Class 12

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 6 Application of Derivatives Ex 6.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5

Question 1.
Find the maximum and minimum values, if any, of the following functions given by
(i) = (2x – 1)² + 3
(ii) f(x) = 9x² + 12x + 2
(iii) f(x) = – (x – 1)² + 10
(iv) g(x) = x³ + 1
Solution.
(i) The given function is f(x) = (2x – 1)² + 3.
It can be observed that (2x – 1)² ≥ 0 for every x ∈ R.

Therefore, f(x) = (2x – 1)² + 3 ≥ 3 for every x ∈ R.
The minimum value of f is attained when 2x – 1 = 0
⇒ x = \(\frac{1}{2}\)
∴ Minimum value of f = f(\(\frac{1}{2}\))
= (2 . \(\frac{1}{2}\) – 1)² + 3 = 3
Hence, function f does not have a maximum value.

(ii) The given function is f(x) = 9x² + 12x + 2
= (3x + 2)² – 2
It can be observed that (3x + 2)² ≥ 0 for every x ∈ R.
Therefore, f(x) = (3x + 2)² – 2 ≥ – 2 for every x ∈ R.
The minimum value of f is attained when 3x +2 = 0
⇒ x = \(\frac{-2}{3}\)
∴ Minimum value of f = f(\(\frac{-2}{3}\))= (3 . (\(\frac{- 2}{3}\)) + 2)² – 2 = – 2.
Hence, function f does not have a maximum value.

(iii) The given function is f(x) = -(x – 1)² + 10.
It can be observed that (x – 1)² ≥ 0 for every x ∈ R.
Therefore, f(x) = – (x – 1)² + 10 ≤ 10 for every x ∈ R.
The minimum value off is attained when (x – 1) = 0
⇒ x = 1
∴ Minimum value of f = f(1) = -(1 – 1)² + 10 = 10
Hence, function f does not have a minimum value.

(iv) The given function is g(x) = x³ + 1.
We observe that the value of f(x) increases when the value of x increases and f(x) can be made as large as we please by giving large value to x. So, f(x) does not have the maximum value.
Similarly, f(x) can be made as small as we please by giving smaller values of x.
So, f(x) does not have the minimum value.

Question 2.
Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = |x + 2| – 1
(ii) g(x) = – |x + 1| + 3
(iii) h(x) = sin (2x) + 5
(iv) f(x) = |sin 4x| + 3|
(v) h(x) = x + 1, x e (-1, 1)
Solution.
(i) Given, f(x) = |x + 2 | – 1
We know that, | x + 2 | ≥ 0 for every x ∈ R.
Therefore, f(x) = |x + 2| – 1 ≥ – 1 for every x ∈ R.
The minimum value of / is attained when | x + 2| = 0
⇒ x = – 2
∴ Minimum value of f = f(- 2) = = |- 2 + 2| – 1 = – 1
Hence, function f does not have a maximum value.

(ii) Given, g(x) = – | x + 1| + 3
We know that -| x + 1| < 0 for every x ∈ R.
Therefore, g(x) = -|x + 1| + 3 ≤ 3 for every x ∈ R.
The maximum value of g is attained when |x + 1| = 0
⇒ x = – 1
∴ Maximum value of g = g(- 1) = -|- 1 + 1| + 3 = 3
Hence, function g does not have a minimum value.

(iii) Given, h(x) = sin 2x + 5
We know that, – 1 ≤ sin 2x ≤ 1
⇒ – 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5
⇒ 4 ≤ sin 2x + 5 ≤ 6
Hence, the maximum and minimum values of h are 6 and 4, respectively.

(iv) Given,f(x) = |sin 4x + 3|
We know that, – 1 < sin 4x < 1
⇒ 2 ≤ sin 4x + 3 ≤ 4
2 ≤ |sin 4x + 3| ≤ 4
Hence, the maximum and minimum values of f are 4 and 2, respectively.

(v) h(x) = x + 1, x ∈ (- 1, 1)
Here, if a point x₀ is closest to – 1, then we find \(\frac{x_{0}}{2}\) + 1 < x₀ ∈ (- 1, 1).
Also, if x₁ is closest to 1, then
x₁ + 1 < \(\frac{x_{1}+1}{2}\) + 1 for all x₁ ∈ (- 1, 1).
Hence, function h(x) has neither maximum nor minimum value in (- 1, 1).

Question 3.
Find the local maxima and local minima, if any, of the following functions, Find also the local maximum and the local minimum values, as the case may be :
(i) f(x) = x²
(ii) g(x) = x³ – 3x
(iii) h(x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\)
(iv) f(x) = sin x – cos x, 0 < x < 2π
(v) f(x) = x³ – 6x² + 9x + 15
(vi) g(x) = \(\frac{x}{2}+\frac{2}{x}\), x > 0
(vii) g(x) = \(\frac{1}{x^{2}+2}\)
(viii) f(x) = \(x \sqrt{1-x}\), x > 0
Solution.
(i) Given, f(x) = x²
⇒ f'(x) = 2x
Now, f'(x) = 0
⇒ x = 0
Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.
We have f”(0) = 2, which is positive.
Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f(0) = 0.

(ii) Given, g(x) = x³ – 3x
⇒ g'(x) = 3x² – 3
Now, g'(x) = 0
⇒ 3x² = 3
⇒ x = ± 1
∴ g'(x) = 6x
⇒ g'(1) = 6 > 0
⇒ g'(- 1) = – 6 > 0
By second derivative test x = 1 is a point of local minima and local minimum value of g at x = 1 is
g(1) = 1³ – 3
= 1 – 3 = – 2.
However, x = – 1 is a point of local maxima and local maximum value of g at x = – 1 is
g(- 1) = (- 1)³ – 3(- 1)
= – 1 + 3 = 2.

(iii) Given, h(x) = sinx + cos x, 0 < x < \(\frac{\pi}{2}\)
∴ h'(x) = – sin x – cos x = -(sin x + cos x)
h'(x) = 0
⇒ sin x = cos x
⇒ tan x = 1
⇒ x = \(\frac{\pi}{4}\) ∈ (0, \(\frac{\pi}{2}\))
h”(x) = – sin x – cos x
= – (sin x + cos x)
h(\(\frac{\pi}{4}\)) = – \(\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0\)
Therefore, by second derivative test, x = \(\frac{\pi}{4}\) is a point of local maxima and
and the local maximum value of h at x = \(\frac{\pi}{4}\) is h(\(\frac{\pi}{4}\)) = sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) = √2.

(iv) f(x) = sin x – cos x, 0 < x < 2π
∴ f'(x) = cos x + sin x f'(x) = 0
⇒ cos x = – sin x ⇒ tan x = 1
⇒ x = \(\frac{3 \pi}{4}\), \(\frac{7 \pi}{4}\) ∈ (0, 2π)
f”(x) = – sin x + cos x f”(\(\frac{3 \pi}{4}\)) = \(-\sin \frac{3 \pi}{4}+\cos \frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}>0\)

f”(\(\frac{7 \pi}{4}\)) = \(-\sin \frac{7 \pi}{4}+\cos \frac{7 \pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0\)

Therefore, by second derivative test, x = \(\frac{3 \pi}{4}\) is a point of local maxima and the local maximum value of f at x = \(\frac{3 \pi}{4}\) is
f(\(\frac{3 \pi}{4}\)) = sin \(\frac{3 \pi}{4}\) – cos \(\frac{3 \pi}{4}\)
= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) = √2.

However, x = \(\frac{7 \pi}{4}\) is a point of local minima and the local minimum value of f at x = \(\frac{7 \pi}{4}\) is
f(\(\frac{7 \pi}{4}\)) = sin \(\frac{7 \pi}{4}\) – cos \(\frac{7 \pi}{4}\)
= \(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}\)

(v) Given, f(x) = x³ – 6x² + 9x + 15
f'(x) = 3x² – 12x + 9
f'(x) = 0
⇒ 3(x² – 4x + 3) = 0
⇒ 3(x – 1) (x – 3) = 0
⇒ x = 1, 3
Now, f'(x) = 6x – 12 = 6 (x – 2)
f'(1) = 6(1 – 2) = – 6 < 0 ⇒ f'(3) = 6(3 – 2) = 6 > 0
Therefore, by second derivative test, x = 1 is a point of local maxima and the local maximum value of f at x = 1 is
f(1) = 1 – 6 + 9 + 15 = 19.
However, x = 3 is a point of local minima and the local minimum value of f at x = 3 is
f(3) = 27 – 54 + 27 + 15 = 15

(vi) Given, g(x) = \(\frac{x}{2}+\frac{2}{x}\), x > 0

∴ g'(x) = \(\frac{1}{2}-\frac{2}{x^{2}}\)

Now, g'(x) = 0 gives \(\frac{2}{x^{2}}=\frac{1}{2}\)
⇒ x² = 4
⇒ x = ± 2
Therefore, by second derivative test, x = 2 is a point of local minima and the local minimum value of g at x = 2 is
g(2) = \(\frac{2}{2}+\frac{2}{2}\) = 1 + 1 = 2.

(vii) Given, g(x) = \(\frac{1}{x^{2}+2}\)

∴ g'(x) = \(\frac{-(2 x)}{\left(x^{2}+2\right)^{2}}\)
Now, g'(x) = 0
⇒ \(\frac{-2 x}{\left(x^{2}+2\right)^{2}}\) = 0
⇒ x = 0
Now, for values close to x = 0 and to the left of 0, g’(x) > 0.
Also, for values close to x = 0 and to the right of 0, g’(x) < 0.
Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of g(0) is \(\frac{1}{0+2}=\frac{1}{2}\).

(viii) Given, f(x) = x\(\sqrt{1-x}\)

Therefore, by second derivative test, x = \(\frac{2}{3}\) is a point of local maxima and the local maximum value of f at x = \(\frac{2}{3}\) is
f(\(\frac{2}{3}\)) = \(\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \sqrt{\frac{1}{3}}=\frac{2}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{9}\).

Question 4.
Prove that the following functions do not have maxima or minima:
(i) f(x) = e^x
(ii) g(x) = log x
(iii) h(x) = x³ + x² + x + 1
Solution.
(i) Given function is f(x) = e^x
⇒ f'(x) = e^x
Now, if f'(x) = 0, then e^x = 0.
But, the exponential function can never assume 0 for any value of x. ,
Therefore, there does not exist c ∈ R such that f'(c) = 0.
Hence, function f does not have maxima or minima.

(ii) Given function is g(x) = log x
⇒ g'(x) = \(\frac{1}{x}\)
Since, log x is defined for a positive number x, g’ (x) > 0 for any x.
Therefore, there does not exist ce R such that g’ (c) = 0.
Hence, function g does not have maxima or minima.

(iii) Given function is h(x) = x³ + x² + x + 1
⇒ h'(x) = 3x + 2x + 1
Now, h(x) = 0
⇒ 3x² + 2x +1 = 0
⇒ x = \(\frac{-2 \pm 2 \sqrt{2} i}{6}=\frac{-1 \pm \sqrt{2} i}{3}\) ∉ R
Therefore, there does not exist c ∈ R such that h'(c) = 0.
Hence, function h does not have maxima or minima.

Question 5.
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals :
(i) f(x) = x³, x ∈ [- 2, 2]
(ii) f(x) = sin x + cos x, x ∈ [0, π]
(iii) f(x) = 4x – \(\frac{1}{2}\) x², x ∈ [- 2, \(\frac{9}{2}\)]
(iv) f(x) = (x – 1)² + 3, x ∈ [- 3, 1]
Solution.
(i) The given function is f(x) = x³
∴ f'(x) = 3x²
Now, f(x) = 0
⇒ x = 0
Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [- 2, 2].
∴ f(0) = 0; f(- 2) = (- 2) 3 = – 8; 012f(2) = (2)³ – 8
Hence, we can conclude that the absolute maximum value of f on [- 2, 2] is 8 occurring at x = 2.
Also, the absolute minimum value of f on [- 2, 2] is – 8 occurring at x = – 2.

(ii) The given function is f(x) = sin x + cos x.
f'(x) = cos x – sin x
Now, f'(x) = 0
⇒ sin x = cos x
⇒ tan x = 1
⇒ x = \(\frac{\pi}{4}\)
Then, we evaluate the value of f at critical point x = \(\frac{\pi}{4}\) and at the end points of the interval [0, π].

∴ 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(0) = sin 0 + cos 0 = 0 + 1 = 1
f(π) = sin π + cos π = 0 – 1 = – 1
Hence, we can conclude that the absolute maximum value of f on [0, π] is 2 occurring at x = \(\frac{\pi}{4}\) and the absolute minimum value of f on [0, π] is – 1 occurring at x = π.

(iii) The given function is f(x) = 4x – \(\frac{1}{2}\) x²
f(x) = 4 – \(\frac{1}{2}\) (2x) = 4 – x
Now, f'(x) = 0
⇒ x = 4
Then, we evaluate the value of f at critical point x = 4 and at the end [- 2, \(\frac{9}{2}\)]
∴ f(4) = 16 – \(\frac{1}{2}\) (16) = 16 – 8 = 8

f(- 2) = – 8 – \(\frac{1}{2}\) (4) = – 8 – 2 = – 10

f(\(\frac{1}{2}\)) = \(4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^{2}=18-\frac{81}{8}\)

= 18 – 10.125 = 7.875
Hence, we can conclude that the absolute maximum value of f on is 8 occurring at x = 4 and the absolute minimum value of f on – 10 occurring at x = – 2.

(iv) The given function is f(x) = (x – 1)² + 3
∴ f'(x) = 2(x – 1)
Now, f'(x) = 0
⇒2 (x – 1) = 0
⇒ x = 1
Then, we evaluate the value of f at critical point x = 1 and at the end points of the interval [- 3, 1].
∴ f(1) = (1 – 1)² + 3 = 0 + 3 = 3;
f(- 3) = (- 3 – 1)² + 3 = 16 + 3 = 19
Hence, we can conclude that the absolute maximum value of / on [- 3, 1] is 19 occurring at x = – 3 and the minimum value of f on [- 3, 1] is 3 occurring at x = 1.

Question 6.
Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 – 72x – 18x²
Solution.
The profit function is given as p(x) = 41 – 72x – 18x²
On differentiating both sides w.r.t. x, we get
p'(x) = – 72 – 36x = – 36 (2 + x)
Again, differentiating both sides w.r.t. x, wt ,et
p”(x) = – 36
For maxima or minima, put p(x) = 0
⇒ – 36 (2 + x) = 0
⇒ x + 2 = 0
⇒x = – 2
At x = – 2, p'(-2) = – 36 < 0
∴ x = – 2 is a point of maxima.
Maximum profit, p(- 2) = 41 – 72 (- 2) – 81 (- 2)²
= 41 + 144 – 72 = 113
Hence, the maximum profit that a company can make, is 113 units.

Question 7.
Find both the maximum value and the minimum value of 3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0, 3].
Solution.
Let f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25
⇒ f'(x) = 12x³ – 24x² + 24x – 48
= 12 (x³ – 2x² + 2x – 4)
= 12 (x² (x – 2) + 2 (x – 2))
= 12(x – 2) (x² + 2)
For maxima or minima put f'(x) = 0
12(x – 2) (x² + 2) = 0
⇒ if x – 2 = 0 ⇒ x = 2 e [0, 3]
And if, x² + 2 = 0
⇒ x² = – 2
⇒x = √- 2
Hence, the only real root is x = 2 which is considered as critical point.
Now, we evaluate the value of f at critical point x = 2 and at the end point of the interval [0, 3].
At x = 2, f(2) = 3 × 2⁴ – 8 × 2³ + 12 × 2² – 48 × 2 + 25
= 48 – 64 + 48 – 96 + 25 = – 39

At x = 0, f(0) = 0 – 0 + 0 – 0 + 25 = 25
At x = 3, f(3) = 3 × 3⁴ – 8 × 3³ + 12 × 3² – 48 × 3 + 25
= 243 – 216 + 108 – 144 + 25 = 16
Hence, we can conclude that the absolute maximum value of f is 25 at x = 0 and the absolute minimum value of f is – 39 at x = 2.

Question 8.
At what points in the interval [0, 2π], does the function sin2x attain its maximum value?
Solution.
Let f(x) = sin 2x
⇒ f'(x) = 2 cos 2x
Now, f'(x) = 0
⇒ cos 2x = 0
⇒ 2x = \(\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}\)

⇒ x = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\)

Then, we evaluate the value off at critical points x = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\) and at the end points of the interval [0, 2π].
f(\(\frac{\pi}{4}\)) = sin \(\frac{\pi}{2}\) = 1;

f(\(\frac{3 \pi}{4}\)) = sin \(\frac{3 \pi}{2}\) = – 1;

f(\(\frac{5 \pi}{4}\)) = sin \(\frac{5 \pi}{2}\) = 1;

f(\(\frac{7 \pi}{4}\)) = sin \(\frac{7 \pi}{2}\) = – 1;

f(0) = sin 0 = 0;
f(2π) = sin 2π = 0
Hence, we can conclude that the absolute maximum value of f on [0, 2π] is occurring at x = \(\frac{\pi}{4}\) and x = \(\frac{5 \pi}{4}\).

Question 9.
What is the maximum value of the function sin x + cos x?
Solution.
Let f(x) = sin x + cos x
f”(x) = – sin x – cos x = – (sin x + cos x)
Now, f”(x) will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both positive.
Also, we know that, sin x and cos x both are positive in the first quadrant.
Then, f”(x) will be negative when x ∈ (0, \(\frac{\pi}{2}\))
Thus, we consider x = \(\frac{\pi}{4}\)

f”(\(\frac{\pi}{4}\)) = – (sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\))

= –\(\left(\frac{2}{\sqrt{2}}\right)\) = – √2
∴ By second derivative test, f will be the maximum at x = \(\frac{\pi}{4}\) and the maximum value of f is
f(\(\frac{\pi}{4}\)) = sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\).

Question 10.
Find the maximum value of 2x³ – 24x + 107in the interval [1, 3]. Find the maximum value of the same function in [- 3, – 1].
Solution.
Let f(x) = 2x³ – 24x + 107
⇒ f(x) = 6x² – 24
= 6(x² – 4)
Now, f'(x) = 0
⇒ 6 (x² – 4) = 0
⇒ x² = 4
⇒ x = ± 2
We first consider the interval [1, 3]
Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at the end points of the interval [1, 3].
f(2) = 2(8) – 24(2) +107 = 16 – 48 +107 = 75
f(1) = 2(1)-24(1)+ 107 = 2 – 24 +107 = 85
f(3) = 2(27) – 24(3) +107 = 54 – 72 +107 = 89
Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.
Next, we consider the interval [- 3, – 1],
Evaluate the value of f at the critical point x = – 2 ∈ [- 3, – 1] and at the end points of the interval [1, 3].
f(- 3) = 2(- 27) – 24(- 3) +107 = – 54 + 72 +107 = 125
f(- 1) = 2(- 1) – 24(- 1) + 107 = – 2 + 24 +107 = 129
f(- 2) = 2(- 8) – 24(- 2) + 107 = – 16 + 48 +107 = 139
Hence, the absolute maximum value of f(x) in the interval [- 3, – 1] is 139 occurring at x = – 2.

Question 11.
It is given that at x = 1, the function x⁴ – 62x² + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.
Solution.
Let f(x) = x⁴ – 62x² + ax + 9
f'(x) = 4x³ – 124x + a
It is given that function f attains its maximum value on the interval [0, 2] at x = 1.
∴ f'(1) = 0
⇒ 4 – 124 + a = 0
⇒ a = 120
Hence, the value of a is 120.

