

Question 8.
Verify the conditions of the Lagrange’s mean value theorem for the following functions. In each case find a point ‘c’ in the interval as stated by the theorem
i) x² -1 on [2, 3]
Solution:
Solution:
Let f(x) = x² – 1
f is continous on [2, 3]
and f is differentiable
Given f(x) = x² – 1
f'(x) = 2x
By Lagrange’s mean value theorem ∃ C ∈ (2, 3) such there

ii) sin x – sin 2x on [0, π]
Solution:
Let f(x) = sin x – sin 2x
f is continuous on [0, π] and f is differentiable
Given f(x) = sin x – sin 2x
f'(x) = cos x – 2 cos 2x
By Lagrange’s mean value than ∃ C ∈ (0, π) such there
f'(c) = \(\frac{f(\pi)-f(0)}{\pi-0}\)
cosc – 2 cos 2c = 0
cosc 2(2cos² – 1) = 0
cosc – 4 cos²c + 2 = 0
4 cos² c – cos c – 2 = 0

iii) log x on [1, 2].
Solution:
Let f(x) = log x
f is continuous on [1, 2] and f is differentiable
Given f(x) = log x
f'(x) = \(\frac{1}{x}\)
By Lagrange’s mean value theorem ∃ c ∈ (1, 2) such that

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(e) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(e)

I.

Question 1.
At time t, the distance s of a particle moving in a straight line is given by s = -4t² + 2t. Find the average velocity between t = 2 sec and t = 8 sec.
Solution:
s = -4t² + 2t ds
v = \(\frac{ds}{dt}\) = -8t + 2 dt
Velocity at t = 2 is v = (\(\frac{ds}{dt}\))_t=2
v = -16 + 2 = -14 units/sec.
Velocity at t = 8 is v = (\(\frac{ds}{dt}\))_t=8
v = -64 + 2 = -62
Average velocity = \(\frac{-62-14}{2}\) = -38 units/sec.

Question 2.
If y = x⁴ then find the average rate of change of y between x = 2 and x = 4.
Solution:
y = x⁴ ⇒ \(\frac{dy}{dt}\) = 4x³
(\(\frac{dy}{dt}\))_x=2 = 32
(\(\frac{dy}{dt}\))_x=4 = 256
Average rate of change = \(\frac{256+32}{2}\) = 144.

Question 3.
A particle moving along a straight line has the relation s = t³ + 2t + 3, connecting the distance s describe by the particle in time t. Find the velocity and acceleration of the particle of t = 4 sec.
Solution:
s = t³ + 2t + 3
\(\frac{ds}{dt}\) = 3t² + 2, velocity v = \(\frac{ds}{dt}\) = 3t² + 2
Velocity at t = 4
⇒ (\(\frac{ds}{dt}\))_t=4 = 48 + 2 = 50 units/sec
v = 3t² + 2
\(\frac{dv}{dt}\) = 6t ⇒ a = (\(\frac{dv}{dt}\))_t=4 = 24 units/sec².

Question 4.
The distance – time formula for the motion of a particle along a straight line is s = t³ – 9t² + 24t – 18. Find when and where the velocity is zero.
Solution:
Given s = t³ – 9t² + 24t – 18
v = \(\frac{ds}{dt}\) = 3t² – 18t + 24
v = 0 ⇒ 3(t² – 6t + 8) = 0
∴ (t – 2) (t – 4) = 0
∴ t = 2 or 4
The velocity is zero after 2 and 4 seconds.

Case (i):
t = 2
s = t³ – 9t² + 24t – 18
= 8 – 36 + 48 – 18 = 56 – 54 = 2

Case (ii) :
t = 4 ; s = t³ – 9t² + 24t – 18
= 64 – 144 + 96 – 18
= 160 – 162 = -2
The particle is at a distance of 2 units from the starting point ‘O’ on either side.

Question 5.
The displacement s of a particle travelling in a straight line in t seconds is given by s = 45t + 11t² – t³. Find the time when the particle comes to rest.
Solution:
s = 45t + 11t² – t³
v = \(\frac{ds}{dt}\) = 45 + 22t – 3t²
If a particle becomes to rest
⇒ v = 0 ⇒ 45 + 22t – 3t² = 0
⇒ 3t² – 22t – 45 = 0
⇒ 3t² – 27t + 5t – 45 = 0
⇒ (3t + 5) (t – 9) = 0
∴ t = 9 or t = – \(\frac{5}{3}\)
∴ t = 9
∴ The particle becomes to rest at t = 9 seconds.

II.

Question 1.
The volume of a cube is increasing at the rate of 8 cm³/sec. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution:
Suppose ‘a’ is the edge of the cube and v be the volume of the cube.
v = a³ ……………….. (1)
\(\frac{dv}{dt}\) = 8 cm³/sec.
a = 12 cm
Surface Area of cube. S = 6a²
\(\frac{ds}{dt}\) = 12a\(\frac{da}{dt}\) ………….. (2)

Question 2.
A stone is dropped into a quiet lake and ripples move in circles at the speed of 5 cm/sec. At the instant when the radius of circular ripple is 8 cm., how fast is the enclosed area increases?
Solution:
Suppose r is the value of the outer ripple and A be its area
Area of circle A = πr²
\(\frac{dA}{dt}\) = 2πr \(\frac{dr}{dt}\)
Given r = 8, \(\frac{dr}{dt}\) = 5
\(\frac{dA}{dt}\) = 2π (8) (5)
= 80π cm²/sec.

Question 3.
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:
\(\frac{dr}{dt}\) = 0.7 cm/sec
Circumference of a circle, c = 2πr
\(\frac{dc}{dt}\) = 2π \(\frac{dr}{dt}\)
= 2π(0.7)
= 1.4π cm/sec.

Question 4.
A balloon which always remains spherical on inflation is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of balloon increases when the radius in 15 cm.
Solution:
\(\frac{dv}{dt}\) = 900 c.c/sec
r = 15 cm
Volume of the sphere v = \(\frac{4}{3}\) πr³

Question 5.
The radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/sec. At what rate is the volume of the bubble increasing when the radius is 1 cm.?
Solution:
\(\frac{dr}{dt}=\frac{1}{2}\) cm/sec
radius r = 1 cm
Volume sphere v = \(\frac{4}{3}\) πr³
\(\frac{dv}{dt}\) = 4πr² \(\frac{dr}{dt}\)
= 4π(1)²\(\frac{1}{2}\)
= 2π cm³/sec.

Question 6.
Assume that an object is launched upward at 980 m/sec. Its position would be given by s = 4.9 t² + 980 t. Find the maximum height attained by the object.
Solution:
s = – 4.9 t² + 980 t
\(\frac{ds}{dt}\) = -9.8 t + 980
v = -9.8 t + 980
for max. height, v = 0
-9.8 t + 980 = 0
980 = 9.8 t
\(\frac{980}{9.8}\) = t
100 = t
s = -4.9(100)² +980(100)
s = -49000 + 98000
s = 49000 units.

Question 7.
Let a kind of bacteria grow in such a way that at time t sec. there are t^(3/2) bacteria. Find the rate of growth at time t = 4 hours.
Solution:
Let g be the amount of growth of bacteria at t then g(t) = t^3/2
The growth rate at time t is given by
g'(t) = \(\frac{3}{2}\)t^1/2
given t = 4hr
g'(t) = \(\frac{3}{2}\) (4 × 60 × 60)^1/2
= \(\frac{3}{2}\) (2 × 60) = 180

Question 8.
Suppose we have a rectangular aquarium with dimensions of length 8m, width 4 m and height 3 m. Suppose we are tilling the tank with water at the rate of 0.4 m³/sec. How fast is the height of water changing when the water level is 2.5 m?
Solution:
Length of aquarium l = 8 m
Width of aquarium b = 4 m
Height of aquarium h = 3
\(\frac{dv}{dt}\) = 0.4 m³/sec.
v = lbh
= 8(4)(3)
= 96
v = lbh
⇒ log v = log l + log b + log h

Note : Text book Ans. \(\frac{1}{80}\) will get when h = 3.