Question 12.
Find the maximum and minimum values of x + sin 2x on [0, 2π].
Solution.
Let f(x) = x + sin 2x
⇒ f'(x) = 1 + 2 cos 2x
Now, f'(x) = 0
⇒ cos 2x = – \(\frac{1}{2}\) = – cos \(\frac{\pi}{3}\)

= cos (π – \(\frac{\pi}{3}\))

= cos \(\frac{2 \pi}{3}\)

2x = 2n ± \(\frac{2 \pi}{3}\), n ∈ Z

x = nπ ± \(\frac{\pi}{3}\), n ∈ Z
⇒ x = \(\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}\) ∈[0, 2π]

Then, we evaluate the value of f at critical points x = \(\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}\) and at the end points of the interval [0, 2π],
∴ f(\(\frac{\pi}{3}\)) = \(\frac{\pi}{3}+\sin \frac{2 \pi}{3}=\frac{\pi}{3}+\frac{\sqrt{3}}{2}\);

f(\(\frac{2 \pi}{3}\)) = \(\frac{2 \pi}{3}+\sin \frac{4 \pi}{3}=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\);

f(\(\frac{4 \pi}{3}\)) = \(\frac{4 \pi}{3}+\sin \frac{8 \pi}{3}=\frac{4 \pi}{3}+\frac{\sqrt{3}}{2}\);

f(\(\frac{5 \pi}{3}\)) = \(\frac{5 \pi}{3}+\sin \frac{10 \pi}{3}=\frac{5 \pi}{3}-\frac{\sqrt{3}}{2}\);
f(0) = 0 + sin 0 = 0;
f(2π) = 2π + sin 4π = 2π + 0 = 2π
Thus, maximum value is 2π at x = 2π and minimum value is 0 at x = 0.

Question 13.
Find two numbers whose sum is 24 and whose product is as large as possible.
Solution.
Let one number be x.
Then, the other number is (24 – x).
Let f(x) denote the product of the two numbers.
Thus, we have P(x) = x(24 – x) = 24x – x²
P'(x) = 24 – 2x;
P”(x) = – 2
Now, P’ (x) = 0
⇒ x =12
Also, P”(12) = – 2 < 0
By second derivative test, x = 12 is the point of local maxima of P.
Hence, the product of the number is the maximum when the numbers are 12 and 24 – 12 = 12.

Question14.
Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
Solution.
The two numbers are x and y such that x + y – 60
⇒ y = 60 – x
Let f(x) = xy³
⇒ f(x) = x(60 – x)³
⇒ f'(x) = (60 – x)³ – 3x (60 – x)²
= (60 – x)² [60 – x – 3x]
= (60 – x)² (60 – 4x)
Also, f(x) = – 2 (60 – x) (60 – 4x) – 4 (60 – x)²
= – 2 (60 – x) [60 – 4x + 2(60 – x)]
= – 2 (60 – x) (180 – 6x)
= -12(60 – x) (30 – x)
Now, f'(x) = 0
⇒ x = 60 or x = 15
When, x = 60 then, f”(x) = 0
When x = 15, then f'(x) – 12(60 – 15) (30 – 15) = – 12 × 45 × 15 < 0 .
∴ By second derivative test, x = 15 is a point of local maxima of f.
Thus, function xy³ is maximum when x = 15 and y = 60 -15 = 45.
Hence, the required numbers are 15 and 45.

Question 15.
Find two positive numbers x and y such that their sum is 35 and the product x²zy⁵ is a maximum.
Solution.
Let one number be x. Then, the other number is y = (35 – x).
Let P(x) = x²y⁵, then
P(x) = x² (35 – x)⁵
∴ P'(x) = 2x (35 – x)⁵ – 5x² (35 – x)⁴
= x (35 – x)⁴ [2(35 – x) – 5x]
= x (35 – x)⁴ (70 – 7x)
= 7x (35 – x)4^{4/sup> (10 – x)}

And p”(x) = 7(35 – x)⁴ (10 – x) + 7x [- (35 – x)⁴ – 4(35 – x)³ (10 – x)]
= 7 (35 – x)⁴ (10 – x) – 7x (35 – x)⁴ – 28x (35 – x)³ (10 – x)
= 7 (35 – x)³ [(35 – x)(10 – x) – x(35 – x) – 4x(10 – x)]
= 7 (35-x)³ [350 – 45x + x² – 35x + x² – 40x + 4x²]
= 7 (35 – x)³ (6x² – 120x + 350)
Now, P'(x) = 0
⇒ x = 0, x = 35, x = 10
When, x = 35, f'(x) = f(x) = 0 and y = 35 – 35 = 0.
This will make the product x²y⁵ equal to 0.
When x = 0, then y = 35 -0 = 35 and the product x2y2 will be 0.
∴ x = 0 and x = 35 cannot be the possible values of x.
When x = 10 then, we have
P'(x) = 7(35 – 10)³ (6 × 100 – 120 × 10 + 350)
= 7(25)³ (- 250) < 0
∴ By second derivative test, P(x) will be the maximum when x = 10 and y = 35 – 10 = 25.
Hence, the required numbers are 10 and 25.

Question 16.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Solution.
Let one number be x.
Then, the other number is (16 – x).
Let the sum of the cubes of these numbers be denoted by S(x). Then, S(x) = x3 + (16 -x)3
∴ S'(x) = 3x² – 3(16 – x)²;
S'(x) = 6x + 6(16 – x)
Now, S'(x) = 0
⇒ 3x² – 3(16 – x)² = 0
⇒ x² – (16 – x)² = 0
⇒ x² – 256 – x² + 32x = 0
⇒ x =\(\frac{256}{32}\) = 8
Now, S'(8) = 6(8) + 6(16 -8) = 48 + 48 = 96 > 0.
By second derivatives test, x = 8 is the point of local minima of S.
Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 – 8 = 8.

Question 17.
A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Solution.

Let the side of the square to be cut-off be x cm (0 < x < 9).
Then, the length and the breadth of the box will be (18 – 2x) cm each and the height of the box is x cm.
Let V the volume of the open box formed by folding up the flaps, then
V = x (18 – 2x) (18 – 2x)
= 4x (9 – x)²
= 4x (81 + x² – 18x)
= 4(x³ – 18x² + 81x)
On differentiating twice w.r.t. x, we get
\(\frac{d V}{d x}\) = 4(3x² – 36x + 81)
= 12(x² – 12x+27)
and \(\frac{d^{2} V}{d x^{2}}\) = 12 (2x – 12) = 24 (x – 6)
For maxima put \(\frac{d V}{d x}\) = 0
⇒ 12(x – 12x + 27) = 0
⇒ x² – 12x + 27 = 0
⇒ (x – 3) (x – 9) = 0
⇒ x = 3, 9
But x = 9 is not possible,
∵ 2x = 2 × 9 = 18
Which is equal to side of square piece.
At x = 3,
\(\left(\frac{d^{2} V}{d x^{2}}\right)_{x=3}\) = 24 (3 – 6) = – 72 < 0

∴ By second derivative test, x = 5 is the point of maxima.
Hence, the side of the square to be cut-off to make the volume of the box maximum possible is 5 cm.

Question 18.
A rectangular sheet of tin 45cm by 24cm is to be made into a box without top, by cutting off square from each comer and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Solution.

Let the side of the square to be cut off be x cm.
Then, the height of the box is x, the length is 45 – 2x and the breadth is 24 f'(x) 2x.
Let V be the corresponding volume of the box then,
V = x (24 – 2x) (45 – 2x)
V = x (4x² – 138x + 1080)
=4x³ – 138 x² + 1080x
On differentiating twice w.r.t. x,

\(\frac{d V}{d x}\) = 12x² – 276x + 1080

\(\frac{d^{2} V}{d x^{2}}\) = 24x – 276
For maxima put \(\frac{d V}{d x}\) = o
⇒ 12x² – 276x + 1080 = 0
⇒ x² – 23x + 90 = 0
= (x – 18) (x – 5) = 0
⇒ x = 5, 18
It is not possible to cut-off a square of side 18 cm from each corner of the rectangular sheet.
Thus, x cannot be equal to 18.
At x = 5,
\(\left(\frac{d^{2} V}{d x^{2}}\right)_{x=5}\) = 24 × 5 – 276
= 120 – 276
= – 156 < 0
∴ By second derivative test, x = 5 is the point of maxima.
Hence, the side of the square to be cut-off to make the volume of the box maximum possible is 5 cm.

Question 19.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Solution.
Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.
Then, the diagonal passes through the centre and is of length 2a cm.

Now, by applying the Pythagoras theorem, we have (2 a)² = l² + b²
⇒ b² = 4 a² – l²
⇒ b = \(\sqrt{4 a^{2}-l^{2}}\)
∴ Area of rectangle, A = l \(\sqrt{4 a^{2}-l^{2}}\)

∴ By the second derivative test, when l = √2a then the area of the rectangle is the maximum.
Since, l = b = √2a, therefore the rectangle is a square.
Hence, it has been proved that of all the rectangles inscribed in the given fixed circle the square has the maximum area.

Question 20.
Show that th right circular cylinder of given surface and maximum volume Is such that its heigh is equal to the diameter of the base.
Solution.

Let r and h be the radius and height of the cylinder, respectively.
Then, the surface area (S) of the cylinder is given by
S = 2πr² + 2πrh
⇒ h = \(\frac{S-2 \pi r^{2}}{2 \pi r}=\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r\)

Let V be the volume of the cylinder. Then,

V = πr²h
= πr² \(\left[\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r\right]=\frac{S r}{2}-\pi r^{3}\)

On differentiating w.r.t. x, we get
\(\frac{d V}{d r}=\frac{S}{2}-3 \pi r^{2}\)

For maxima or mmima put \(\frac{d V}{d r}\) = 0
\(\frac{S}{2}\) = 3πr²
r² = \(\frac{S}{6 \pi}\)

Now, \(\frac{d^{2} V}{d r^{2}}\) = – 6π \(\left(\sqrt{\frac{S}{6 \pi}}\right)\) < 0

∴ By second derivative test, the volume is the maximum when r² = \(\frac{S}{6 \pi}\).

Now, when r² = \(\frac{S}{6 \pi}\), then h = \(\frac{6 \pi r^{2}}{2 \pi}\left(\frac{1}{r}\right)\) – \(\frac{1}{r}\) = 3r – r = 2r.
Hence, the volume is the maximum when the height is twice the radius si.e., when the height is equal to the diameter.

Question 21.
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic cm, find the dimensions of the can which has the minimum surface area?
Solution.

 

Let r and h be the radius and height of the cylinder, respectively.
Then, volume (V) of the cylinder is given by
V = πr²h = 100
∴ h = \(\frac{100}{\pi r^{2}}\)
Surface area (S) of the cylinder is given by S = 2πr² + 2πrh = 2πr² + \(\frac{200}{r}\)
On differentiating w.r.t. x, we get h
∴ \(\frac{d S}{d r}\) = 4πr – \(\frac{200}{r^{2}}\)
Now, for maxima or minima put \(\frac{d S}{d r}\) = 0

⇒ 4πr = \(\frac{200}{r^{2}}\)

⇒ r³ = \(\frac{200}{4 \pi}=\frac{50}{\pi}\)

⇒ r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\)

Again, differentiating w.r.t. x, we get.
\(\frac{d^{2} S}{d r^{2}}\) = 4π + \(\frac{400}{r^{3}}\)

Now, it is observed that when r = latex]\left(\frac{50}{\pi}\right)^{\frac{1}{3}}[/latex], \(\frac{d^{2} S}{d r^{2}}\) > 0.

∴ By second derivative test, the surface area is the minimum when the radius of the cylinder is \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm.

when r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\), then h = \(\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{2}{3}}}=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm

Hence, the required dimensions of the can which has the minimum surface area, is given by radius = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm and height = 2 \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm.

Question 22.
A wire of length 28 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 length of the two pieces so that the combined area of the square and the circle is minimum?
Solution.
Let a piece of length l be cut from the given wire to make a square.
Then, the other piece of wire to be made into a circle is of length (28 – l) m.
Now, side of square = \(\frac{l}{4}\)
Let r be the radius of the circle. Then, 2πr = 28 – l
The combined area of the square and the circle (A) is given by A = (side of the square)² + πr²

∴ By second derivative test, the area (A) is the minimum when l = \(\frac{112}{\pi+4}\)

Hence, the combined area is the minimum when the length of the wire in making the square is \(\frac{112}{\pi+4}\) cm while the length of the wire in making the circle is 28 – \(\frac{112}{\pi+4}=\frac{28 \pi}{\pi+4}\) cm.

Question 23.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R, is \(\frac{8}{27}\) of the volume of the sphere.
Solution.
Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.
Let V be the volume of the cone.
Then, V = \(\frac{1}{3}\) πr²h
Height of the cone is given by h = R + AB
= R + \(\sqrt{R^{2}-r^{2}}\) [∵ ABC is right angled triangle]

For maxima put
\(\frac{d V}{d r}\) = 0

⇒ \(\frac{2}{3}\) πrR = \(\frac{3 \pi r^{3}-2 \pi r R^{2}}{3 \sqrt{R^{2}-r^{2}}}\)

⇒ 2R = \(\frac{3 r^{2}-2 R^{2}}{\sqrt{R^{2}-r^{2}}}\)

⇒ 2R \(\sqrt{R^{2}-r^{2}}\) = (3r² – 2R²)²
⇒ 4R² (R² – r²)
= (3r² – 2R²)²
⇒ 4R⁴ – 4R²r² = 9r⁴ + 4R⁴ – 12r²R²
⇒ 9r⁴ = 8R²r²
⇒ r² = \(\frac{8}{9}\) R²

∴ By second derivative test, the volume of the cone is the maximum when r² = \(\frac{8}{9}\) R².

When r² = \(\frac{8}{9}\) R², then
h = R + \(\sqrt{R^{2}-\frac{8}{9} R^{2}}=R+\sqrt{\frac{1}{9} R^{2}}=R+\frac{R}{3}=\frac{4}{3} R\)

Therefore, V = \(\frac{1}{3} \pi\left(\frac{8}{9} R^{2}\right)\left(\frac{4}{3} R\right)=\frac{8}{27}\left(\frac{4}{3} \pi R^{3}\right)\)

= \(\frac{8}{27}\) × (volume of the sphere)
Hence, the volume of the largest cone that can be inscribed in the sphere, is \(\frac{8}{27}\) of the volume of the sphere.

Question 24.
Show that the right circular cone of least curved surface and given volume has an altitude equal toJ time the radius of the base.
Solution.
Let r be the radius of the base, h be the height, V be the volume and S be the curved surface area of the cone.
Then, V = \(\frac{1}{3}\) πr²h
⇒ 3V = πr²h
⇒ 9V² = π² r⁴ h²
⇒ h² = \(\frac{9 V^{2}}{\pi^{2} r^{4}}\) ……………..(i)
And S = πrl
⇒ S = πr \(\sqrt{r^{2}+h^{2}}\) (∵ l = \(\sqrt{h^{2}+r^{2}}\))
⇒ S² = π² r² (r² + h²)
= π² r² (\(\frac{9 V^{2}}{\pi^{2} r^{4}}\) + r²) [Using Eq. (i)]

⇒ S² = \(\frac{9 V^{2}}{r^{2}}\) + π² r⁴ …………(ii)

Hence, S² and therefore S is minimum when 9V² = 2π²r⁶
On putting 9 V² = 2π²r⁶ in Eq. (i) we get
2π²r⁶ = π²r⁴h²
⇒ 2r² = h²
⇒ h = √2r
Hence, altitude of right circular cone is √2 times the radius of the base.

Question 25.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^-1 √2.
Solution.

Let θ be the semi-vertical angle of the cone.
It is clear that θ ∈ [0, \(\frac{\pi}{2}\)]
Let r, h and l be the radius, height and the slant height of the cone, respectively.
The slant height of the cone is given as constant.
Now, r = l sin θ and h = l cos θ
The volume (V) of the cone is given by V = \(\frac{1}{3}\) πr² h
⇒ \(\frac{1}{3}\) π (l² sin² θ) (l cos θ)
= \(\frac{1}{3}\) π l² sin² θ cos θ
On differentiating w.r.t. θ, we get
∴ \(\frac{d V}{d \theta}\) = \(\frac{l^{2} \pi}{3}\) [sin² θ (- sin θ) + cos θ (2 sin θ cos θ)]

= \(\frac{l^{3} \pi}{3}[\) [- sin³ θ + 2sin θ cos² θ]

Again, differentiating w.r.t. θ, we get
\(\frac{d^{2} V}{d \theta^{2}}\) = \(\frac{l^{3} \pi}{3}\) [- 3 sin² θ cos θ + 2 cos³ θ – 4 sin² θ cos θ]

= \(\frac{l^{3} \pi}{3}\) [2 cos³ θ – 7 sin² θ cos θ]
For maxima put \(\frac{d V}{d \theta}\) = 0
⇒ sin³ θ = 2 sin θ cos θ v d0
⇒ tan² θ = 2
⇒ tan θ = √2
⇒ θ = tan^-1 √2
Now, when θ = tan^-1 √2, then tan² θ = 2 or sin² θ = 2 cos² θ
Then, we have

\(\frac{d^{2} V}{d \theta^{2}}\) = \(\frac{l^{3} \pi}{3}\) [2 cos³ θ – 14 cos³ θ]
= – 4πl³ cos³ θ < 0 for θ ∈ [0, \(\frac{\pi}{2}\)]

∴ By second derivative test, the volume (V) is the maximum when θ = tan^-1 √2.
Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is tan^-1 √2.

Question 26.
Show that the semi-vertical angle of right circular cone of given surface area and maximum volume is sin^-1 (\(\frac{1}{3}\)).
Solution.
With usual notation, given that total surface area S = πrl + πr²

Direction (27 – 29): Choose the correct answer.

Question27.
The point on the curve x² = 2y which is nearest to the point (0, 5) is
(A) (2√2, 4)
(B) (2√2, 0)
(C) (0, 0)
(D) (2, 2)
Solution.
Let d be the distance of the point (x, y) on x² = 2y from the point (0, 5),
then
d = \(\sqrt{(x-0)^{2}+(y-5)^{2}}=\sqrt{x^{2}+(y-5)^{2}}\) ……………(i)

= \(\sqrt{2 y+(y-5)^{2}}\) [Putting x² = 2y]

= \(\sqrt{y^{2}-8 y+25}\)

= \(\sqrt{y^{2}-8 y+4^{2}+9}=\sqrt{(y-4)^{2}+9}\)

d is least when (y – 4)² = 0 i.e., when y = 4
when y = 4 then x² = 2 × 4
⇒ x = ± √8 = ± 2√2
∴ The points (2√2, 4) and (- 2√2, 4) on the given curve are nearest to the point (0, 5).
The correct answer is (A).

Question 28.
For all real values of x, the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}\) is
(A) 0
(B) 1
(C) 3
(D) \(\frac{1}{3}\)
Solution.

The correct answer is (D).

Question 29.
The maximum value of [x (x – 1) + 1]^{\(\frac{1}{3}\)},0 < x < 1 is
(A) \(\left(\frac{1}{3}\right)^{\frac{1}{3}}\)

(B) \(\frac{1}{2}\)

(C) 1
(D) 0
Solution.
Let f(x) = [x (x – 1) + 1]^{\(\frac{1}{3}\)}

∴ f'(x) = \(\frac{2 x-1}{3[x(x-1)+1]^{\frac{2}{3}}}\)

Now, f'(x) = 0
⇒ x = \(\frac{1}{2}\)
Then, we evaluate the value of f at critical point x = \(\frac{1}{2}\) and at the point of the interval [0, 1]
{i.e., at x = 0 and x = 1}.
f(0) = [0 (0 – 1) + 1]^{\(\frac{1}{3}\)} = 1;

f(1) = [1 (1 – 1) + 1]^{\(\frac{1}{3}\)} = 1

f(\(\frac{1}{2}\)) = \(\left[\frac{1}{2}\left(\frac{-1}{2}\right)+1\right]^{\frac{1}{3}}=\left(\frac{3}{4}\right)^{\frac{1}{3}}\)

Hence, we can conclude that the maximum value off in the interval [0, 1] is 1.
The correct answer is (C).

Punjab State Board Syllabus A Rainbow of English 12 Class PSEB Solutions Pdf, General English Class 12 PSEB Solutions, English Guide for Class 12 PSEB Pdf Free Download is part of PSEB Solutions for Class 12.