Question 9.
A container is in the shape of an inverted cone has height 8m and radius 6m at the top. If it is filled with water at the rate of 2m³/minute, how fast is the height of water changing when the level is 4m?
Solution:

h = 8m = OC
r = 6m = AB
\(\frac{dv}{dt}\) = 2 m³/minute
∆ OAB and OCD are similar angle then
\(\frac{CD}{AB}=\frac{OC}{OA}\)
\(\frac{r}{6}=\frac{h}{8}\)
r = h \(\frac{3}{4}\)
Volume of cone v = \(\frac{1}{3}\)πr²h

Question 10.
The total cost C(x) in rupees associated with the production of x units of an item is given by C(x) 0.007x³ – 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.
Solution:
Let m represents the marginal cost, then
M = \(\frac{dc}{dx}\)
Hence
M = \(\frac{d}{dx}\)(0.007x³ – 0.003x² + 15x + 4000)
= (0.007) (3x²) – (0.003) (2x) + 15 /.
∴ The marginal cost at x = 17 is
(M)_m=17 = (0.007) 867 – (0.003) (34) + 15
= 6.069-0.102+ 15
= 20.967.

Question 11.
The total revenue in rupees received from the sale of x units of a produce is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.
Solution:
Let m denotes the marginal revenue. Then
M = \(\frac{dR}{dx}\)
Similar R(x) = 13x² + 26x +15
∴ m = 26x + 26
The marginal revenue at x = 7
(M)_{x = 7} = 26(7) + 26
= 208.

Question 12.
A point P is moving on the curve y = 2x². The x co-ordinate of P is increasing at the rate of 4 units per second. Find the rate at which y co-ordinate is increasing when the point is (2, 8).
Solution:
Given y = 2x²
\(\frac{dy}{dx}\) = 4x. \(\frac{dx}{dt}\)
Given x = 2, \(\frac{dx}{dt}\) = 4.\(\frac{dy}{dt}\)
= 4(2).4 = 32
y co-ordinate is increasing at the rate of 32 units/sec.

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(d) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(d)

I. Find the angle between the curves given below.

Question 1.
x + y + 2 = 0 ; x² + y² – 10y = 0.
Solution:
x + y + 2 = 0 ⇒ x = -(y + 2)
x² + y² – 10y = 0
(y + 2)² + y² – 10y = 0
y² + 4y + 4 + y² – 10y = 0
2y² – 6y + 4 = 0
y² – 3y + 2 = 0
(y + 1) (y – 2) = 0
y = 1 or y – 2
x = – (y + 2)
y = 1 ⇒ x = -(1 + 2) = -3
y = 2 ⇒ x = -(2 + 2) = -4
The points of intersection are P(-3, 1) and Q(-4, 2), equation of the curve is
x² + y² – 10y = 0
Differentiating w.r.to x.
2x + 2y\(\frac{dy}{dx}\) – 10 \(\frac{dy}{dx}\) = 0
2\(\frac{dy}{dx}\)(y – 5) = -2x
\(\frac{dy}{dx}\) = –\(\frac{x}{y-5}\)
f'(x₁) = –\(\frac{x}{y-5}\)
Equation of the line is x + y + 2 = 0
1 + \(\frac{dy}{dx}\) = 0 ⇒ \(\frac{dy}{dx}\) = -1
g'(x) = -1

Case (i):
At P(-3, 1), f'(x₁) = \(\frac{3}{1-5}=-\frac{3}{4}\), g'(x₁) = -1

Case (ii):