A Rainbow of English 12 Class Solutions PSEB | General English Class 12 PSEB Solutions

English Guide for Class 12 PSEB Pdf Free Download | General English Class 12 PSEB Book

A Rainbow of English 12 Class PSEB Solutions Pdf

A Rainbow of English 12 Class Guide Prose

• Chapter 1 Hassan’s Attendance Problem
• Chapter 2 The March King
• Chapter 3 Thinking Out of the Box: Lateral Thinking
• Chapter 4 Robots and People
• Chapter 5 On Giving Advice
• Chapter 6 On Saying “Please”
• Chapter 7 The Story of My Life
• Chapter 8 Two Gentlemen of Verona
• Chapter 9 In Celebration of Being Alive
• Chapter 10 Ghadari Babas in Kalapani Jail

A Rainbow of English 12 Class Solutions PSEB Poetry

• Poem 1 Prayer of the Woods
• Poem 2 On Friendship
• Poem 3 The Echoing Green
• Poem 4 Once upon a Time
• Poem 5 Cheerfulness Taught By Reason
• Poem 6 Father Returning Home
• Poem 7 The Road Not Taken
• Poem 8 On His Blindness

General English Class 12 PSEB Solutions Supplementary Reading

• Chapter 1 The School for Sympathy
• Chapter 2 A Chameleon
• Chapter 3 Bholi
• Chapter 4 The Gold Frame
• Chapter 5 The Barber’s Trade Union
• Chapter 6 The Bull beneath the Earth

PSEB 12th Class English Grammar & Composition

• Reading Comprehension Unseen Passages Type-I
• Reading Comprehension Unseen Passages Type-II

PSEB 12th Class English Grammar

• Determiners
• Use of Non-finites (Infinitives, Gerunds and Participles)
• Transformation of Sentences (Simple, Compound & Complex)
• Active and Passive Voice
• Narration / Direct and Indirect Speech
• Translation

PSEB 12th Class English Composition

• Precis Writing
• Application Writing
• Letter Writing Letters to Editor
• Letter Writing Business Letters
• E-Mail Writing
• Explaining Newspaper Headlines

General English Class 12 PSEB Syllabus

Class – XII
General English
(2021-22)

Time: 3hrs

Theory: 80 Marks
IA: 20 Marks
(Listening and Speaking skills-based practical: 18 marks and Book bank: 2 marks)
Total: 100 Marks

Syllabus

Unseen Passages For Testing Reading Skills
Text Book

Section A (Lessons for Intensive Study)

1. Hassan’s Attendance Problem – Sudha Murthy
2. The March King – Katherine Little Bakeless
3. Thinking Out of the Box: Lateral Thinking – Adapted from the article from the Internet
4. On Saying ‘Please’ – A. G. Gardiner
5. The Story of My Life – Helen Keller
6. Two Gentlemen of Verona – A. J. Cronin
7. In Celebration of Being Alive – Dr. Christian Barnard
8. Gadari Babas in Kalapani Jail – Dr. Harish Puri

Section B (Poetry)

1. Prayer of the Woods – Anonymous
2. On Friendship – Khalil Gibran
3. The Echoing Green – William Blake
4. Once upon a Time – Gabriel Okara
5. Father Returning home – Dilip Chitre
6. The Road Not Taken – Robert Frost
7. On His Blindness – John Milton

Section C (Lessons for Extensive Study)

1. The School for Sympathy – E. V. Lucas
2. A Chamelon – Anton Chekhov
3. Bholi – K.A. Abbas
4. The Gold Frame – R.K. Luxman
5. The Barber’s Trade Union – Mulk Raj Anand
6. The Bull beneath the Earth – K.S. Virk

Grammar, Composition & Translation

Grammar

1. Determiners
2. Use of Non-finites (Infinitives, Gerunds, Participles)
3. Transformation of Sentences
4. Voice
5. Narration

Composition

1. Precis writing
2. Letter writing (Official/Business/To Editors)
3. Applications
4. Explaining Newspaper Headlines
5. E-Mail writing

Translation from English to Hindi/Punjabi and Translation from Hindi/ Punjabi to English.

From Chapter 18 The Art of Translation given in the book English Grammar And Composition for XI and XII

Note: Following two lessons & one poem has been deleted from the syllabus from the academic session 2020-21 onwards.

• Robots and People – Isaac Asimov
• On Giving Advice – Joseph Addison
• Cheerfulness Taught by Reason – Elizabeth Barret Browning

Punjab State Board Syllabus PSEB 12th Class Economics Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 12.

PSEB 12th Class Economics Guide | Economics Guide for Class 12 PSEB

Economics Guide for Class 12 PSEB | PSEB 12th Class Economics Book Solutions

PSEB 12th Class Economics Book Solutions in Hindi Medium

• Chapter 1 समष्टि अर्थशास्त्र
• Chapter 2 समष्टि अर्थशास्त्र में मूल धारणाएं
• Chapter 3 आय का चक्रीय प्रवाह
• Chapter 4 राष्ट्रीय आय की अवधारणाएं
• Chapter 5 राष्ट्रीय आय का माप
• Chapter 6 मुद्रा, मुद्रा की पूर्ति तथा कार्य
• Chapter 7 बैंकिंग
• Chapter 8 कुल मांग
• Chapter 9 उपभोग प्रवृत्ति और बचत प्रवृत्ति
• Chapter 10 आय तथा रोजगार का निर्धारण
• Chapter 11 निवेश गुणक
• Chapter 12 अधिक तथा अभावी माँग की समस्याएँ
• Chapter 13 अभावी तथा अधिक मांग को ठीक करने के लिए उपाय
• Chapter 14 सरकारी बजट तथा अर्थव्यवस्था
• Chapter 15 भुगतान सन्तुलन
• Chapter 16 विदेशी विनिमय दर
• Chapter 17 स्वतन्त्रता के समय भारतीय अर्थव्यवस्था
• Chapter 18 आर्थिक नियोजन के उद्देश्य
• Chapter 19 भारत में कृषि
• Chapter 20 भारत में औद्योगिक विकास नीति तथा लाइसेसिंग
• Chapter 21 1991 से आर्थिक सुधार अथवा नई आर्थिक नीति
• Chapter 22 निर्धनता की समस्या
• Chapter 23 ग्रामीण ऋण
• Chapter 24 कृषि उपज का मण्डीकरण
• Chapter 25 सहकारिता, कृषि में विभिन्नता और जैविक खेती
• Chapter 26 आर्थिक विकास में मानव पूँजी का योगदान
• Chapter 27 भारत में शिक्षा, स्वास्थ्य तथा बेरोजगारी
• Chapter 28 बुनियादी ढांचा और ऊर्जा
• Chapter 29 धारणीय आर्थिक विकास, आर्थिक विकास का वातावरण पर प्रभाव तथा ग्लोबल वार्मिंग
• Chapter 30 सह-सम्बन्ध
• Chapter 31 सूचकांक

PSEB 12th Class Economics Book Solutions: Punjab Economy

• Chapter 32 पंजाब की मानवीय शक्ति और भौतिक साधन
• Chapter 33 1966 से पंजाब में कृषि का विकास
• Chapter 34 1966 से पंजाब में औद्योगिक विकास
• Chapter 35 पंजाब सरकार की आर्थिक स्थिति

PSEB 12th Class Economics Book Solutions in English Medium

• Chapter 1 Macro Economics
• Chapter 2 Basic Concepts in Macro Economics
• Chapter 3 Circular Flow of Income
• Chapter 4 Concepts of National Income
• Chapter 5 Measurement of National Income
• Chapter 6 Money, Supply of Money and Functions
• Chapter 7 Banking
• Chapter 8 Aggregate Demand
• Chapter 9 Propensity to Consume and Propensity to Save
• Chapter 10 Determination of Income and Employment
• Chapter 11 Investment Multiplier
• Chapter 12 Problems of Excess and Deficient Demand
• Chapter 13 Measures to Correct Deficient and Excess Demand
• Chapter 14 Government Budget and the Economy
• Chapter 15 Balance of Payments
• Chapter 16 Foreign Exchange Rate
• Chapter 17 Indian Economy on the Eve of Independence
• Chapter 18 Objectives of Economic Planning
• Chapter 19 Agriculture in India
• Chapter 20 Industrial Development Policy and Licensing in India
• Chapter 21 Economic Reforms since 1991 or New Economic Policy
• Chapter 22 Problem of Poverty
• Chapter 23 Rural Credit
• Chapter 24 Marketing of Agricultural Produce
• Chapter 25 Co-operatives, Diversification in Agriculture and Organic Farming
• Chapter 26 Role of Human Capital in Economic Development
• Chapter 27 Education, Health and Unemployment in India
• Chapter 28 Infrastructure and Energy
• Chapter 29 Sustainable Economic development, Effect of Economic Development on Environment and Global Warming
• Chapter 30 Correlation
• Chapter 31 Index Numbers

Punjab Economy

• Chapter 32 Man Power and Physical Resources of Punjab
• Chapter 33 Agriculture Development of Punjab Since 1966
• Chapter 35 Industrial Development of Punjab Since 1966
• Chapter 35 Financial Position of Punjab Government

PSEB 12th Class Economics Syllabus

Class – XII (PB.)
Economics
Time Allowed: 3 Hours

Theory: 80 Marks
Internal Assessment: 20 Marks
Marks Total: 100 Marks

Part – A
Introductory Macro Economics

Unit 1 National Income and Related Aggregates
What is Macro Economics? Classical and Keynesian views about Macro Economics. Scope, Importance, and Limitations of Macro Economics. Concept of Equilibrium: Partial Equilibrium and General Equilibrium. Basic concepts in Macro Economics: consumption goods, capital goods, final goods, intermediate goods, stock and flow variables, etc. Circular flow of income and output (two-sector economy model). Real flow and Monetary flow. Concept of Injections and withdrawals in Circular flow of Income and Output. Aggregates related to National Income: Gross National Product (GNP), Net National Product (NNP), Gross and Net Domestic Product (GDP and NDP) – at market price, at factor cost. Methods of calculating National Income – Value Added or Product method, Expenditure method, Income method with numerical questions.

Unit 2 Determination of Income and Employment
Aggregate Demand-Aggregate Supply and their components. Consumption function, Saving function, Investment function. Propensity to consume and propensity to save (average and marginal). Short-run equilibrium output. Meaning full employment and involuntary unemployment. Investment multiplier and its mechanism. Problems of excess demand and deficient demand. Measures to correct excess and deficient demand through Monetary and Fiscal policies of the government along with the instruments of monetary and fiscal policies.

Unit 3 Money and Banking
Money – Barter System of Exchange: Meaning and Limitations. Money: Meaning, Importance, and Functions of Money. Concept of Supply of Money and its measurement.
Banking: Meaning and Functions of Commercial Banks. Meaning and functions of Central bank (example of the Reserve Bank of India). Control of Credit by Central Bank through quantitative and qualitative measures.

Unit 4 Government Budget and the Economy
Government Budget – meaning, objectives, and components. Classification of receipts – revenue receipts and capital receipts, classification of expenditure – revenue expenditure and capital expenditure. Financial Position of Punjab Government. Measures of government deficit – revenue deficit. fiscal deficit, primary deficit their meaning,

Unit 5 Foreign Exchange Rate and Balance of Payments
Foreign exchange rate – Meaning of fixed and flexible rates and methods of their determination along with their advantages and limitations. Foreign Exchange market – Meaning and Functions. Balance of Payments – Meaning and components. Various types of accounts in Balance of Payment. The deficit in Balance of Payment: Meaning and measures to correct it.

Part – B
Indian Economic Development

Unit 6 Development Experience (1947-90) and Economic Reforms Since 1991
A brief introduction of the state of the Indian economy on the eve of independence. Five Year Plans and NITI Aayog; the rationale behind the adoption of five years economic plans, common goals of five-year economic plans with their success and failures. NITI AAYOG; A brief introduction, structure, and it’s working. Agriculture: Meaning, Importance, main features, problems and policies of agriculture (institutional aspects and new agricultural strategy), Agriculture Development of Punjab since 1966. Industry: Meaning, Importance, problems, and policies for industrial development industrial licensing, etc.), Industrial Development of Punjab since 1966. Economic Reforms since 1991: Features of Liberalisation, Globalisation, and Privatisation (LPG policy)

Unit 7 Current challenges Faced by the Indian Economy
Poverty – absolute and relative. Causes of Poverty and main programmes for poverty alleviation: A critical assessment.
Unemployment: Meaning, types, and causes. Main programmes for the solution of the problem of unemployment in India.
Rural development: Key issues – credit and marketing – the role of cooperatives; agricultural diversification; organic farming.
Human Capital Formation: How people become resources; Role of human capital in economic development; Growth of Education Sector in India. Manpower Resources of Punjab.
Infrastructure: Meaning and Types; Energy and Health; Problems and Policies; A critical assessment. Physical Resources of Punjab.
Sustainable Economic Development: Meaning, Effects of Economic Development on Resources and Environment, including global warming.

Part – C
Statistics in Economics

Unit 8 Correlation and Index Numbers
Coefficient of Correlation – meaning and properties. Methods for the measurement of coefficient of correlation: scatter diagram method, Karl Pearson’s method (only by direct method) (two variables ungrouped data) Spearman’s rank correlation (in case of untied ranks only).
Index Numbers – Meaning, methods of constructing; Unweighted Index (Simple aggregative and simple average of price relative method). Weighted Index Numbers (Weighted aggregative methods including only Laspeyre’s, Pasche’s, and Fisher’s Index Numbers). Wholesale price index, Consumer price index, and index of industrial production. Uses of index numbers: Inflation and index numbers.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 6 Application of Derivatives Ex 6.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4

(i) \(\sqrt{\mathbf{2 5 . 3}}\)

(ii) \(\sqrt{49.5}\)

(iii) \(\sqrt{0.6}\)

(iv) \((0.009)^{\frac{1}{3}}\)

(v) \((0.999)^{\frac{1}{10}}\)

(vi) \((15)^{\frac{1}{4}}\)

(vii) \((26)^{\frac{1}{3}}\)

(viii) \((255)^{\frac{1}{4}}\)

(ix) \((82)^{\frac{1}{4}}\)

(x) \((401)^{\frac{1}{2}}\)

(xi) \((0.0037)^{\frac{1}{2}}\)

(xii) \((\mathbf{2 6 . 5 7})^{\frac{1}{3}}\)

(xiii) \((81.5)^{\frac{1}{4}}\)

(xiv) \((3.968)^{\frac{3}{2}}\)

(xv) \((32.15)^{\frac{1}{5}}\)

Solution.
(i) \(\sqrt{25.3}\)
Consider y = √x. Let x = 25 and ∆x = 0.3.
Then ∆y = \(\sqrt{x+\Delta x}-\sqrt{x}\)

= \(\sqrt{25.3}-\sqrt{25}=\sqrt{25.3}-5\)

⇒ √253 = ∆y + 5
Now, dy is approximately equal to ∆y and is given by
dy = \(\left(\frac{d y}{d x}\right)\) ∆x

= \(\frac{1}{2 \sqrt{x}}\) (0.3) = 0.03
Hence, the approximate value of √25.3 is 0.03 + 5 = 5.03.

(ii) √49.5
Consider y = √x.
Let x = 49 and ∆x = 0.5
Then, ∆y = \(\sqrt{x+\Delta x}-\sqrt{x}\)
= \(\sqrt{49.5}-\sqrt{49}=\sqrt{49.5}-7\)
⇒ √49.5 = 7 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(0.5)\) [as y = √x]

= \(\frac{1}{2 \sqrt{49}}\) (0.5)

= \(\frac{1}{14}\) (0.5) = 7.035.

(iii) √0.6
Consider y = √x.
Let x = 1 and ∆x = – 0.4.
Then, ∆y = \(\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{0.6}-1\)
⇒ √0.6 = 1 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = (\(\frac{d y}{d x}\)) ∆x = \(\frac{1}{2 \sqrt{x}}\) (∆x) [as y = √x]
= \(\frac{1}{2}\) (- 0.4) = – 0.2
Hence, the approximate value of √0.6 is 1 + (- 0.2) = 1 – 0.2 = 0.8.

(iv) \((0.009)^{\frac{1}{3}}\)
Consider y = \((x)^{\frac{1}{3}}\).
Let x = 0.008 and ∆x = 0.001.
Then, ∆y = (x + ∆x)^{\(\frac{1}{3}\)} – (x)^{\(\frac{1}{3}\)}
= (0.009)^{\(\frac{1}{3}\)} – (0.008)^{\(\frac{1}{3}\)}
= (0.009)^{\(\frac{1}{3}\)} – 0.2
⇒ (0.009)^{\(\frac{1}{3}\)} = 0.2 + ∆Y
Now, dy is approximately equal ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x

= \(\frac{1}{3(x)^{\frac{2}{3}}}\) ∆x [as y = x^{\(\frac{1}{3}\)}]

= \(\frac{1}{3 \times 0.04}(0.001)=\frac{0.001}{0.12}\) = 0.008

Hence, the approximate value of \((0.009)^{\frac{1}{3}}\) is 0.2 + 0.008 = 0.208.

(v) \((0.999)^{\frac{1}{10}}\)
Consider y = \((x^{\frac{1}{10}}\).
Let x = 1 and ∆x = – 0.001.
Then, ∆y = (x + ∆x)^{\(\frac{1}{10}\)} – (x)^{\(\frac{1}{10}\)}
= (0.999)^{\(\frac{1}{10}\)} – 1
= (0.999)^{\(\frac{1}{10}\)}
=1 + ∆y
Now, dy is approximately equal to ∆y and is given by

dy = \(\frac{d y}{d x}\) ∆x [as y = x^{\(\frac{1}{10}\)}]

= \(\frac{1}{10(x)^{\frac{9}{10}}}\) (∆x)

= \(\frac{1}{10}\) (- 0.001)
= – 0.0001
Hence, the approximate value of (0.999)^{\(\frac{1}{10}\)} is 1 + (- 0.0001) = 0.0999.

(vi) \((15)^{\frac{1}{4}}\)
Consider y = (x)^{\(\frac{1}{4}\)}.
Let x = 16 and ∆x = – 1.
Then, ∆y = (x + ∆x)^{\(\frac{1}{4}\)} – (x)^{\(\frac{1}{4}\)}
= (15)^{\(\frac{1}{4}\)} – (16)^{\(\frac{1}{4}\)}
= (15)^{\(\frac{1}{4}\)} – 2
⇒ (15)^{\(\frac{1}{4}\)} = 2 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x^{\(\frac{1}{4}\)}]

= \(\frac{1}{4(16)^{\frac{3}{4}}}(-1)=\frac{-1}{4 \times 8}=\frac{-1}{32}\)

= – 0.03125
Hence, the approximate value of \((15)^{\frac{1}{4}}\) is 2 + (- 0.03125) = 1.96875.

(vii) \((26)^{\frac{1}{3}}\)
Consider y = (x)^{\(\frac{1}{3}\)}.
Let x = 27 and ∆x = – 1
Then, ∆y= (x + ∆x)^{\(\frac{1}{3}\)} – (x)^{\(\frac{1}{3}\)}
= (26)^{\(\frac{1}{3}\)} – (27)^{\(\frac{1}{3}\)}
= (26)^{\(\frac{1}{3}\)} – 3
⇒ (26)^{\(\frac{1}{3}\)} = 3 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x^{\(\frac{1}{3}\)}]

= \(\frac{1}{3(x)^{\frac{2}{3}}}\) (∆x)

= \(\frac{1}{3(27)^{\frac{2}{3}}}(-1)=\frac{-1}{27}\)
= – 0.0370.
Hence, the approximate value of \((26)^{\frac{1}{3}}\) is 3 + (- 0.03125) = 2.9629.

(viii) \((255)^{\frac{1}{4}}\)
Consider y = (x)^{\(\frac{1}{4}\)}.
Let x = 256 and ∆x = – 1.
Then, ∆y = (x + ∆x)^{\(\frac{1}{4}\)} – (x)^{\(\frac{1}{4}\)}
= (255)^{\(\frac{1}{4}\)} – (256)^{\(\frac{1}{4}\)}
= (255)^{\(\frac{1}{4}\)} – 4
⇒ (255)^{\(\frac{1}{4}\)} = 4 + ∆y
Now, dy is approximately equal to zy and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x^{\(\frac{1}{4}\)}]

= \(\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x)\)

= \(\frac{1}{4(256)^{\frac{3}{4}}}(-1)=\frac{-1}{4 \times 4^{3}}\)

= – 0.0039.
Hence, the approximate value of (255)^{\(\frac{1}{4}\)} is 4 + (- 00039) = 3.9961.