Question 2.
y² = 4x, x² + y² = 5.
Solution:
Eliminating y; we get x² + 4x = 5
x² + 4x – 5 = 0
(x – 1) (x + 5) = 0
x – 1 = 0 or x + 5 = 0
x = 1 or -5
Now y² = 4x
x = 1 ⇒ y² = 4
y = ±2
x = -5 ⇒ y is not real.
∴ Points of interstection of P(1, 2) and Q(1, -2) equation of the first curve is y² = 4x
2y.\(\frac{dy}{dx}\) = 4
\(\frac{dy}{dx}\) = \(\frac{4}{2y}\)
f(x) = \(\frac{2}{y}\)
Equation of the second curve is x² + y² = 5 dy
2x + 2y \(\frac{dy}{dx}\) = 0 dx
2.\(\frac{dy}{dx}\) = -2x

Question 3.
x² + 3y = 3 ; x² – y² + 25 = 0.
Solution:
x² = 3 – 3y; x² – y² + 25 = 0
3 – 3y – y² + 25 = 0
y² + 3y – 28 = 0
(y – 4) (y + 7) = 0
y – 4 = 0 (or) y + 7 = 0
y = 4 or – 7
x² = 3 – 3y
y = 4 ⇒ x² = 3 – 12 = – 9
⇒ x is not real
y = -7 ⇒ x² = 3 + 21 = 24
⇒ x = ± √24 = ± 2√6
Points of intersection are
P(2√6, -7), Q(-2√6, -7)
Equation of the first cur ve is x² + 3y = 3
3y = 3 – x²
3.\(\frac{dy}{dx}\) = -2x
\(\frac{dy}{dx}\) = –\(\frac{2x}{3}\) i.e, f'(x₁) = –\(\frac{2x}{3}\)
Equation of the second curve is
x² – y² + 25 = 0
y² = x² + 25
2y.\(\frac{dy}{dx}\) = 2x ⇒ \(\frac{dy}{dx}=\frac{2x}{2y}=\frac{x}{y}\)

Question 4.
x² = 2(y + 1), y = \(\frac{8}{x^{2}+4}\).
Solution:
x² = 2(\(\frac{8}{x^{2}+4}\) + 1) = \(\frac{16+2x^{2}+8}{x^{2}+4}\)
x²(x² + 4) = 2x² + 24
x⁴ + 4x² – 2x² – 24 = 0
x⁴ + 2x² – 24 = 0
(x² + 6) (x² – 4) = 0
x² = -6 or x² = 4
x² = -6 ⇒ x is not real
x² = 4 ⇒ x = ±2
y = \(\frac{8}{x^{2}+4}=\frac{8}{4+4}=\frac{8}{8}\) = 1
∴ Points of intersection are P(2, 1) and Q(-2, 1)
Equation of the first curve is x² = 2(y + 1)
2x = 2.\(\frac{dy}{dx}\) ⇒ \(\frac{dy}{dx}\) = x
f'(x₁) = x₁
Equation of the second curve is y = \(\frac{8}{x^{2}+4}\)

∴ The given curves cut orthogonally
i.e., θ = \(\frac{\pi}{2}\)
At Q (-2, -1),f'(x₁) = -2, g'(x₁) = \(\frac{32}{64}=\frac{1}{2}\)
f'(x₁) g'(x₁) = -2 × \(\frac{1}{2}\) = -1
∴ The given curves cut orthogonally
⇒ θ = \(\frac{\pi}{2}\)

Question 5.
2y² – 9x = 0, 3x² + 4y = 0 (in the 4^th quadrant).
Solution:
2y² – 9x = 0 ⇒ 9x = 2y²
x = \(\frac{2}{9}\)y²
3x² + 4y = 0
⇒ 3.\(\frac{4}{8}\) y⁴ + 4y = 0
\(\frac{4y^{2}+108y}{27}\) = 0
4y(y³ + 27) = 0
y = 0 or y³ = -27 ⇒ y = -3
9x = 2y² 2 × 9 ⇒ x = 2
Point of intersection (in 4^th quadrant) is P(2, -3)
Equation of the first curve is 2y² = 9x
4y\(\frac{dy}{dx}\) = 9 ⇒ \(\frac{dy}{dx}=\frac{9}{4y}\)
f'(x₁) = \(\frac{9}{4y}\)
At P(2, -3), f'(x₁) = \(\frac{9}{-12}=-\frac{3}{4}\)
Equation of the second curve is
3x² + 4y = 0
4y = -3x²
4.\(\frac{dy}{dx}\) = -6x