(ix) \((82)^{\frac{1}{4}}\)
Consider y = (x)^{\(\frac{1}{4}\)}.
Let x = 81 and ∆x = 1.
Then, ∆y = (x + ∆x)^{\(\frac{1}{4}\)} – (x)^{\(\frac{1}{4}\)}
= (82)^{\(\frac{1}{4}\)} – (81)^{\(\frac{1}{4}\)}
= (82)^{\(\frac{1}{4}\)} – 3
(82)^{\(\frac{1}{4}\)} = 3 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x^{\(\frac{1}{4}\)}]

= \(\frac{1}{4(x)^{\frac{3}{4}}}\) (∆x)

= \(\frac{1}{4(81)^{\frac{3}{4}}}(1)=\frac{1}{4(3)^{3}}=\frac{1}{108}\)
= 0.009
Hence, the approximate value of \((82)^{\frac{1}{4}}\) is 3 + 0.009 = 3.009.

(x) \((401)^{\frac{1}{2}}\)
Consider y = (x)^{\(\frac{1}{4}\)}.
Let x = 400 and ∆x = 1.
Then, ∆y = \(\sqrt{x+\Delta x}-\sqrt{x}\)

= \(\sqrt{401}-\sqrt{400}=\sqrt{401}-20\)
⇒ √401 = 20 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(\Delta x)\) [as y = x^{\(\frac{1}{4}\)}]

= \(\frac{1}{2 \times 20}(1)=\frac{1}{40}\)
= 0.025
Hence, the approximate value of √401 is 20 + 0.025 = 20.025.

(xi) \((0.0037)^{\frac{1}{2}}\)
Consider y = (x)^{\(\frac{1}{2}\)}.
Let x = 0.0036 and ∆x = 0.0001.
Then,
∆y = (x + ∆x)^{\(\frac{1}{2}\)} – (x)^{\(\frac{1}{2}\)}
= (0.0037)^{\(\frac{1}{2}\)} – (0.0036)^{\(\frac{1}{2}\)}
= (0.0037)^{\(\frac{1}{2}\)} – 0.06
(0.0037)^{\(\frac{1}{2}\)} = 0.06 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x^{\(\frac{1}{4}\)}]

= \(\frac{1}{2 \sqrt{x}}(\Delta x)\)

= \(\frac{1}{2 \times 0.06}(0.0001)\)

= \(\frac{0.0001}{0.12}\) = 0.00083

Hence, the approximate value of \((0.0037)^{\frac{1}{2}}\) is 0.06 + 0.00083 = 0.06083.

(xii) (26.57)^{\(\frac{1}{3}\)}
Consider y = (x)^{\(\frac{1}{3}\)}.
Let x = 27 and ∆x = – 0.43.
Then, ∆y = (x + ∆x)^{\(\frac{1}{3}\)} – (x)^{\(\frac{1}{3}\)}
= (26.57)^{\(\frac{1}{3}\)} – (27)^{\(\frac{1}{3}\)}
= (26.57)^{\(\frac{1}{3}\)} – 3
⇒ (26.57)^{\(\frac{1}{3}\)} = 3 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x^{\(\frac{1}{4}\)}]

= \(\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x)\)

= \(\frac{1}{3(9)}(-0.43)=\frac{-0.43}{27}\) = – 0.016
Hence, the approximate value of (26.57)^{\(\frac{1}{3}\)} is 3 + (- 0.0 16) = 2.984.

(xiii) \((81.5)^{\frac{1}{4}}\)
Consider y = (x)^{\(\frac{1}{4}\)}.
Let x = 81 and Ax 0.5.
Then, Ay = (x + &)^{\(\frac{1}{4}\)} – (x)^{\(\frac{1}{4}\)}
= (81.5)^{\(\frac{1}{4}\)} – (81)^{\(\frac{1}{4}\)}
= (81.5)^{\(\frac{1}{4}\)} – 3
⇒ \((81.5)^{\frac{1}{4}}\) = 3 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x^{\(\frac{1}{4}\)}]

= \(\frac{1}{4(x)^{4}}(\Delta x)\)

= \(\frac{1}{4(3)^{3}}(0.5)=\frac{0.5}{108}\)

= 0.0046
Hence, the approximate value of (81.5)^{\(\frac{1}{4}\)} is 3 + (0.0046) = 3.0046.

(xiv) \((3.968)^{\frac{3}{2}}\)
Consider y = (x)^{\(\frac{3}{2}\)}.
Let x = 4 and ∆x = – 0.032.
Then, ∆y = (x + ∆x)^{\(\frac{3}{2}\)} – (x)^{\(\frac{3}{2}\)}
= (3.968)^{\(\frac{3}{2}\)} – (4)^{\(\frac{3}{2}\)}
= (3.968)^{\(\frac{3}{2}\)} – 8
⇒ (3.968)^{\(\frac{3}{2}\)} = 8 + ∆y
Now, dy is approximately equal ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x^{\(\frac{1}{4}\)}]
= \(\frac{3}{2}\) (2) (- 0.032)
= – 0.096
Hence, the approximate value of \((3.968)^{\frac{3}{2}}\) is 8 + (- 0.096) = 7.904.

(xv) \((32.15)^{\frac{1}{5}}\)
Consider y = (x)^{\(\frac{1}{5}\)}.
Let x = 32 and ∆x = 0.15
Then, ∆y = \((x+\Delta x)^{\frac{1}{5}}-(x)^{\frac{1}{5}}\)

= \((32.15)^{\frac{1}{5}}-(32)^{\frac{1}{5}}=(32.15)^{\frac{1}{5}}-2\)
⇒ (32.15)^{\(\frac{1}{5}\)} = 2 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x^{\(\frac{1}{5}\)}]

= \(\frac{1}{5(x)^{\frac{4}{5}}}(\Delta x)\)

= \(\frac{1}{5 \times(2)^{4}}(0.15)=\frac{0.15}{80}\)

= 0.00187
Hence, the approximate value of \((32.15)^{\frac{1}{5}}\) is 2 + 0.00187 = 2.00187.

Question 2.
Find the approximate value of f(2.01), where f(x) = 4x² + 5x + 2.
Solution.
Let x = 2 and ∆x = 0.01
Then, we have f(2.01) = f(x + ∆x)
= 4(x + ∆x)² + 5(x + ∆x) + 2
Now, ∆y = f(x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆y
~ f(x) + f'(x) . ∆x [as dx = ∆x]
⇒ f(2.01) = (4x² + 5x + 2) + (8x + 5)∆x
= [4(2)² + 5(2) + 2] + [8(2) + 5] (0.01) [as x = 2, ∆x = 0.01]
= (16 + 10 + 2) + (16 + 5) (0.01)
= 28 + (21) (0.01)
= 28 + 0.21 = 28.21
Hence, the approximate value of f(2.01) is 28.21.

Question 3.
Find the approximate value of f(5.001), where f(x) = x³ – 7x² +15. ,
Solution.
Let x = 5 and ∆x = 0.001
Then, we have f(5.001) = f(x + ∆x)
= (x + ∆x)³ – 7(x + ∆x)² + 15
Now, ∆y = f(x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆y
~ f(x) + f'(x) . ∆x [as dx = ∆x]
⇒ f(5.001) = (x³ – 7x² + 15) + (3x² – 14x)∆x
= [(5)³ – 7(5)² + 15] + [3(5)² – 14(5)] (0.001) [as x = 5, ∆x = 0.001]
= (125 – 175 + 15) + (75 – 70) (0.001)
= – 35 + (5) (0.001)
= – 35 + 0.005
= – 34.995
Hence, the approximate value of f(5.001) is – 34.995.

Question 4.
Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.
Solution.
The volume of a cube (V) of side x is given by V = x³.
dV = (\(\frac{d V}{d x}\)) ∆x
= (3x²) ∆x
= (3x²) (0.01 x) [as 1% of x = 0.01 x]
= 0.03 x³
Hence, the approximate change in the volume of the cube is 0.03 x³ m³.

Question 5.
Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.
Solution.
The surface area of a cube (S) of side x is given by S = 6x².
∴ \(\frac{d S}{d x}=\left(\frac{d S}{d x}\right) \Delta x\)
= (12 x) ∆x
= (12 x) (- 0.01 x)[as 1% of x = 0.01 x]
= – 0.12 x²
Hence, the approximate change in the surface area of the cube is 0.12x² m².

Question 6.
If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its
volume.
Solution.
Let r be the radius of the sphere and ∆r be the error in measuring the radius.
Then, r = 7 m and ∆r = 0.02 m
Now, the volume V of the sphere is given by

V = \(\frac{4}{3}\) πr³
⇒ \(\frac{d V}{d r}\) = 4πr²
⇒ dV = (\(\frac{d V}{d r}\)) ∆r
= (4πr²) ∆r
= 4π(7)² (0.02) m³
= 3.92 π m³
Hence, the approximate error in calculating the volume is 3.92 π m³.

Question 7.
If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.
Solution.
Let r be the radius of the sphere and ∆r be the error in measuring the radius.
Then, r = 9 m and ∆r = 0.03 m
Now, the surface area of the sphere (S) is given by
S = 4πr²
⇒ \(\frac{d S}{d r}\) = 8πr
⇒ dS = (\(\frac{d S}{d r}\)) ∆r
= (8πr) ∆r
= 8π (9) (0.03) m²
= 2.16K m²
Hence, the approximate error in calculating the surface area is 2.16 π m².

Question 8.
If fix) = 3x² + 15x + 5, then the approximate value of f(3.02) is
(A) 47.66
(B) 57.66
(C) 67.66
(D) 77.66
Solution.
Let x = 3 and ∆x = 0.02. Then, we have
f(3.02) = f(x + ∆x)
= 3(x + ∆x)² + 15(x + ∆x) + 5
Now, ∆y = f(x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆y
≈ f(x) + f'(x)∆x [as dx = ∆x]
⇒ f(3.02) = (3x² + 15x + 5) + (6x + 15) ∆x
= [3(3)² +15(3) + 5] + [6(3) +15] (0.02) [as x = 3, ∆x = 0.02]
= (27 + 45 + 5) + (18 +15) (0.02)
= 77 + (33) (0.02) = 77 + 0.66 = 77.66
Hence, the approximate value of f(3.02) is 77.66.
The correct answer is (D).

Question 9.
The approximate change in the volume of a cube of side x metres caused by increasing the side of 3% is
(A) 0.06 x³ m³
(B) 0.6 x³ m³
(C) 0.09 x³ m³
(D) 0.9 x³ m³
Solution.
The volume of a cube (V) is side x is given by V = x³.
∴ dV = (\(\frac{d V}{d x}\)) ∆x
= (3x²) ∆x
= (3x²) (0.03 x) [as 3% of x = 0.03x]
= 0.09 x³ m³.
Hence, the approximate change in the volume of the cube is 0.09 x³ m³3.
The correct answer is (C).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 6 Application of Derivatives Ex 6.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3

Question 1.
Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4.
Solution.
The given curve is y = 3x⁴ – 4x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (3x⁴ – 4x)
⇒ \(\frac{d y}{d x}\) = 3 × 4x³ – 4 × 1
= 12x³ – 4

The slope of the tangent to the given curve at x = 4 is given by
\(\frac{d y}{d x}\) = 12(4)³ – 4
= 12(64) – 4 = 764.

Question 2.
Find the slope of the tangent of the curve y = \(\frac{x-1}{x-2}\), x ≠ 2 at x = 10.
Solution.
The given curve is y = \(\frac{x-1}{x-2}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{(x-2)(1)-(x-1)(1)}{(x-2)^{2}}\)

= \(\frac{x-2-x+1}{(x-2)^{2}}=\frac{-1}{(x-2)^{2}}\)

The slope of the tangent at x = 10 is given by
\(\left[\frac{d y}{d x}\right]_{x=10}=\frac{-1}{(10-2)^{2}}=\frac{-1}{8^{2}}=\frac{-1}{64}\)
Hence, the slope of the tangent at x = 10 is \(\frac{-1}{64}\).

Question 3.
Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x-coordinate is 2.
Solution.
The given curve is y = x³ – x +1
\(\frac{d y}{d x}\) = 3x² – 1
The slope of the tangent to a curve at (x₀, y₀) is \(\left[\frac{d y}{d x}\right]_{\left(x_{0}, y_{0}\right)}\).

It is given that x₀ = 2.
Hence, the slope of the tangent at the point where the x-coordinate is 2, is
\(\left[\frac{d y}{d x}\right]_{\left(x_{0}, y_{0}\right)}\) = 3(2)² – 1 = 12 – 1 = 11.

Question 4.
Find the slope of the tangent to the curve y = x³ – 3x + 2 at the point whose x-coordinate is 3.
Solution.
The given curve is y = x³ – 3x + 2
∴ \(\frac{d y}{d x}\) = 3x² – 3

The slope of the tangent to a curve at (x₀, y₀) is \(\left[\frac{d y}{d x}\right]_{\left(x_{0}, y_{0}\right)}\).

Hence, the slope of the tangent at the point where the x-coordinate is 3, is
\(\left[\frac{d y}{d x}\right]_{\left(x_{0}, y_{0}\right)}\) = 3 (3)² – 3 = 27 – 3 = 24

Question 5.
Find the slope of the normal to the curve x = a cos³ θ, y = a sin³ θ at θ = -.
Solution.
It is given that x = a cos³ θ and y = a sin³ θ
Differentiating x and y both w.r.t. θ, we get

\(\frac{d x}{d \theta}\) = 3a cos² θ (- sin θ)
= – 3a cos² θ sin θ

\(\frac{d y}{d \theta}\) = 3a sin² θ cos θ

\(\frac{d y}{d \theta}\) = \(\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{3 a \sin ^{2} \theta \cdot \cos \theta}{-3 a \cos ^{2} \theta \cdot \sin \theta}=-\frac{\sin \theta}{\cos \theta}\) = – tan θ

Therefore, the slope of the tangent at θ = \(\frac{\pi}{4}\) is given by

\(\left[\frac{d y}{d x}\right]_{\theta=\frac{\pi}{4}}\) = – tan \(\frac{\pi}{4}\) = – 1
Hence, the slope of the normal at θ = \(\frac{\pi}{4}\) is given by \(-\frac{1}{\text { slope of the tangent at }\left(\theta=\frac{\pi}{4}\right)}=\frac{-1}{-1}\) = 1.

Question 6.
Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos² θ at θ = \(\frac{\pi}{2}\).
Solution.
It is given that x = 1 – a sin θ and y = b cos² θ
Differentiating x and y both w.r.t. θ, we get
\(\frac{d x}{d \theta}\) = – a cos θ and
\(\frac{d y}{d \theta}\) = 2b cos θ (- sin θ) = – 2b sin θ cos θ

∴ \(\frac{d y}{d x}\) = \(\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{-2 b \sin \theta \cos \theta}{-a \cos \theta}=\frac{2 b}{a} \sin \theta\)

Therefore, the slope of the tangent at θ = \(\frac{\pi}{2}\) is given by
\(\left[\frac{d y}{d x}\right]_{\theta=\frac{\pi}{2}}=\frac{2 b}{a} \sin \frac{\pi}{2}=\frac{2 b}{a}\)

Hence, the slope of the normal at θ = \(\frac{\pi}{2}\) is given by
\(\frac{1}{\text { slope of the tangent at }\left(\theta=\frac{\pi}{4}\right)}=\frac{-1}{\left(\frac{2 b}{a}\right)}=-\frac{a}{2 b}\)

Question 7.
Find points at which the tangent to the curve y = x³ – 3x² – 9x + 7 is parallel to the x-axis.
Solution.
The equation of the given curve is y = x³ – 3x² – 9x + 7
Differentiating w.r.t. x, we get
∴ \(\frac{d y}{d x}\) = 3x² – 6x – 9
Now, the tangent is parallel to the x-axis if the slope of the tangent is zero.
∴ 3x² – 6x – 9 = 0
⇒ x² – 2x – 3 = 0
⇒ (x – 3)(x + 1) = 0
⇒ x = 3 or x = – 1
When x = 3, y = (3)³ – 3(3)² – 9(3) + 7
= 27 – 27 – 27 + 7 = – 20
When x = – 1, then y = (- 1)³ – 3 (- 1)² – 9 (- 1) + 7
= – 1 – 3 + 9 + 7 = 12
Hence, the point at which the tangent is parallel to the x-axis, are (3,- 20) and (- 1, 12).

Question 8.
Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Solution.
If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then slope of the tangent = Slope of the chord

The slope of the chord = \(]\frac{4-0}{4-2}=]\frac{4}{2}\) = 2

Now, the slope of the tangent to the given curve at a point (x, y) is given by
\(\frac{d y}{d x}\) = 2(x – 2) dx
∵ Slope of the tangent = Slope of the chord
∴ 2 (x – 2) = 2
⇒ x – 2 = 1
⇒ x = 3
When x = 3, then y = (3 – 2)² = 1
Hence, the required point is (3, 1).

Question 9.
Find the point on the curvey x³ – 11x + 5 at which the tangcnt is y = x – 11.
Solution.
Equation of the given curve is y = x³ – 11x + 5
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = 3x² – 11 ……………..(i)
Also the slope of the tangent y = x – 11
coefficient of x= 1 …………(ii)
Equating Eqs. (i) and (ii), we get
3x² – 11 = 1
⇒ 3x² + 12
x² = 4, x = ± 2
When x = 2,then y = x – 11 = 2 – 11 = – 9
When x = – 2, then y = x – 11 = – 2 -11 = – 13
But (- 2, – 13) does not lie on the curve
Hence, y = x – 11 is the tangent at(2, – 9).

Question 10.
Find the equation of all lines having slope – 1 that are tangents to the curve y = \(\frac{1}{x-1}\), x ≠ 1.
Solution.
The equation of the given curve is y = \(\frac{1}{x-1}\), x ≠ 1

The slope of the tangents to the given curve at any point (x, y) is given by \(\frac{d y}{d x}=\frac{-1}{(x-1)^{2}}\)

If the slope of the tangent is – 1, then we have
⇒ \(\frac{-1}{(x-1)^{2}}\) = – 1
⇒ (x – 1)² = ± 1
⇒ x – 1 = ± 1
⇒ x = 2, 0
When, x = 0, then y = – 1 and when x = 2, then y = 1.
Thus, there are two tangents to the given curve having slope – 1.
These are passing through the points (0, – 1) and (2, 1).
∴ The equation of the tangent through (0,- 1) is given by
y – (- 1) = – 1 (x – 0)
⇒ y + 1 = – x
⇒ y + x + 1 = 0
∴ The equation of the tangent through (2, 1) is given by
y – 1 = – 1 (x – 2)
⇒ y – 1 = – x + 2
⇒ y + x – 3 = 0
Hence, the equations of the required lines are y + x + 1 = 0 and y + x – 3 = 0.

Question 11.
Find the equation of all lines having slope 2 which are tangents to the curve y = \(\frac{1}{x-3}\), x ≠ 3.
Solution.
The equation of the given curve is y = \(\frac{1}{x-3}\).

The slope of the tangent to the given curve at any point (x, y) is given by
\(\frac{d y}{d x}=\frac{-1}{(x-3)^{2}}\)

If the slope of the tangent is 2, then we have
\(\frac{-1}{(x-3)^{2}}\) = 2
⇒ 2 (x – 3)² = – 1
⇒ (x – 3)² = \(\frac{-1}{4}\)
This is not possible since the LH.S. is positive while the R.H.S. is negative.
Hence, there is no tangent to the given curve having slope 2.

Question 12.
Find the equation of all lines having slope O which are tangent to the curve y = \(\frac{1}{x^{2}-2 x+3}\)
Solution.
The equation of the given curve is y = \(\frac{1}{x^{2}-2 x+3}\)

The slope of the tangent to the given curve at any point (x, y) is given by

\(\frac{d y}{d x}=\frac{-(2 x-2)}{\left(x^{2}-2 x+3\right)^{2}}=\frac{-2(x-1)}{\left(x^{2}-2 x+3\right)^{2}}\)

If the slope of the tangent is 0, then we have
\(\frac{-2(x-1)}{\left(x^{2}-2 x+3\right)^{2}}\) = 0

⇒ – 2 (x – 1) = 0
⇒ x = 1
When x = 1, then y = \(\frac{1}{1-2+3}=\frac{1}{2}\)
∴ The equation of the tangent through (1, \(\frac{1}{2}\)) is given by
y – \(\frac{1}{2}\) = 0 (x – 1)
⇒ y – \(\frac{1}{2}\) = 0
⇒ y = \(\frac{1}{2}\)
Hence, the equation of the required line is y = \(\frac{1}{2}\).

Question 13.
Find points on the curve \(\frac{x^{2}}{9}+\frac{y^{2}}{16}\) = 1 at which the tangents are
(i) parallel to x-axis
(ii) parallel to y-axis
Solution.
The equation of the given curve is \(\frac{x^{2}}{9}+\frac{y^{2}}{16}\) = 1.
On differentiating both sides w.r.t. x, we have
\(\frac{2 x}{9}+\frac{2 y}{16} \cdot \frac{d y}{d x}\) = 0

⇒ \(\frac{d y}{d x}=\frac{-16 x}{9 y}\)

(i) The tangent is parallel to the x-axis, if the slope of the tangent is 0 i.e., \(\frac{-16 x}{9 y}\) = 0
which is possible if x = 0.
Then, \(\frac{x^{2}}{9}+\frac{y^{2}}{16}\) = 1 for x = 0;
⇒ y² = 16
⇒ y = ± 4
Hence, the points at which the tangents are parallel to the x-axis are (0, 4) and (0, – 4).

(ii) The tangent is parallel to the y-axis if the slope of the normal is 0, which gives
\(\frac{-1}{\left(\frac{-16 x}{9 y}\right)}=\frac{9 y}{16 x}\) = 0
⇒ y = 0
Then, \(\frac{x^{2}}{9}+\frac{y^{2}}{16}\) = 1 for y = 0
⇒ x = ± 3
Hence, the points at which the tangents are parallel to the y-axis are (3, 0) and (- 3, 0).

Question 14.
Find the equations of the tangent and normal to the given curves at the indicated points :
(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)
(ii) y = x⁴ – 6x³ + 13x² – 10x + 5 at (1, 3)
(iii) y = x³ at (1, 1)
(iv) y = x² at (0, 0)
(v) x – cos t and y = sin t at t = \(\frac{\pi}{4}\)
Solution.
(i) The equation of the given curve is
y = x⁴ – 6x³ + 13x² – 10x + 5
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = 4x³ – 18x² + 26 x – 10
Putting x = 0, \(\frac{d y}{d x}\) at (0, 5) = – 10
∴ Slope of tangent at (0, 5) = – 10
Thus, the equation of tangent at P(0, 5) is
y – y₁ = \(\left(\frac{d y}{d x}\right)_{a t P}\) (x – x₁)

⇒ y – 5 = – 10 (x – 0)
⇒ y + 10x – 5 = 0
and the equation of normal at P(0, 5)
(x – x₁) + \(\left(\frac{d y}{d x}\right)_{a t P}\) (y – y₁) = 0
⇒ (x – 0) + (- 10)(y – 5) = 0
⇒ x – 10y + 50 = 0.

(ii) The equation of the given curve is y = x⁴ – 6x³ + 13x² – 10x + 5
Differentiating w.r.t. x, we get
∴ \(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10
Putting x = 1,
\(\frac{d y}{d x}\) = 4 – 18 + 26 – 10 = 2
∴ Slope of tangent at (1, 3) is 2.
∴ Equation of tangent at (1, 3) is
y – 3 = 2 (x – 1)
⇒ y – 3 = 2x – 2
⇒ y = 2x + 1
and, equation of normal is
(x – 1) + 2 . (y – 3) = 0
⇒ x – 1 + 2y – 6 = 0
⇒ x + 2y – 7 = 0

(iii) The equation of the given curve is y = x³
∴ \(\frac{d y}{d x}\) = 3x²
Now, \(\frac{d y}{d x}\) at (1, 1) dx = 3 (1)² = 3
i.e., slope of tangent at (1, 1) is 3
Equation of the tangent at (1, 1) is
y – 1 = 3(x – 1)
⇒ y – 1 = 3x – 3
⇒ y = 3x – 2
and, equation of normal at (1, 1) is
(x – 1) + 3 (y – 1) – 0
⇒ x – 1 + 3y – 3 = 0
⇒ x + 3y – 4 = 0

(iv) The equation of the given curve is y = x²
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = 2x
Now, \(\frac{d y}{d x}\) at (0, 0) is 0
i.e., slope of tangent at (0, 0) is 0
∴ Equation of tangent at (0, 0) is
y – 0 = 0 (x – 0)
⇒ and Equation of normal at (0, 0) is
(x – 0) + 0 (y – 0) = 0
⇒ x = 0

(v) The equation of the given curve is
x = cos t ……………(i)
y = sin t …………….(ii)
From Eqs. (i) and (ii), we have

Question 15.
Find the equation of the tangent line to the curve y = x² – 2x + 7 which is
(i) parallel to the line 2x – y + 9 = 0.
(ii) perpendicular to the line 5y – 15x = 13.
Solution.
The equation of the given curve is y = x² – 2x + 7
On differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = 2x – 2

(i) The equation of the line is
2x – y + 9 = 0
⇒ y = 2x + 9
This is of the form y = mx + c.
∴ Slope of the line = 2
If a tangent is parallel to the line 2x – y + 9 = 0, then the slope of the tangent is equal to the slope of the line.
Therefore, we have 2 = 2x -2
⇒ 2x = 4
⇒ x = 2
When, x = 2, then y = 4 – 4 + 7 = 7
Thus, the equation of the tangent passing through (2, 7) is given by y – 7 = 2 (x – 2)
⇒ y – 2x – 3 = 0
Hence, the equation of the tangent line to the given curve (which is parallel to line(2x – y + 9 = 0) is y – 2x – 3 = 0.

(iii) The equation of the line is 5y – 15x = 13
⇒ y = 3x + \(\frac{13}{5}\)
This is of the form y = mx + c
∴ Slope of the line = 3
If a tangent is perpendicular to the line 5y – 15x = 13,
then the slope of the tangent is \(\frac{-1}{\text { Slope of the line }}=\frac{-1}{3}\)
⇒ 2x – 2 = \(-\frac{1}{3}\)

⇒ 2x = \(-\frac{1}{3}\) + 2

⇒ 2x = \(\frac{5}{3}\)

⇒ x = \(\frac{5}{6}\)

When, x = \(\frac{5}{6}\), then y = \(\frac{25}{36}-\frac{10}{6}+7\)

= \(\frac{25-60+252}{36}\frac{217}{36}\)

Thus, the equation of the tangent passing through (\(\frac{5}{6}\), \(\frac{217}{36}\)) is given by
y – \(\frac{217}{36}\) = – \(\frac{1}{3}\) (x – \(\frac{5}{6}\))

⇒ \(\frac{36 y-217}{36}\) = – \(\frac{1}{18}\) (6x – 5)
⇒ 36y – 217 = – 2 (6x – 5)
⇒ 36y – 217 = – 12x + 10
⇒ 36y + 12x – 227 = 0
Hence, the equation of the tangent line to the given curve (which is perpendicular to line 5y -15x = 13 is 36y + 12x – 227 = 0.

Question 16.
Show that the tangents to the curve y = 7x³ + 11 at the points where x = 2 and x = – 2, are parallel.
Solution.
The equation of the given curve is y = 7x³ + 11.
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = 21x²
The slope of the tangent to a curve at (x₀, y₀) is \(\left[\frac{d y}{d x}\right]_{\left(x_{0}, y_{0}\right)}\)
Therefore, the slope of the tangent at the point where x = 2, is given by
\(\left[\frac{d y}{d x}\right]_{x=2}\) = 21 (2)² = 84
Also, the slope of the tangent at the point where x = -2, is given by dv
\(\left[\frac{d y}{d x}\right]_{x=-2}\) = 21 (- 2)² = 21 × 4 = 84
It is observed that the slopes of the tangents at the points where x = 2 and x = – 2 are equal.
Hence, the two tangents are parallel.

Question 17.
Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinate of the point.
Solution.
The equation of the given curve is y = x³
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = 3x²

The slope of the tangent at the point (x, y) is given by
\(\left[\frac{d y}{d x}\right]_{(x, y)}\) = 3x²

When the slope of the tangent is equal to the y-coordinate of the point.
Then, y = 3x²
Also, we have y = x³
3x² = x³
⇒ x² (x – 3) = 0
⇒ x = 0, x = 3
When, x = 0, then y = 0 and when x = 3, then y = 3(3)² = 27.
Hence, the required points are (0, 0) and (3, 27).

Question 18.
For the curve y = 4x³ – 2x², find all the points at which the tangent passes through the origin.
Solution.
The equation of the given curve is y = 4x³ – 2x².
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = 12x² – 10x⁴
Therefore, the slope of the tangent at a point (x, y) is 12x² – 10x⁴.
The equation of the tangent at (x, y) is given by
Y – y = (12x² – 10x⁴) (X – x) ……………(i)
When the tangent passes through the origin (0, 0), then X = Y = 0
Therefore, Eq.(i) reduces to
– y = (12x² – 10x⁴) (- x)
⇒ y = 12x³ – 10x⁵
Also, we have y = 4x³ – 2x²
∴ 12x³ – 10x⁵ = 4x³ – 2x⁵
⇒ 8x⁵ – 8x³ = 0
⇒ x⁵ – x³ = 0
⇒ x³ (x² – 1) = 0
⇒ x = 0, ± 1
When x = 0, then y 4 (0)³ – 2(0)⁵ = 0
When, x = 1, then y = 4 (1)³ – 2(1)⁵ = 2
When x = – 1, then y = 4(- 1)³ – 2(-l)⁵ = – 2
Hence, the required points are (0, 0), (1, 2) and (- 1, – 2).

Question 19.
Find the points on the curve x² + y² – 2x – 3 = 0 at which the tangents are parallel to the x-axis.
Solution.
The equation of the given curve is x² + y² – 2x – 3 = 0
On differentiating both sides w.r.t. x, we have
2x + 2y \(\frac{d y}{d x}\) – 2 = 0
⇒ y \(\frac{d y}{d x}\) = 1 – x

⇒ \(\frac{d y}{d x}=\frac{1-x}{y}\)
Now, the tangents are parallel to the x- = axis if the slope of the tangent is 0.
∴ \(\frac{1-x}{y}\) = 0
⇒ 1 – x = 0
⇒ x = 1
But, x² + y² – 2x – 3 = 0 for x = 1.
⇒ y² = 4
⇒ y = ± 2
Hence, the points at which the tangents are parallel to the x-axis, are (1, 2) and (1, – 2).

Question 20.
Find the equation of the normal at the point (am², am³) for the curve ay² = x³.
Solution.
The equation of the given curve is ay² = x³.
On differentiating both sides w.r.t. x, we have
2ay \(\frac{d y}{d x}\) = 3x²

⇒ \(\frac{d y}{d x}=\frac{3 x^{2}}{2 a y}\)

The slope of a tangent to the curve at (x₀, y₀) is \(\left[\frac{d y}{d x}\right]_{\left(x_{0}, y_{0}\right)}\).

⇒ The slope of the tangent to the given curve at (am², am³) is
\(\left[\frac{d y}{d x}\right]_{\left(a m^{2}, a m^{3}\right)}=\frac{3\left(a m^{2}\right)^{2}}{2 a\left(a m^{3}\right)}=\frac{3 a^{2} m^{4}}{2 a^{2} m^{3}}=\frac{3 m}{2}\)

∴ Slope of normal at (am², am³)
\(\frac{-1}{\text { Slope of the tangent at }\left(a m^{2}, a m^{3}\right)}=\frac{-2}{3 m}\)

Slope of the tangent at (am², am³)
Hence, the equation of the normal at (am2, am3) is given by
y – am³ = \(\frac{-2}{3 m}\) (x – am²)

⇒ 3my – 3am⁴ = – 2x + 2am²
2x + 3my – am² (2 + 3m²)= 0.

Question 21.
Find the equation of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Solution.
The equation of the given curve is y = x³ + 2x + 6
The slope of the tangent to the given curve at any point (x, y) is given by
\(\frac{d y}{d x}\) = 3x² + 2
∴ Slope of normal at any point is given at any point (x, y)
= Slope of the tangent at the point (x, y) = \(\frac{-1}{3 x^{2}+2}\)
The equation of the given line is
x + 14 y + 4 = 0
⇒ y = \(-\frac{1}{14} x-\frac{4}{14}\)
(Which is of the form y = mx + c)
∴ Slope of the given line = \(\frac{-1}{14}\)
If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line.
∴ \(\frac{-1}{3 x^{2}+2}=\frac{-1}{14}\)
⇒ 3x² + 2 = 14
= 3x² = 12
x² = 4
When x = 2, then y = 8 + 4 + 6=18
When x = – 2, then y = – 8 – 4 + 6 = – 6
Therefore, there are two normals to the given curve with slope \(\frac{-1}{14}\) and passing through the points (2, 18) and (- 2, – 6).
Thus, the equation of the normal through (2, 18) is given by y – 18 = \(\frac{-1}{14}\) (x – 2)
⇒ 14y – 252 = – x + 2
⇒ x + 14y – 254 = 0
And. the equation of the normal through (- 2, – 6) is given by
y – (- 6) = \(\frac{-1}{14}\) [x – (- 2)]
y + 6 = – (x + 2)
⇒ 14y + 84 = – x – 2
⇒ x + 14y + 86 = 0
Hence, the equation of the normals to the given curve (which are parallel to the given line) are x + 14y – 254 = 0 and x + 14y + 86 = 0.

Question 22.
Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at², 2at).
Solution.
The equation of the given parabola is y² = 4ax
On differentiating y² = 4ax w.r.t. x, we have
2y \(\frac{d y}{d x}\) = 4a

⇒ \(\frac{d y}{d x}=\frac{2 a}{y}\)
Now, the slope of the normal at (at², 2at) is given by = \(Slope of the tangent at \left(a t^{2}, 2 a t\right)\) = – t
Slope of the tangent at (at², 2at)
Thus, the equation of the normal at (at², 2at) is given as
y – 2at = – t (x – at²)
⇒ y – 2at = – tx + at³
y = – tx + 2at + at³

Question 23.
Prove that the curves x = y² and xy = k cut a right angles, if 8k² = 1.
[Note: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.]
Solution.
The equations of the given curves are x = y² and xy = k
Putting x = y² in xy = k, we get
y³ = k
⇒ y = k^1/3
∴ x = k^2/3
Thus, the point of intersection of the given curves is (k^2/3, k^1/3)
On differentiating x = y² with respect to x, we get

1 = 2y \(\frac{d y}{d x}\)
⇒ \(\frac{d y}{d x}=\frac{1}{2 y}\)

Therefore, the slope of the tangent to the curve x = y² at (k^2/3, k^1/3) is
\(\left[\frac{d y}{d x}\right]_{\left(k^{2 / 3}, k^{1 / 3}\right)}=\frac{1}{2 k^{1 / 3}}\)

∴ Slope of the tangent to the curve xy = k at (k 2/3, k”3) is given by
\(\left[\frac{d y}{d x}\right]_{\left(k^{2 / 3} \cdot k^{1 / 3}\right)}=\left[\frac{-y}{x}\right]_{\left(k^{2 / 3} \cdot k^{1 / 3}\right)}\)

= \(-\frac{k^{1 / 3}}{k^{2 / 3}}=-\frac{-1}{k^{1 / 3}}\)

We know that two curves intersect at right angles if the tangents to the curve at the point of intersection i.e., at (k^2/3, k^1/3) are perpendicular to each other.
This implies that we should have the product of the tangents as – 1.
Thus, the given two curves cut at right angles, if the product of the slopes of their respective tangents at (k^2/3, k^1/3) is – 1.
i.e., \(\left(\frac{1}{2 k^{1 / 3}}\right)\left(\frac{-1}{k^{1 / 3}}\right)\) = – 1

⇒ 2k^2/3 = 1

⇒ (2k^2/3)³ = (1)³

⇒ 8k² = 1
Hence, the given two curves cut at right angles, if 8k² = 1.

Question 24.
Find the equations of the tangent and normal to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 at the point (x₀, y₀).
Solution.
The equation of the hyperola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
Differentiating w.r.t x, we get

Question 25.
Find the equation of the tangent to the curve y = \(\sqrt{3 x-2}\), which is parallel to the line 4x – 2y + 5 = 0.
Solution.
The equation of the given curve is y = \(\sqrt{3 x-2}\)
The slope of the tangent to the given curve at any point (x, y) is given by
\(\frac{d y}{d x}=\frac{3}{2 \sqrt{3 x-2}}\)
The equation of the given line is 4x – 2y + 5 = 0
⇒ y = 2x + (which is of the form y = mx + c)
∴ Slope of the line = 2
Now, the tangent to the given curve is parallel to the line 4x – 2y – 5 = 0 .
If the slope of the tangent is equal to the slope of the line.

Hence, the equation of the required tangent is 48x – 24y = 23.

Direction (26 – 27):
Choose the correct answer.

Question 26.
The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is
(A) 3
(B) \(\frac{1}{3}\)
(C) – 3
(D) = – \(\frac{1}{3}\)
Solution.
The equation of the given curve is y = 2x² + 3 sin x
Slope of the tangent to the given curve at x = 0 is given by \(\left[\frac{d y}{d x}\right]_{x=0}\) = 0 + 3 cos 0 = 3
Hence, the slope of the normal to the given curve at x = 0 is given by \(\frac{-1}{\text { Slope of the tangent at }(x=0)}=\frac{-1}{3}\)
The correct answer is (D).

Question 27.
The line y = x + 1 is a tangent to the curve y² = 4x at the point
(A) (1, 2)
(B) (2, 1)
(C) (1, – 2)
(D) (- 1, 2)
Solution.
The equation of the given curve is y² = 4x …………..(i)
Differentiating w.r.t x, we get
2y \(\frac{d y}{d x}\) = 4

∴ \(\frac{d y}{d x}=\frac{4}{2 y}=\frac{2}{y}\)

The given line is y = x +1 (which is of the form y = mx + c)
∴ Slope of this line is 1.
The line y = x +1 is a tangent to the given curve, if the slope of the line is equal to the slope of the tangent.
Also, the line must intersect the curve.
Thus, we must have \(\frac{2}{y}\) = 1
⇒ y = 2
On putting y = 2 in Eq. (i), we get
2² = 4x
⇒ x = 1
Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2).
So, the correct answer is (A).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 6 Application of Derivatives Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2

Question 1.
Show that the function given by f(x) = 3x +17 is strictly increasing on R.
Solution.
Let x₁ and x₂ by any two numbers in R, where x₁ < x₂ Then, we have
3x₁ < 3x₂
⇒ 3₁ + 17 < 3x₂ +17
⇒ f(x₁) < f(x₂)
Hence, f is strictly increasing on R.

Alternate Method:
Given, f(x) = 3x +17
On differentiating w.r.t. x, we get f'(x) = 3 > 0, in every interval of R.
Thus, the function is strictly increasing on R.

Question 2.
Show that the function given by f(x) = e^2x is strictly increasing on R.
Solution.
Let x₁ and x₂ be any two numbers in R, where x₁ < x₂ Given, f(x) = e^2x Then, we have
x₁ < x₂
⇒ 2x₁ < 2x₂
⇒ e^2x₁ < e^2x₂
⇒ f(x₁) < f(x₂)
Hence, f is strictly increasing on R.

Question 3.
Show that the function given y f(x) = sin x is
(a) strictly increasing in (0, \(\frac{\pi}{2}\))
(b) strictly decreasing in (\(\frac{\pi}{2}\), π)
(c) neither increasing nor decreasing in (0, π)
Solution.
The given function is f(x) = sin x.
∴ f'(x) = cos x

(a) Since, for each x ∈ (0, \(\frac{\pi}{2}\)), cos x > 0, we have f'(x) > 0.
Hence, f is strictly increasing in (o, \(\frac{\pi}{2}\)).

(b) Since for each x ∈ (\(\frac{\pi}{2}\), π), cos x < 0, we have f'(x) < 0.
Hence, f is strictly decreasing in (\(\frac{\pi}{2}\), π)

(c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor decreasing in (0, π).