Question 6.
y² = 8x, 4x² + y = 32
Solution:
4x² + 8x = 32 ⇒ x² + 2x = 8
x² + 2x – 8 = 0
(x – 2) (x + 4) = 0
x = 2 or -4
y² = 8x
x = -4 ⇒ y² is not real
x = 2 ⇒ y² = 16 ⇒ y = ±4
Point of intersection are P(2, 4), Q(2, -4)
Equation of the first curve is y² = 8x
2y.\(\frac{dy}{dx}\) = 8 ⇒ \(\frac{dy}{dx}=\frac{8}{2y}=\frac{4}{y}\)
f'(x₁) = \(\frac{4}{y}\)
Equation of the second curve is
4x² + y² = 32
8x + 2y.\(\frac{dy}{dx}\) = 0

Question 7.
x²y = 4, y(x² + 4) = 8.
Solution:
x²y = 4 ⇒ x = \(\frac{4}{y}\)
y(x² + 4) = 8
y(\(\frac{4}{y}\) + 4y) = 8
y\(\frac{(4+4y)}{y}\) = 8
4y – 4 ⇒ y = 1
x² = 4 ⇒ x = ±2
Points of intersection are P(2, 1), Q(-2, 1)
x²y = 4 ⇒ y = \(\frac{4}{x^{2}}\)
\(\frac{dy}{dx}\) = –\(\frac{8}{x^{3}}\) ⇒ f'(x₁) = –\(\frac{8}{x^{3}}\)
y(x² + 4) = 8 ⇒ y = \(\frac{8}{x^{2}+4}\)

Question 8.
Show that the curves 6x² – 5x + 2y = 0 and 4x² + 8y² = 3 touch each other at (\(\frac{1}{2}\), \(\frac{1}{2}\))
Solution:
Equation of the first curve is
6x² – 5x + 2y = 0
2y = 5x – 6x²
2.\(\frac{dy}{dx}\) = 5 – 12x
\(\frac{dy}{dx}=\frac{5-12x}{2}\)

Equation of the second curve is 4x² + 8y² = 3

∴ f'(x₁) = g'(x₁)
The given curves touch each other at P(\(\frac{1}{2}\), \(\frac{1}{2}\)).

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c)

I.

Question 1.
Find the lengths of subtangent and sub-normal at a point of the curve
Solution:
Equation of the curve is y = b. sin \(\frac{x}{a}\)
\(\frac{dy}{dx}\) = b. cos \(\frac{x}{a}.\frac{1}{a}=\frac{b}{a}\). cos \(\frac{x}{a}\)
Length of the sub-tangent = |\(\frac{y_{1}}{f^{\prime}\left(x_{1}\right)}\)|
= \(\frac{b \cdot \sin \frac{x}{a}}{\frac{b}{a} \cdot \cos \frac{x}{a}}=\left|a \cdot \tan \frac{x}{a}\right|\)
Length of the sub-normal = |y₁, f'(x₁)|

Question 2.
Show that the length of sub normal at any point on the curve xy = a2 varies as the cube of the ordinate of the point.
Solution:
Equation of the curve is xy = a²
y = \(\frac{a^{2}}{x}\)
\(\frac{dy}{dx}=\frac{-a^{2}}{x^{2}}\)
Length of the sub-normal = |y₁, f’ (x₁)|

\(\frac{y_{1}^{3}}{a^{2}}\) ∝ y³₁ = cube of the ordinate.