Question 4.
Find the intervals in which the function f given by
f(x) = 2x² – 3x is
(a) strictly increasing
(b) strictly decreasing
Solution.
The given function is f(x) = 2x² – 3x.
Differentiating w.r.t. x, we get
∴ f'(x) = 4x – 3
On putting f'(x) = 0, we get
4x – 3 = 0
⇒ x = \(\frac{3}{4}\)

Now, the point \(\frac{3}{4}\) divides the real line into two disjoint intervals i.e., (- ∞, \(\frac{3}{4}\)) and (\(\frac{3}{4}\), ∞)

In Interval (- ∞, \(\frac{3}{4}\)),
f'(x) = 4x – 3 < 0 Hence, the given function (f) is strictly decreasing in interval (- ∞, \(\frac{3}{4}\)). In interval, (\(\frac{3}{4}\), ∞), f'(x) = 4x – 3 > 0.
Hence, the given function (f) is stricdy increasing in interval (\(\frac{3}{4}\), ∞).

Question 5.
Find the intervals in which the function f given by f(x) = 2x³ – 3x² – 36x + 7 is
(a) strictly increasing
(b) strictly decreasing
Solution.
The given function is f(x) = 2x³ – 3x² – 36x + 7
∴ f'(x) = 6x² – 6x – 36
= 6(x² – x – 6)
= 6(x + 2) (x – 3)
∴ f'(x) = 0
⇒ x = – 2, 3.
The points x = – 2 and x = 3 divide the real line into three disjoint intervals i.e., (- ∞, – 2), (- 2, 3) and (3, ∞).

In intervals (- ∞, – 2) and (3, ∞), f'(x) is positive while in interval (- 2 , 3), f’ (x) is negative.
Hence, the given function (f) is strictly increasing in intervals (- ∞, – 2) and (3, ∞), while function (f) is strictly decreasing in interval (- 2, 3).

Question 6.
Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x² + 2x – 5
(b) 10 – 6x – 2x²
(c) -2x³ – 9x² – 12x + 1
(d) 6 – 9x – x²
(e) (x + 1)³ (x – 3)³
Solution.
(a) we have f(x) = x² + 2x – 5
∴ f'(x) = 2x + 2
= 2 (x + 1)
The function f(x) will be increasing if f'(x) > 0
i.e., if 2(x + 1) > 0
⇒ x + 1 > 0
⇒ x > – 1
The function f(x) will be decreasing if f'(x) < 0
i.e., 2(x + 1) < 0
⇒ (x + 1) < 0
⇒ x < – 1
Hence, f(x) is increasing on (- 1, ∞) and decreasing on (- ∞, – 1).

(b) We have, f(x) = 10 – 6x – 2x²
Differentiating w.r.t. x, we get
f'(x) = – 6 – 4x
= – 2 (3 + 2x)
Now, f(x) is increasing, if f'(x) > 0
i.e., – 6 – 4x > 0
i.e., – 4x > 6
⇒ x < \(-\frac{3}{2}\)
and f'(x) is decreasing if f'(x) < 0
i.e., if – 6 – 4x < 0
i.e., – 4x < 6 ⇒ x > \(-\frac{3}{2}\)
Hence, f(x) is increasing for x < \(-\frac{3}{2}\) i.e., in the interval (- ∞, \(-\frac{3}{2}\)) and decreasing on x > \(-\frac{3}{2}\) i.e., (\(-\frac{3}{2}\), ∞).

(c) Let f(x) = – 2x³ – 9x² – 12x + 1
Differentiating w.r.t. x, we get
f'(x) = – 6x² – 18x – 12
= – 6(x² + 3x + 2)
= – 6 (x +1) (x + 2)
f'(x) = 0 gives x = – 1 or x = – 2
The points x = – 2 and x = – 1 (arranged in ascending order) divide the real line into three disjoint intervals namely, (- ∞, – 2), (- 2, – 1) and (- 1, ∞).
In the interval (- ∞, – 2) i.e., – ∞ < x < – 2 (x +1) and (x + 2) are negative
∴ f'(x) = (-) (-) (-) = – ve
⇒ f(x) is decreasing in (- ∞, – 2)
In the interval (- 2, – 1) i.e., – 2 < x < – 1 (x + 1) is -ve and (x + 2) is positive
∴ f'(x) = (-) (-) (+) = +ve
⇒ f(x) is increasing in (- 2, – 1)
In the interval (- 1, ∞) i.e., – 1 < x < ∞
(x + 1) and (x + 2) are both positive
∴ f'(x) = (-) (+) (+) = – ve
⇒ f{x) is decreasing in (- 1, ∞)
Hence, f(x) is increasing for – 2 < x < -1 and decreasing for x < – 2 and x > – 1.

(d) We have, f(x) = 6 – 9x – x²
Differentiating w.r.t. x, we get
f'(x) = – 9 – 2x
Now, f(x) is increasing if f'(x) > 0 i.e., if – 9 – 2x > 0
i.e., if – 2x > 9
⇒ x < \(-\frac{9}{2}\) and f(x) is decreasing if f'(x) > 0 i.e., if – 9 – 2x < 0
i.e., – 2x < 9 ⇒ x > \(-\frac{9}{2}\)
Hence, f(x) is increasing for x < \(-\frac{9}{2}\) and decreasing for x > \(-\frac{9}{2}\).

(e) We have, f(x) = (x + 1)³ (x – 3)³
Differentiating w.r.t x, we get
f'(x) = 3(x + 1)² [\(\frac{d}{d x}\) (x + 1)] + (x – 3)³ + (x + 1)³ . 3 (x – 3)² . \(\frac{d}{d x}\) (x – 3)
= 3(x + 1)² (x – 3)³ + 3 (x + 1)³ (x – 3)²
= 3(x + 1)² (x – 3)² (x – 3 + x +1)
= 6(x + 1)² (x – 3)² (x – 1)

For f(x) to be increasing :
f'(x) > 0
⇒ 6(x + 1)² (x – 3)² (x – 1) > 0
⇒ (x – 1) > 0 [∵ 6(x + 1)² (x – 3)² > 0]
⇒ x > 1
But f'(x) = 0 at x = 3
⇒ f is increasing in (1, 3) and (3, 0)
∴ f(x) is increasing on (1, ∞).

For f(x) to be decreasing :
f'(x) < 0
⇒ 6(x + 1)² (x – 3)² (x – 1) < 0
(x – 1) < 0 [∵ 6 (x +1)² (x – 3)² > 0]
⇒ x < 1 But f'(x) = 0 at x = – 1 Hence, f(x) is strictly decreasing in (- ∞, 1), (- 1, 1).

Question 7.
Show that y = log (1 + x) – \(\frac{2 x}{(2+x)}\) , x > -1, is an increasing function of x throughout its domain.
Solution.
Given, y = log(1 + x) – \(\frac{2 x}{(2+x)}\)
On differentiating, we get \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [log (1 + x) – \(\frac{2 x}{(2+x)}\)]

y’ > 0, when x > – 1
Hence, y is increasing function throughout (x > – 1) its domain.

Question 8.
Find the values of x for which y = [x(x – 2)]² is an increasing function.
Solution.
We have, y = [x(x – 2)]²
= [x² – 2x]²
∴ \(\frac{d y}{d x}\) = y’
= 2 (x² – 2x) (2x – 2)
= 4x (x – 2) (x – 1)
∴ \(\frac{d y}{d x}\) = 0
⇒ x = 0, x = 2, x = 1
The points x = 0, x = 1 and x = 2 divide the real line into four disjoint intervals i.e., (- ∞, 0), (0, 1) (1, 2) and (2, ∞).
∴ In intervals (- ∞, 0) and (1, 2), \(\frac{d y}{d x}\) < 0. ∴ y is strictly decreasing in intervals (- ∞, 0) and (1, 2). However, in intervals (0, 1) and (2, ∞), \(\frac{d y}{d x}\) > 0.
∴ y is strictly increasing in intervals (0, 1) and (2, ∞).
∴ y is strictly increasing for 0 < x < 1 and x > 2.

Question 9.
Prove that y = \(\frac{4 \sin \theta}{(2+\cos \theta)}\) – θ is an increasing function of θ in [0,
\(\frac{\pi}{2}\)].
Solution.
We have, y = \(\frac{4 \sin \theta}{(2+\cos \theta)}\) – θ
Differentiating w.r.t θ, we get

For the function to be increasing:
\(\frac{d y}{d \theta}\) > 0
⇒ \(\frac{\cos \theta\left(4-\cos ^{2} \theta\right)}{(2+\cos \theta)^{2}}\) > 0
⇒ cos θ (4 – cos² θ) > 0
⇒ cos θ > 0, [(4 – cos² θ) > 0 as cos² θ is not > 1]
∴ θ ∈ (0, \(\frac{\pi}{2}\))
Hence, given function is increasing in interval [0, \(\frac{\pi}{2}\)].

Question 10.
Prove that the logarithmic function is strictly increasing on (0, ∞).
Solution.
The given function is f(x) = log x
Differentiating w.r.t. x, we get
f'(x) = \(\frac{1}{x}\)
It is clear that for x > 0, f’ (x) = \(\frac{1}{x}\) > 0.
Hence, f(x) = log x is strictly increasing in interval (0, ∞).

Question 11.
Prove that the function f given by f(x) = x² – x + 1 is neither strictly increasing nor strictly decreasing on (- 1, 1).
Solution.
The given function is f(x) = x² – x + 1
Differentiating w.r.t. x, we get f'(x) = 2x – 1
Now, f'(x) = 0
⇒ x = \(\frac{1}{2}\)
The point \(\frac{1}{2}\) divides the interval (- 1, 1) into two disjoint intervals i.e., (- 1, \(\frac{1}{2}\)) and (\(\frac{1}{2}\), 1)

Now, in interval (- 1, \(\frac{1}{2}\)), f'(x) = 2x – 1 < 0 Therefore, f is strictly decreasing in interval (- 1, \(\frac{1}{2}\)) However, in interval (\(\frac{1}{2}\), 1). f’(x) = 2x – 1 > 0.
Therefore, f is strictly increasing in interval (\(\frac{1}{2}\), 1)
Hence, f is neither strictly increasing nor strictly decreasing in interval (- 1, 1).

Question 12.
Which of the following functions are strictly decreasing on
(A) cos x
(B) cos 2x
(C) cos 3x
(D) tan x
Solution.
(A) Let, f(x) = cos x then f’(x) = – sin x
For 0 < x < \(\frac{\pi}{2}\), sin x > 0
∴ f’(x)= – sin x < 0 in (o, \(\frac{\pi}{2}\))
∴ f(x) is a decreasing function.

(B) Let, f(x) = cos 2x then f’(x) = – 2 sin 2x
For 0 < x < \(\frac{\pi}{2}\)
or 0 < 2x < π, sin 2x is positive
∴ f(x) is a decreasing function.

(C) Let, f(x) = cos 3x then f’(x) = – 3 sin 3x
For 0 < x < \(\frac{\pi}{2}\) ⇒ 0 < 3x < \(\frac{\pi}{2}\)
sin 3x is positive in 0 < 3x < π
∴ f’(x) < 0
⇒ f(x) is decreasing
And sin 3x is negative in π < 3x < \(3 \frac{\pi}{2}\) ∴ f’(x) > 0
⇒ f(x) is increasing
∴ f(x) is neither increasing nor decreasing in (o, \(\frac{\pi}{2}\))
Hence, f(x) is not a decreasing function in (o, \(\frac{\pi}{2}\)).

(D) Let, f(x) = tan x then f’(x) = sec² x
In interval x ∈ (o, \(\frac{\pi}{2}\))f’(x) > 0
∴ f(x) is an increasing function.
Thus, (A) cos x (B) cos 2x are strictly decreasing functions on (o, \(\frac{\pi}{2}\)).

Question 13.
On which of the following intervals is the function f given by f(x) = x¹⁰⁰ + sin x – 1 strictly decreasing?
(A) (0, 1)

(B) (\(\frac{\pi}{2}\), π)

(C) (0, \(\frac{\pi}{2}\))

(D) None of these
Solution.
Given, f(x) = x¹⁰⁰ + sin x – 1
⇒ f'(x) = 100 x⁹⁹ + cos x
In interval (0, 1), cos x > 0 and 100 x⁹⁹ > 0.
f'(x) > 0
Thus, function f is strictly increasing in interval (0, 1).
In interval (\(\frac{\pi}{2}\), π), cos x < 0 and 100 x⁹⁹ < 0.
also, 100 x⁹⁹ > cos x
∴ f'(x) > 0 in (\(\frac{\pi}{2}\), π)
In interval (0, \(\frac{\pi}{2}\)), cos x > 0 and 100 x⁹⁹ > 0
∴ 100 x⁹⁹ + cos x > 0
⇒ f'(x) > 0 on (0, \(\frac{\pi}{2}\))
∴ f is strictly increasing in interval (0, \(\frac{\pi}{2}\))
Hence, function f is strictly decreasing in the given intervals.
The correct answer is (D).

Question 14.
Find the least value of a such that the function f given by fix) – x2 + ax + 1 is strictly increasing on (1, 2).
Solution.
Given, f(x) = x² + ax +1
⇒ f'(x) = 2x + a
Now, function f will be increasing in (1, 2), if f'(x) > 0 in (1, 2).
∴ f'(x) > 0
⇒ 2x + a > 0
⇒ 2x > – a
⇒ x > – \(\frac{a}{2}\)
Therefore, we have to find the least value of a such that x > – \(\frac{a}{2}\), when x ∈ (1, 2).
⇒ x > – \(\frac{a}{2}\) (when 1 < x < 2)
Thus, the least value of a for f to be increasing on (1, 2) given by
– \(\frac{a}{2}\) = 1
⇒ a = – 2
Hence, the least value of a is – 2.

Question 15.
Let I be any interval disjoint from (- 1, 1). Prove that the function f given by f(x) = x + \(\frac{1}{x}\) is strictly increasing on I.
Solution.
Given, f(x) = x + \(\frac{1}{x}\)
∴ f'(x) = 1 – \(\frac{1}{x^{2}}\)

= \(\frac{x^{2}-1}{x^{2}}\)
Now, x ∈ I ⇒ x ∉ (- 1, 1)
⇒ x ≤ – 1 or x ≥ 1
⇒ x² ≥ 1
⇒ x² – 1 ≥ 0
⇒ \(\frac{x^{2}-1}{x^{2}}\) ≥ 0 [∵ x² ≥ 1 > 0]
∴ f'(x) ≥ 0
Thus, f'(x) ≥ 0 for all x ∈ I.
Hence, f'(x) is strictly increasing on I.

Question 16.
Prove that the function f given hy f(x) = log sin x is strictly increasing on (0, \(\frac{\pi}{2}\)) and strictly decreasing on (\(\frac{\pi}{2}\), π).
Solution.
Given, f(x) = log sin x
∴ f'(x) = \(\frac{1}{\sin x}\) . cos x = cot x

In interval (0, \(\frac{\pi}{2}\)), f'(x) = cot x > 0

∴ f is strictly increasing in (0, \(\frac{\pi}{2}\))

In inerval (\(\frac{\pi}{2}\), π), f'(x) = cot x < 0 ∴ f is strictly increasing on (\(\frac{\pi}{2}\), π).

Question 17.
Prove that the function f given by f(x) = log (cos x) is strictly decreasing on (0, \(\frac{\pi}{2}\)) and strictly increasing on (\(\frac{\pi}{2}\), π). Solution. Given, f(x) = log (cos x) ⇒ f’(x) = \(\frac{1}{\cos x}\) (- sin x) = – tan x In interval (0, \(\frac{\pi}{2}\)), tan x > 0
⇒ – tan x < 0
∴ f’(x) < 0 on (0, \(\frac{\pi}{2}\))
∴ f is strictly decreasing on (o, \(\frac{\pi}{2}\))
In interval (\(\frac{\pi}{2}\), π) tan x < 0 ⇒ – tan x > 0.
∴ f’(x) > 0 on (\(\frac{\pi}{2}\), π)
∴ f is strictly increasing on (\(\frac{\pi}{2}\), π).

Question 18.
Prove that the function given by f(x) = x³ – 3x² + 3x – 100 is increasing in R.
Solution.
Given, f(x) = x³ – 3x² + 3x – 100
f'(x) = 3x² – 6x + 3
= 3(x² – 2x + 1)
= 3(x – 1)²
For any x ∈ R, (x – 1)² > 0.
Thus, f(x) is always positive in R.
Hence, the given function (f) is increasing in R.

Question 19.
The interval in which y = x² e^-x is increasing, is
(A) (- ∞, ∞)
(B) (- 2, 0)
(C) (2, ∞)
(D) (0, 2)
Solution.
Given, y = x² e^-x
⇒ \(\frac{d y}{d x}\) = 2xe^-x – x²e^-x
= xe^-x (2 – x)
Now, \(\frac{d y}{d x}\) = 0
⇒ x = 0 and x = 2
The points x = 0 and x = 2 divide the real line into three disjoint intervals
i.e., (- ∞, 0), (0, 2) and (2, ∞).
In intervals (- ∞, 0) and (2, ∞), f(x) < 0 and e~x is always positive. ∴ f is decreasing on (- ∞, 0) and (2, ∞). In interval (0, 2), f'(x) > 0
∴ f is strictly increasing on (0, 2).
Hence, f is strictly increasing in interval (0, 2).
The correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 5 Continuity and Differentiability Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise

Direction (1 – 11) :
Differentiate the following functions with respect to x.
Question 1.
(3x² – 9x + 5)⁹
Solution.
Let y = (3x² – 9x + 5)⁹
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (3x² – 9x + 5)⁹
= 9(3x² – 9x + 5)⁸ – (3x² – 9x + 5) dx
= 9(3x² – 9x + 5)⁸ . ⁸ (6x – 9)
= 9(3x² – 9x + 5)⁸ . 3(2x – 3)
= 27(3x² – 9x + 5)⁸ (2x – 3)

Question 2.
sin³ x + cos⁶ x
Solution.
Let y = sin³ x + cos⁶ x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (sin³ x) + \(\frac{d}{d x}\) (cos⁶ x)
= 3 sin² x \(\frac{d}{d x}\) (sin x) + 6 cos⁵ x . \(\frac{d}{d x}\) (cos x)
= 3 sin² x cos x + 6 cos⁵ x . (- sin x)
= 3 sin x cos x (sin x – 2 cos⁴ x).

Question 3.
(5x)^{3 cos 2x}
Solution.
Let y = (5x)^{3 cos 2x}
Taking logarithm on both sides, we get
log y = 3 cos 2x log 5x
Differentiating on both sides w.r.t. x, we get
\(\frac{1}{y}\) \(\frac{d y}{d x}\) = 3 [log 5x . \(\frac{d}{d x}\) (cos 2x) + cos 2x . \(\frac{d}{d x}\) (log 5x)]

⇒ \(\frac{d y}{d x}\) = 3y [log 5x (- sin 2x) . \(\frac{d}{d x}\) (2x) + cos 2x . \(\frac{1}{5 x}\) . \(\frac{d}{d x}\) (5x)]

⇒ \(\frac{d y}{d x}\) = 3y [- 2 sin 2x log 5x + \(\frac{\cos 2 x}{x}\)]

⇒ \(\frac{d y}{d x}\) = y [\(\frac{3 \cos 2 x}{x}\) – 6 sin 2x log 5x]

∴ \(\frac{d y}{d x}\) = (5x)^{3 cos 2x} [\(\frac{3 \cos 2 x}{x}\) – 6 sin 2x log 5x]

Question 4.
sin^-1(x√x), 0 ≤ x ≤ 1
Solution.
Let y = sin^-1 (x√x)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) sin^-1 (x√x)
= \(\frac{1}{\sqrt{1-(x \sqrt{x})^{2}}} \times \frac{d}{d x}(x \sqrt{x})\)

= \(\frac{1}{\sqrt{1-x^{3}}} \cdot \frac{d}{d x}\left(x^{\frac{3}{2}}\right)\)

= \(\frac{1}{\sqrt{1-x^{3}}} \times \frac{3}{2} \cdot x^{\frac{1}{2}}\)

= \(\frac{3 \sqrt{x}}{2 \sqrt{1-x^{3}}} \Rightarrow \frac{3}{2} \sqrt{\frac{x}{1-x^{3}}}\)

Question 5.
\(\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}\), – 2 < x < 2
Solution.
Let y = \(\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}\)
Differentiating w.r.t. x, we get

Question 6.
cot^-1 \(\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]\), 0 < x < \(\frac{\pi}{2}\)
Solution.

Therefore, Eq. (i) becomes
y = cot^-1 (cot \(\frac{x}{2}\))
⇒ y = \(\frac{x}{2}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{1}{2}\) \(\frac{d}{d x}\) (x)
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2}\).

Question 7.
(log x)^{log x}, x > 1
Solution.
Let y = (log x)^{log x}
Taking logarithm on both sides, we get
log y = log (log x)log x
⇒ log y = log x . log(log x)
Differentiating both sides w.r.t. x, we get
\(\frac{1}{y}\) . \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [log x . log(log x)] y
⇒ \(\frac{1}{y}\) . \(\frac{d y}{d x}\) = log (log x) . \(\frac{d}{d x}\) (log x) + log x . \(\frac{d}{d x}\) [log (log x)]

⇒ \(\frac{d y}{d x}\) = y [log (log x) . \(\frac{1}{x}\) + log x . \(\frac{1}{\log x}\) . \(\frac{d}{d x}\) (log x)]

⇒ \(\frac{d y}{d x}\) = y [\(\frac{1}{x}\) log (log x) + \(\frac{1}{x}\)]

∴ \(\frac{d y}{d x}\) = (log x)^{log x} [\(\frac{1}{x}\) log (log x) + \(\frac{1}{x}\)]

Question 8.
cos(a cos x + b sin x), for some constant a and b.
Solution.
Let y = cos(acosx + bsinx)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) cos(a cos x + b sin x)
\(\frac{d y}{d x}\) = – sin(a cos x + b sin x) . \(\frac{d}{d x}\) (a cos x + b sin x)
= – sin(a cos x + b sin x) . [a (- sin x) + b cos x]
= (a sin x – b cos x) . sin (a cos x + b sin x).

Question 9.
(sin x – cos x)^{(sin x – cos x)}, \(\frac{\pi}{4}<x<\frac{3 \pi}{4}\)
Solution.
Let y = (sin x – cos x) (sin x – cos x)
Taking logarithm on both sides, we get
log y = log (sin x – cos x)[(sin x – cos x)^{(sin x – cos x)}]
⇒ log y = (sin x – cos x) . log(sin x – cos x)
Differentiating on both sides w.r.t. x, we get
\(\frac{1}{y}\) . \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [(sin x – cos x) log(sin x – cos x)]

⇒ \(\frac{1}{y}\) . \(\frac{d y}{d x}\) = log (sin x – cos x) . \(\frac{d}{d x}\) log (sin x – cos x) – (sin x – cos x) + (sin x – cos x) . \(\frac{d}{d x}\) log (sin x – cos x)

⇒ \(\frac{d y}{d x}\) = (sin x – cos x)^{(sin x – cos x)} [(cos x + sin x) . log (sin x – cos x) + (cos x – sin x)]

∴ \(\frac{d y}{d x}\) = (sin x – cos x)^{(sin x – cos x)} (cos x + sin x) [1 + log (sin x – cos x)]

Question 10.
x^x + x^a + a^x + a^a, for some fixed a > 0 and x > 0.
Solution.
Let y = x^x + x^a + a^x + a^a
Also, let x^x = u, x^a = v, a^x = w, and a^a = s
∴ y = u + v + w + s
Differentiating w.r.t. x, we get

\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}+\frac{d s}{d x}\) ……………(i)

Then, u = x^x
⇒ log u = log x^x (Taking log on both sides)
⇒ log u = x logx
Differentiating on both sides w.r.t. x, we get
\(\frac{1}{u}\) \(\frac{d u}{d x}\) = log x . \(\frac{d}{d x}\) (x) + x . \(\frac{d}{d x}\) (log x)
⇒ \(\frac{d u}{d x}\) = u [log x . 1 + x . \(\frac{1}{x}\)]

⇒ \(\frac{d u}{d x}\) = x^x [log x + 1]
= x^x (1 + log x) …………..(ii)

v = x^a
Differentiating w.r.t. x, we get
\(\frac{d v}{d x}\) = \(\frac{d}{d x}\) (x^a)

⇒ \(\frac{d v}{d x}\) = a x^{a – 1} ………….(iii)

w = a^x
⇒ log w = log a^x
⇒ log w = a log x
Differentiating on both sides w.r.t. x, we get
⇒ \(\frac{d w}{d x}\) = log a . \(\frac{d}{d x}\) (x)

⇒ \(\frac{d w}{d x}\) = a^x log a ………………(iv)

s = a^a
Since, a is constant, therefore a^a is also a constant.
∴ \(\frac{d s}{d x}\) = 0 …………..(v)
From Eqs. (i), (ii), (iii), (iv) and (v), we get
\(\frac{d y}{d x}\) = x^x (1 + log x) + a x^{a – 1} + a^x log a + 0
= xx^x (1 + log x) + a x^{a – 1} + a^x log a.

Question 11.
x^{x2 – 3} + (x – 3)^x2, for x > 3.
Solution.
Let y = x^{x2 – 3} + (x – 3)^x2
Also, let u = x^{x2 – 3} and v = (x – 3)^x2
∴ y = u + v
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
Then, u = x^{x2 – 3}
Taking log on both sides, we get
log u = log x^{x2 – 3}
= (x² – 3) log x
Differentiating w.r.t. x, we get
\(\frac{1}{u}\) \(\frac{d u}{d x}\) = 2x log x + (x² – 3) log x × \(\frac{1}{x}\)

∴ \(\frac{d u}{d x}\) = u (2x log x + \(\frac{x^{2}-3}{x}\))
= x^{x2 – 3} (2x log x + \(\frac{x^{2}-3}{x}\)) …………….(ii)
Also, v = (x – 3)^{x2
Differentiating on both sides, we get
\(\frac{1}{v}\) \(\frac{d v}{d x}\) = \(\frac{d v}{d x}\) [2x log (x – 3) + \(\frac{x^{2}-3}{x}\)] ……………..(iii)
From Eqs. (i), (ii) and (iii), we get
\(\frac{d y}{d x}\) = xx2 – 3 \(\left(2 x \log x+\frac{x^{2}-3}{x}\right)+(x-3)^{x^{2}}\left[2 x \log (x-3)+\frac{x^{2}}{x-3}\right]\)}

Question 12.
Find \(\frac{d y}{d x}\), if y = 12 (1 – cos t), x = 10 (t – sin t), \(-\frac{\pi}{2}<t<\frac{\pi}{2}\)
Solution.
Given, y = 12 (1 – cos t) and x = 10 (t – sin t)
Differentiating w.r.t t, we get
\(\frac{d x}{d t}\) = \(\frac{d}{d t}\) [10 (t – sin t)]
= 10 . \(\frac{d}{d t}\) (t – sin t)
= 10 (1 – cos t)

and \(\frac{d y}{d t}\) = \(\frac{d}{d t}\) [12 (1 – cos t)]
= 12 . \(\frac{d}{d t}\) (1 – cos t)
= 12 . [0 – (- sin t)] = 12 sin t

Now, \(\frac{d y}{d x}\) = \(\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{12 \sin t}{10(1-\cos t)}\)

= \(\frac{12 \cdot 2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{10 \cdot 2 \sin ^{2} \frac{t}{2}}=\frac{6}{5} \cot \frac{t}{2}\).

Question 13.
Find \(\frac{d y}{d x}\), if y = sin^-1 x + sin^-1 \(\sqrt{1-x^{2}}\), – 1 ≤ x ≤ 1.
Solution.
Given, y = sin^-1 x + sin^-1 \(\sqrt{1-x^{2}}\)
putting x = sin θ in above eq., we get
y = sin^-1 sin θ + sin^-1 \(\sqrt{1-(\sin \theta)^{2}}\)
y = θ + sin^-1 \(\sqrt{1-(\sin \theta)^{2}}\)
= θ + sin^-1 (cos θ)
= θ + sin^-1 sin (\(\frac{\pi}{2}\) – θ)
= θ + \(\frac{\pi}{2}\) – θ
Differentiating w.r.t x, we get
∴ \(\frac{d y}{d x}\) = 0

Question 14.
If x \(\sqrt{1+y}\) + y \(\sqrt{1+x}\) = 0, for – 1 < x < 1, prove that \(\frac{d y}{d x}=\frac{1}{(1+x)^{2}}\).
Solution.
Given, x \(\sqrt{1+y}\) + y \(\sqrt{1+x}\) = 0
⇒ x \(\sqrt{1+y}\) = – y \(\sqrt{1+x}\)
On squaring bothsides, we get
x² (1 + y) = y² (1 + x)
⇒ x² + x²y = y² + xy²
⇒ x² – y² = xy² – x²y
⇒ x² – y² = xy (y – x)
⇒ (x + y) (x – y) = xy (y – x)
∴ x + y = -xy
⇒ (1 + x) y = – x
⇒ y = \(\frac{-x}{(1+x)}\)
Differentiating on bothsides w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{(1+x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+x)}{(1+x)^{2}}\)

= \(-\frac{(1+x)-x}{(1+x)^{2}}=-\frac{1}{(1+x)^{2}}\)

Hence proved.

Question 15.
If (x – a)² + (y – b)² = c², for some c > 0, prove that \(\frac{\left(1+\left(\begin{array}{l}
d y \\
d x
\end{array}\right)^{2}\right]^{3}}{d^{2} y} \frac{d x^{2}}{}\) is a constant independent of a and b.
Solution.
Given, (x – a)² + (y – b)² = c²
Differentiating on both sides w.r.t. x. we get
\(\frac{d}{d x}\) [(x – a)²] + \(\frac{d}{d x}\) [(y – b)²] = \(\frac{d}{d x}\) (c²)
⇒ 2 (x – a) . \(\frac{d}{d x}\) (x – a) + 2 (y – b) . \(\frac{d}{d x}\) (y – b) = 0
⇒ 2 (x – a) . 1 + 2(y – b) . \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{-(x-a)}{y-b}\) ………………(i)
Again, differentiating w.r.t. x, we get
∴ \(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\left[\frac{-(x-a)}{y-b}\right]\)

= – \(\left[\frac{(y-b) \cdot \frac{d}{d x}(x-a)-(x-a) \cdot \frac{d}{d x}(y-b)}{(y-b)^{2}}\right]\)

Hence proved.

Question 16.
If cos y = x cos (a + y), with cos a ≠ ± 1, prove that \(\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}\).
Solution.
Given, cos y = x cos(a + y)
Differentiating w.r.t. x, we get
\(\frac{d}{d x}\) [cos y] = \(\frac{d}{d x}\) [x cos (a + y)]
⇒ – sin y = cos (a + y) . \(\frac{d}{d x}\) (x) + x . \(\frac{d}{d x}\) [cos (a + y)]
⇒ – sin y \(\frac{d y}{d x}\) = cos (a + y) + x [- sin (a + y)]
⇒ [x sin (a + y) – sin y] \(\frac{d y}{d x}\) = cos (a + y) ……………(i)
∵ cos y = x cos (a + y), x = \(\frac{\cos y}{\cos (a+y)}\)
Then, Eq. (i) reduces to
[\(\frac{\cos y}{\cos (a+y)}\) . sin (a + y) – sin y] \(\frac{d y}{d x}\) = cos (a + y)
⇒ [cos y . sin (a + y)] – sin y . cos (a + y)] . \(\frac{d y}{d x}\) = cos ² (a + y)
⇒ sin (a + y – y) \(\frac{d y}{d x}\) = cos² (a + y)
⇒ \(\frac{d y}{d x}\) = \(\frac{\cos ^{2}(a+y)}{\sin a}\).
Hence proved.

Question 17.
If x = a(cos t + t sin t) and y = a(sin t – t cos t), find \(\frac{d^{2} y}{d x^{2}}\).
Solution.
Given, x = a (cos t + t sin t) and y = a (sin t – t cos t)
Differentiating both Eqs. w.r.t. t, we get
∴ \(\frac{d x}{d t}\) = a \(\frac{d}{d t}\) (cos t + t sin t)

= a [- sin t + sin t . \(\frac{d}{d x}\) (t) + t . \(\frac{d}{d t}\) (sin t)]

= a [- sin t + sin t + t cos t] = at cost
Also, \(\frac{d y}{d t}\) = a \(\frac{d}{d t}\) (sin t – t cost)

= a [cos t – {cos t . \(\frac{d}{d t}\) (t) + t . \(\frac{d}{d t}\) (cost)}]
= a [cos t – {cos t – t sin t}] = at sin t

\(\frac{d y}{d x}\) = \(\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a t \sin t}{a t \cos t}\) = tan t
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (tan t)

= sec² t . \(\frac{d t}{d x}\)

[∵ \(\frac{d x}{d t}\) = at cos t
⇒ \(\frac{d t}{d x}\) = \(\frac{1}{a t \cos t}\)]

= sec² t . \(\frac{1}{a t \cos t}\)

= \(\frac{\sec ^{3} t}{a t}\), 0 < t < \(\frac{\pi}{2}\)

Question 18.
If f(x) = |x|³, show that f”(x) exists for all real x and find it.
Solution.
Here, f(x) = | x |³
When x > 0, f(x) = |x|³ = x³
Differentiating both sides w.r.t. x, we get
\(\frac{d}{d x}\) [f(x)] = \(\frac{d}{d x}\) (x³)
⇒ f'(x) = 3x²
Again, differentiating both sides w.r.t. x, we get
\(\frac{d}{d x}\) [f'(x)] = \(\frac{d}{d x}\) (3x²)
⇒ f”(x) = 6x
When x < 0,
f(x) = |x|³ = – x³
Differentiating both sides w.r.t. x, we get
\(\frac{d}{d x}\) [f(x)] = \(\frac{d}{d x}\) (- x³)
⇒ f'(x) = – 3x²
Again, differentiating both sides w.r.t. x, we get
\(\frac{d}{d x}\) [f'(x)] = \(\frac{d}{d x}\) (- 3x²) = – 6x
f”(x) = – 6x
Hence,f”(x) =

Question 19.
Using mathematical induction, prove that \(\frac{d}{d x}\) (xⁿ) = n x^{n – 1} for all positive integers n.
Solution.
Let P(n): \(\frac{d}{d x}\) (xⁿ) = nx^{n – 1} for all positive integers n
For n = 1,
P(1): \(\frac{d}{d x}\)(x) = 1 = 1 . x^{1 – 1}
∴ P(n) is true for n = 1.
Let P(k) is true for some positive integer k.
i.e., P(k): \(\frac{d}{d x}\) (x^k) = k x^{k – 1}
It has to be proved that P(k +1) is also true.
Consider \(\frac{d}{d x}\) (x^{k – 1}1) = \(\frac{d}{d x}\) (x . x^k)
= x^k . \(\frac{d}{d x}\) (x) + x . \(\frac{d}{d x}\) (x^k) [Applying product rule]
= (k + 1) . x^k
= (k + 1) . x^{(k+1) – 1}
Thus, P(k + 1) is true whenever P(k) is true.
Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n. Hence proved.

Question 20.
Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Solution.
We have, sin(A + B) = sin A cos B + cos A sin B
Differentiating on both sides w.r.t. x, we get
⇒ \(\frac{d}{d x}\) [sin(A + B)] = \(\frac{d}{d x}\) (sin A cos B) + \(\frac{d}{d x}\) (cos A sin B)

⇒ cos (A + B) . \(\frac{d}{d x}\) (A + B) = cos B . \(\frac{d}{d x}\) (sin A) + sin A . \(\frac{d}{d x}\) (cos B) + sin B . \(\frac{d}{d x}\) (cos A) + cos A . \(\frac{d}{d x}\) (sin B)
⇒ cos (A + B) . \(\frac{d}{d x}\) (A + B) = cos B cos A \(\frac{d A}{d x}\) + sin A(- sin B) \(\frac{d B}{d x}\) + sin B (- sin A) . \(\frac{d A}{d x}\) + cos A cos B \(\frac{d B}{d x}\)
cos (A + B) = cos A cos B – sin A sin B.

Question 21.
Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
Solution.
Yes, we have the continuous function f(x) = |x – 1| + |x – 2|, which is continuous at all x ∈ R but differentiable at all x except 1, 2.
Here, f(x) = |x – 1| + |x – 2|
=

i.e., f(x) =
Since, polynomial function is continuous, so it is clear that f is continuous at all except possible at 1, 2.
Now, we have to check the continuity at 1, 2
At x = 1 f(1) = 1,
LHL = \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1^{-}}\) (- 2x + 3)
= – 2 + 3 = 1

RHL = \(\lim _{x \rightarrow 1^{+}}\) f(x) = \(\lim _{x \rightarrow 1^{+}}\) 1 = 1

⇒ f(1) = \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1^{+}}\) f(x)
⇒ f is continuous at x = 1

At x = 2 f(2) = 1
⇒ LHL = \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{-}}\) (1) = 1

RHL = \(\lim _{x \rightarrow 2^{+}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) (2x – 3)
= 2 × 2 – 3 = 1

f(x) = \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) f(x)

So, f is continuous at x = 2. Thus, f is continuous at all x ∈ R. -2, if x < 1

Lf'(2) ≠ Rf'(2)
⇒ f is not differentiale at 2.
Thus, we see that f(x) = |x – 1| + |x – 2| is continuous everywhere and differentiable also at all x ∈ R except at 1, 2.

Question 22.
If y = \(\begin{array}{ccc}
\boldsymbol{f}(\boldsymbol{x}) & \boldsymbol{g}(\boldsymbol{x}) & \boldsymbol{h}(\boldsymbol{x}) \\
\boldsymbol{l} & \boldsymbol{m} & \boldsymbol{n} \\
\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c}
\end{array}\), prove that \(\frac{d y}{d x}=\left|\begin{array}{ccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
l & m & n \\
a & b & c
\end{array}\right|\).
Solution.

Question 23.
If y = e^{a cos-1 x} ,- 1 ≤ x ≤ 1, show that (1 – x²) \(\frac{d^{2} y}{d x^{2}}\) – x \(\frac{d y}{d x}\) – a² y = 0.
Solution.
Given, y = e^{a cos-1 x}
Taking logarithm on bothsides, we get
⇒ log y = a cos^-1 x log e
⇒ log y = a cos^-1 x [∵ log e = 1]

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 5 Continuity and Differentiability Ex 5.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7

Direction (1 – 10) : Find the second order derivatives of the following functions.

Question 1.
x² + 3x + 2
Solution.
Let y = x² + 3x + 2
Differentiating w.r.t. r, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x²) + \(\frac{d}{d x}\) (3x) + \(\frac{d}{d x}\) (2)
= 2x + 3 + 0
= 2x + 3
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (2x + 3)
= \(\frac{d}{d x}\) (2x) + \(\frac{d}{d x}\) (3)
= 2 + 0 = 2.