Question 3.
Show that at any point (x, y) on the curve y = be^x/a, the length of the sub¬tangent is a constant and the length of the sub normal is \(\frac{y^{2}}{a}\).
Solution:
Equation of the curve is y = b.e^x/a
\(\frac{a^{2}}{x}\) = b.e^x/a.\(\frac{1}{a}=\frac{y}{a}\)
Length of the sub tangent
= |\(\frac{y_{1}}{f^{\prime}\left(x_{1}\right)}\)| = \(\frac{y_{1}}{\left(\frac{y_{1}}{a}\right)}\) = a = constant
Length of the sub-normal
= |y₁f'(x₁)| = |y₁\(\frac{y_{1}}{a}\)| = \(\frac{y_{1}^{2}}{a}\).

II.

Question 1.
Find the value of k so that the length of the sub normal at any point on the curve xy^k = a^{k + 1} is a constant.
Solution:
Equation of the curve is x.y^k = a^{k + 1}
Differentiating w.r.to x
x.k.y^k-1 \(\frac{dy}{dx}\) + y^k.1 = 0
x.k.y^k-1 \(\frac{dy}{dx}\) = -y^k
\(\frac{dy}{dx}=\frac{-y^{k}}{k \cdot x \cdot y^{k-1}}=-\frac{y}{kx}\)
Length of the sub-normal = |y₁f'(x₁|

Length of the sub-normal is constant at any point on the curve is independent of x₁ and y₁
∴ \(\frac{y_{1}^{k+2}}{k \cdot a^{k+1}}\) is independent of x₁, y₁
⇒ k + 2 = 0 ⇒ k = -2

Question 2.
At any point t on the curve x = a (t + sin t), y = a (1 – cos t), find the lengths of tangent, normal, sub tangent and sub
normal.
Solution:
Equation of the curve is x = a (t + sin t), y = a (1 – cost)
\(\frac{dx}{dt}\) = a (1 + cos t), \(\frac{dy}{dt}\) = a. sin t

Length of the tangent = |y₁|\(\sqrt{1+\frac{1}{\left[f^{\prime}\left(x_{1}\right)\right]^{2}}}\)

Length of the normal = |y₁|\(\sqrt{1+\left[f\left(x_{1}\right)\right]^{2}}\)

Length of the sub tangent.

= |a (2 sin t/2. cos t/2|
= |a. sin t|
Length of the sub normal
= |y₁.f'(x₁)| = |a(1 – cos t).\(\frac{\sin t / 2}{\cos t / 2}\)|
= |2a sin² t/2. tan t/2|
= |2a sin² t/2. tan t/2|

Question 3.
Find the lengths of normal and sub normal at a point on the curve y = \(\frac{a}{2}\) (e^x/a + e^-x/a).
Solution:
Equation of the curve is y = \(\frac{a}{2}\) (e^x/a + e^-x/a)
= a. cosh (\(\frac{x}{a}\))
Length of the normal = |y₁|\(\sqrt{1+\left[f\left(x_{1}\right)\right]^{2}}\)
= |a.cosh \(\frac{x}{a}\)|\(\sqrt{1+\sinh ^{2} \frac{x}{a}}\)
= a. cosh \(\frac{x}{a}\). cosh \(\frac{x}{a}\) = a. cosh² \(\frac{x}{a}\)
Length of the sub normal = |y₁ f'(x₁)|

Question 4.
Find the lengths of subtangent, sub normal at a point t on the curve x = a (cos t + t sin t), y=a (sin t – t cos t)
Solution:
Equations of the curve are x = a (cos t + t sin t)
\(\frac{dx}{dt}\) = a (- sin t + sin t +1 cost) = at cos t
y = a (sin t – t cos t)
\(\frac{dy}{dt}\) = a (cos t – cos t +1 sin t) = a t sin t

Length of the sub tangent

= |a cot t (sin t – t cos t)|
Length of the sub-normal = |y₁.f'(x₁)|
= |a (sin t – t cos t) tan t|
= |a tan t (sin t – t cos t)|