Question 2.
x²⁰
Sol.
Let y = x²⁰
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x²⁰)
= 20 x¹⁹
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (20 x¹⁹)
= 20 \(\frac{d}{d x}\) (x¹⁹)
= 20 × 19 × x¹⁸
= 380 × x¹⁸

Question 3.
x . cos x
Solution.
Let y = x cos x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x . cos x)
= cos x . \(\frac{d}{d x}\) (x) + x \(\frac{d}{d x}\) (cos x)
= cos x . 1 + x (- sin x)
= cos x – x sin x
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) [cos x – x sin x]
= \(\frac{d}{d x}\) (cos x ) – \(\frac{d}{d x}\) (x sin x)
= – sin x – [sin x . \(\frac{d}{d x}\) (x) + x . \(\frac{d}{d x}\) (sin x)]
= – sin x – (sin x + x cos x)
= – (x cos x + 2 sin x)

Question 4.
log x
Solution.
Let y = log x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (log x) = \(\frac{1}{x}\)
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (\(\frac{1}{x}\))
= \(\frac{-1}{x^{2}}\)

Question 5.
x³ log x
Solution.
Let y = x³ log x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [x³ log x]
= log x . \(\frac{d}{d x}\) (x³) + x³ . \(\frac{d}{d x}\) (log x)
= log x . 3x² + x³ . \(\frac{1}{x}\)
= log x . 3x² + x²
= x² (1 + 3 log x)
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) [x² (1 + 3 log x)]
= (1 + 3 log x) . \(\frac{d}{d x}\) (x²) + x² \(\frac{d}{d x}\) (1 +3 log x)
= (1 + 3 log x) . 2x + x² . \(\frac{3}{x}\)
= 2x + 6x log x + 3x
= 5x + 6x log x
= x (5 + 6 log x)

Question 6.
e^x sin 5x
Solution.
Let y = e^x sin 5x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (e^x sin 5x)
= sin 5x . \(\frac{d}{d x}\) (e^x) + e^x . cos 5x . \(\frac{d}{d x}\) (5x)
= e^x sin 5x + e^x cos 5x . 5
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) [e^x (sin 5x + 5 cos 5x)]
= (sin 5x + 5 cos 5x) . \(\frac{d}{d x}\) (e^x) + e^x . \(\frac{d}{d x}\) (sin 5x + 5 cos 5x)
= (sin 5x + 5 cos 5x) e^x + e^x [cos 5x . \(\frac{d}{d x}\) (5x) + 5 (- sin 5x) . \(\frac{d}{d x}\) (5x)]
= e^x (sin 5x + 5 cos 5x) + e^x (5 cos 5x – 25 sin 5x)
= e^x (10 cos 5x – 24 sin 5x)
= 2 e^x (5 cos 5x – 12 sin 5x)

Question 7.
e^6x cos 3x
Solution.
Let y = e^6x cos 3x
Differentiating w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (e^6x . cos 3x)

= cos 3x . e^6x \(\frac{d}{d x}\) (6x) + e^6x . (- sin 3x) . \(\frac{d}{d x}\) (3x)

= 6 e^6x cos 3x – 3 e^6x sin 3x …………….(i)
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (6 e^6x cos 3x – 3 e^6x sin 3x)

= 6 . \(\frac{d}{d x}\) (e^6x cos 3x) – 3 \(\frac{d}{d x}\) (e^6x sin 3x)

= 6 . [6e^6x cos 3x – 3 e^6x sin 3x] – 3 . [sin 3x . \(\frac{d}{d x}\) (e^6x) + e^6x \(\frac{d}{d x}\) (sin 3x)]

36e^6x cos 3x – 18 e^6x sin 3x – 3 [sin 3x . e^6x . 6 + e^6x . cos 3x . 3] = 36 e^6x cos 3x – 18 e^6x sin 3x – 18 e^6x sin 3x – 9 e^6x cos 3x
= 27 e^6x cos 3x – 36 e^6x sin 3x
= 9 e^6x (3 cos 3x – 4 sin 3x)

Question 8.
tan^-1 x
Solution.
Let y = tan^-1 x
Differentiating w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (tan^-1 x)

= \(\frac{1}{1+x^{2}}\)

Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (\(\frac{1}{1+x^{2}}\))

= \(\frac{d}{d x}\) (1 + x²)^-1

= (- 1) (1 + x²)^-2 . \(\frac{d}{d x}\) (1 +x²)

= \(\frac{-1}{\left(1+x^{2}\right)^{2}}\) 2x

= \(\frac{-2 x}{\left(1+x^{2}\right)^{2}}\)

Question 9.
log (log x)
Solution.
Lety = log (log x)
Differentiating w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [log (log x)]

= \(\frac{1}{\log x}\) . \(\frac{d}{d x}\) (log x)

= \(\frac{1}{x \log x}\) = (x log x)^-1

Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (x log x)^-1

= (- 1) . (x log x)^-2 . \(\frac{d}{d x}\) (x log x)

= \(\frac{-1}{(x \log x)^{2}} \cdot\left[\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)\right]\)

= \(\frac{-1}{(x \log x)^{2}} \cdot\left[\log x \cdot 1+x \cdot \frac{1}{x}\right]=\frac{-(1+\log x)}{(x \log x)^{2}}\)

Question 10.
sin (log x)
Solution.
Let y = sin (log x)
Differentiating w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [sin (log x)]

= cos log x . \(\frac{d}{d x}\) (log x)

= \(\frac{\cos (\log x)}{x}\)

Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) [latex]\frac{\cos (\log x)}{x}[/latex]

Question 11.
If y = 5 cos x – 3 sin x, prove that \(\frac{d^{2} y}{d x^{2}}\) = 0.
Solution.
Given, y = 5 cos x – 3sin x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (5 cos x) – \(\frac{d}{d x}\) (3 sin x)
= 5 \(\frac{d}{d x}\) (cos x) – 3 \(\frac{d}{d x}\) (sin x)
= 5(- sin x) – 3 cos x
= – (5 sin x + 3 cos x)
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) [- (5 sin x + 3 cos x)]
= – [5 . \(\frac{d}{d x}\) (sin x) + 3 . \(\frac{d}{d x}\) (cos x)]
= – [5 cos x + 3 (- sin x)]
= – [5 cos x – 3 sin x] = – y
\(\frac{d^{2} y}{d x^{2}}\) + y = 0
∴ Hence proved.

Question 12.
If y = cos^-1 x, find \(\frac{d^{2} y}{d x^{2}}\) in terms of y alone.
Solution.
Given, y = cos^-1 x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (cos^-1 x)

= \(\frac{-1}{\sqrt{1-x^{2}}}=-\left(1-x^{2}\right)^{\frac{-1}{2}}\)

Again, differentiating w.r.t. x, we get

Question 13.
If y = 3 cos (log x) + 4 sin (log x), show that x²y₂ + xy₁ + y = 0.
Solution.
Given, y = 3 cos (log x) + 4 sin (log x), show that x²y₂ + xy₁ + y = 0.
Differentiating w.r.t. x, we get
\(\frac{d x}{d y}\) = y₁
= 3 . \(\frac{d}{d x}\) [cos (log x)] + 4 . \(\frac{d}{d x}\) [sin (log x)]

= 3 . [- sin(log x) . \(\frac{d}{d x}\) (log x)] + 4 . [cos (log x) . \(\frac{d}{d x}\) (log x)]

= – sin (log x) – 7 cos (log x) + 4 cos (log x) – 3 sin (log x) + 3 cos (log x) + 4 sin (log x)
= 0
Hence proved.

Question 14.
If y = A e^mx + B e^nx, show that \(\frac{d^{2} y}{d x^{2}}\) – (m + n) \(\frac{d y}{d x}\) + mny = 0.
Solution.
\(\frac{d y}{d x}\) = A . \(\frac{d}{d x}\) (e^mx) + B . \(\frac{d}{d x}\) (e^nx)

= A . e^mx . \(\frac{d}{d x}\) (mx) + B . e^nx . \(\frac{d}{d x}\) (nx)

= Ame^mx + Bne^nx

Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (Ame^mx + Bne^nx)

= Am . \(\frac{d}{d x}\) (e^mx) + Bn . \(\frac{d}{d x}\) (e^nx)

Am . e^mx . \(\frac{d}{d x}\) (mx) + Bne^nx \(\frac{d}{d x}\) (nx)

= Am² e^mx + Bn² e^nx

∴ \(\frac{d^{2} y}{d x^{2}}\) – (m + n) \(\frac{d y}{d x}\) + mny

= Am² e^mx + Bn² e^nx – (m + n) . (Ame^mx + Bne^nx) + mn (Ae^mx + Be^nx)

= Am² e^mx + Bn² e^nx – Am² e^mx – Bmne^nx – Amne^nx – Bn² e^nx + Amne^mx + Bmne^nx = 0

Hence proved.

Question 15.
If y = 500 e^7x + 600 e^-7x, show that \(\frac{d^{2} y}{d x^{2}}\) = 49y.
Solution.
Given, y = 500 e^7x + 600 e^-7x
Differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = 500 . \(\frac{d}{d x}\) (e^7x) + 600 . \(\frac{d}{d x}\) (e^-7x)

= 500 . e^7x \(\frac{d}{d x}\) (7x) + 600 . \(\frac{d}{d x}\) . (- 7x)

Again, differentiating w.r.t. x, we get

\(\frac{d^{2} y}{d x^{2}}\) = 3500 . \(\frac{d}{d x}\) (e^7x) – 4200 \(\frac{d}{d x}\) (e^-7x)

= 3500 . e^7x \(\frac{d}{d x}\) (7x) – 4200 . e^-7x \(\frac{d}{d x}\) (- 7x)

= 7 × 3500 . e^7x + 7 × 4200 . e^-7x
= 49 × 500 e^7x + 49 × 600 ^-7x
= 49 (500 e^7x + 600 e^-7x) = 49 y
Hence proved.

Question 16.
If e^y (x + 1) = 1, show that \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\).
Solution.
Given, e^y (x + 1) = 1

e^y = \(\frac{1}{x+1}\)

Taking logarithm on both sides, we get
y = log \(\frac{1}{x+1}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = (x + 1) \(\frac{d}{d x}\) \(\frac{1}{x+1}\)

= (x + 1) . \(\frac{-1}{(x+1)^{2}}=\frac{-1}{x+1}\)

Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}=-\frac{d}{d x}\left(\frac{1}{x+1}\right)\) = \(-\left[\frac{-1}{(x+1)^{2}}\right]=\frac{1}{(x+1)^{2}}\)

⇒ \(\frac{d^{2} y}{d x^{2}}=\left(\frac{-1}{x+1}\right)^{2}\)

⇒ \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\)
Hence proved.

Question 17.
If y = (tan^-1 x)², show that (x² + 1)² y₂ + 2x (x² + 1) y₁ = 2.
Solution.
Given, y = (tan x)²
Differentiating w.r.t. x, we get
y₁ = 2 tan^-1 x \(\frac{d}{d x}\) (tan^-1 x)
⇒ y₁ = 2 tan^-1 . \(\frac{1}{1+x^{2}}\)

⇒ (1 + x²) y₁ = 2 tan^-1 x
Again, differentiating w.r.t. x, we get

(1 + x²) \(\frac{d y_{1}}{d x}\) + y₁ \(\frac{d}{d x}\) (1 + x²) = \(\frac{2}{1+x^{2}}\)

⇒ (1 + x²) y₂ + y₁ (0 + 2x) = \(\frac{2}{1+x^{2}}\)
[∵ \(\frac{d}{d x}\) (y₁) = y₁]

⇒ (1 + x²)² y₂ + 2x (1 + x²) y₁ = 2

Hence proved.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 6 Application of Derivatives Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1

Question 1.
Find the rate of change of the area of a circle with respect to its radius r when
(a) r = 3 cm
(b) r = 4 cm
Solution.
The area of a circle (A) with radius (r) is given by
A = πr²
Now, the rate of change of the area with respect to its radius is given by
\(\frac{d A}{d r}\) = \(\frac{d}{d r}\) (πr²) = 2πr

(a) When, r = 3 cm
\(\frac{d A}{d r}\) = 2π(3) = 6π
Hence, the area of the circle is changing at the rate of 6π cm²/s when its radius is 3 cm.

(b) When r = 4 cm,
\(\frac{d A}{d r}\) = 2π(4) = 8π
Hence, the area of the circle is changing at the rate of 8π cm²/s when its radius is 4 cm.

Question 2.
The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution.
Let x be the length of a side (edge), V be the volume and S be the surface area of the cube.
Then, V = x³ and S = 6x² where, x is a function of time t.
It is given that \(\frac{d V}{d t}\) = 8 cm³ /s
Then, y using the chain rule, we have

Thus, when x = 12 cm,
\(\frac{d S}{d t}\) = \(\frac{32}{12}\) cm²/s

= \(\frac{8}{3}\) cm²/s

Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \(\frac{8}{3}\) cm²/s.

Question 3.
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Solution.
The area of circle (A) with radius (r) is given by A = πr²
Now, the rate of change of area (A) with respect to time (t) is given by

\(\frac{d A}{d t}=\frac{d}{d t}\left(\pi r^{2}\right) \cdot \frac{d r}{d t}=2 \pi r \frac{d r}{d t}\)

It i s given that,
\(\frac{d r}{d t}\) = 3 cm/s
∴ \(\frac{d A}{d t}\) = 2πr (3) = 6πr
Thus, when r = 10 cm, dA
\(\frac{d A}{d t}\) = 6π(10) = 60π cm²/s dt
Hence, the rate at which the area of the circle is increasing when the radius is 10 cm, is 60π cm²/s.

Question 4.
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Solution.
Let x be the length of a side and V be the volume of the cube.
Then, V = x³
∴ \(\frac{d V}{d t}\) = 3x² . \(\frac{d x}{d t}\)
It is given that,
\(\frac{d x}{d t}\) = 3 cm/sec
⇒ \(\frac{d V}{d t}\) = 3x² (3) = 9x²
Thus, when x = 10 cm,
\(\frac{d V}{d t}\) = 9 (10)² = 900 cm³/s
Hence, the volume of the cube is increasing at the rate of 900 cm³/s when the edge is 10 cm long.

Question 5.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Solution.
The area of a circle (A) with radius (r) is given by A = πr².
Therefore, the rate of change of area (A) with respect to time (t) is given by

\(\frac{d A}{d t}=\frac{d}{d t}\left(\pi r^{2}\right)=\frac{d}{d r}\left(\pi r^{2}\right) \frac{d r}{d t}=2 \pi r \frac{d r}{d t}\)

It is given that \(\frac{d r}{d t}\) = 5 cm/s dt
Thus, when r = 8 cm,
\(\frac{d A}{d t}\) = 2π (8) (5) = 80 π dt
Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm²/s.

Question 6.
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
Solution.
The circumference of a circle (C) with radius (r) is given by C = 2nr
Therefore, the rate of change of circumference (C) with respect to time (t) is given by
\(\frac{d C}{d t}=\frac{d C}{d r} \cdot \frac{d r}{d t}\)

= \(\frac{d}{d r}(2 \pi r) \frac{d r}{d t}=2 \pi \cdot \frac{d r}{d r}\)

It is given that,
\(\frac{d r}{d t}\) = 0.7 cm/s
Hence, the rate of increase of the circumference is 2π (0.7) = 1.4π cm/s.

Question 7.
The length x of a rectangle is decreasing at the rate of 5 cm/min and the width y is increasing at the rate of 4 cm/min. When x = 8 cm and y = 6 cm, find the rates of change of
(a) the perimeter, and
(b) the area of the rectangle.
Solution.
Since, the length (x) is decreasing at the rate of 5 cm/min and the width (y) is increasing at the rate of 4 cm/min, we have
\(\frac{d x}{d t}\) = – 5 cm/min and

\(\frac{d y}{d t}\) = 4 cm/min

(a) The perimeter (P) of a rectangle is given by P = 2 (x + y)
∴ \(\frac{d P}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)\)
= 2 (- 5 + 4) = – 2 cm/min
Hence, the perimeter is decreasing at the rate of 2 cm/min.

(b) The area (A) of a rectangle is given by A = x × y
∴ \(\frac{d A}{d t}=\frac{d x}{d t} \cdot y+x \cdot \frac{d y}{d t}\)
= – 5y + 4x
When x = 8 cm and y = 6 cm,
\(\frac{d A}{d t}\) = – 5 × 6 +4 × 8
= 2 cm²/min
Hence, the area of the rectangle is increasing at the rate of 2 cm²/min.

Question 8.
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic cm of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Solution.
The volume of a sphere (V) with radius (r) is given by

V = \(\frac{4}{3}\) πr³
∴ Rate of change of volume (V) with respect to time (r) is given by
\(\frac{d V}{d t}=\frac{d V}{d r} \cdot \frac{d r}{d t}\)

= \(\frac{d}{d r}\left(\frac{4}{3} \pi r^{3}\right) \cdot \frac{d r}{d t}=4 \pi r^{2} \cdot \frac{d r}{d t}\)

It is given that \(\frac{d V}{d t}{/latex] = 900 cm³/s
∴ 900 = 4πr² . [latex]\frac{d r}{d t}\)

⇒ \(\frac{d r}{d t}\) = \(\frac{900}{4 \pi r^{2}}=\frac{225}{\pi r^{2}}\)
Therefore, when radius = 15 cm,
\(\frac{d r}{d t}=\frac{225}{\pi(15)^{2}}=\frac{1}{\pi}\).

Question 9.
A balloon, which always remains spherical, has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Solution.
The volume of a sphere (V) with radius (r) is gwen by V = \(\frac{4}{3}\) πr³.
Rate of change of volume (V) with respect to its radius (r) is given by
\(\frac{d V}{d r}=\frac{d}{d r}\left(\frac{4}{3} \pi r^{3}\right)\)

= \(\frac{4}{3}\) π (3r²) = 4πr²
Therefore, when radius = 10 cm,
\(\frac{d V}{d r}{/latex] = 4π(10)² = 400π
Hence, the volume of the balloon is increasing at the rate of 400 π cm³/s.
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm, is [latex]\frac{1}{\pi}\) cm/s.

Question 10.
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
Solution.
Let y m be the height of the wall at which the ladder touches.
Also, let the foot of the ladder be x away from the wall.
Then, by Pythagoras theorem, we have
x² + y² = 25 [∵ Length of the ladder = 5 m]
⇒ y = \(\sqrt{25-x^{2}}\)
Then, the rate of change of height (y) with respect to time (t) is given by

\(\frac{d y}{d t}=\frac{-x}{\sqrt{25-x^{2}}} \cdot \frac{d x}{d t}\)

It is given that \(\frac{d x}{d t}\) = 2 cm/s dt

∴ \(\frac{d y}{d t}=\frac{-2 x}{\sqrt{25-x^{2}}}\)

Now, when x = 4 m, we have

∴ \(\frac{d y}{d t}=\frac{-2 \times 4}{\sqrt{25-4^{2}}}=-\frac{8}{3}\)

Hence, the height of the ladder on the wall is decreasing at the rate of \(\frac{8}{3}\) cm/s.

Question 11.
A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y – coordinate is changing 8 times as fast as the x – coordinate.
Solution.
The equation of the curve is given as 6y = x³ + 2

The rate of change of the position of the particle with respect to time (t) is given by

\(6 \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}+0\)

⇒ \(2 \frac{d y}{d t}=x^{2} \frac{d x}{d t}\)

When the y – coordinate of the particle changes 8 times as fast as the x-coordinate i.e., \(\left(\frac{d y}{d t}=8 \frac{d x}{d t}\right)\), we have

Hence, the points required on the curve are (4, 11) and (- 4, \(-\frac{31}{3}\)).

Question 12.
The radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/s; At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution.
The air bubble is in the shape of a sphere.
Now, the volume of an air bubble (V) with radius (r) is given by

V = \(\frac{4}{3}\) πr³.

The rate of change of volume (V) with respect to time (t) is given by
\(\frac{d V}{d t}=\frac{4}{3} \pi \frac{d}{d r}\left(r^{3}\right) \cdot \frac{d r}{d t}\)

= \(\frac{4}{3} \pi\left(3 r^{2}\right) \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t}\)

It is given that \(\frac{d r}{d t}\) = \(\frac{1}{2}\) cm/s
Therefore, when r = 1 cm,
\(\frac{d V}{d t}\) = 4π(1)² (\(\frac{1}{2}\))
= 2π cm³/s
Hence, the rate at which the volume of the bubble increases, is 2π cm³/s.

Question 13.
A balloon, which always remains spherical, has a variable diameter \(\frac{3}{2}\) (2x + 1). Find the rate of change of its volume with respect to x.
Solution.
The volume of a sphere (V) with radius (r) is given by
V = \(\frac{4}{3}\) πr³
It is given that, Diameter = \(\frac{3}{2}\) (2x + 1)
⇒ r = \(\frac{3}{4}\) (2x + 1)
∴ V = \(\frac{4}{3} \pi\left(\frac{3}{4}\right)^{3}(2 x+1)^{3}\)

= \(\frac{9}{16} \pi(2 x+1)^{3}\)

Hence, the rate of change of volume with respect to x is

\(\frac{d V}{d x}=\frac{9}{16} \pi \frac{d}{d x}(2 x+1)\)

= \(\frac{9}{16} \pi \times 3(2 x+1)^{2} \times 2\)

= \(\frac{27}{8} \) π (2x + 1)²

Question 14.
Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Solution.
The volume of a cone (V) with radius (r) and height (h) is given by V = \(\frac{1}{3}\) πr²h
It is given that, h = \(\frac{1}{6}\) r
⇒ r = 6h
⇒ V = \(\frac{1}{3}\) π(6h)² h
= 12 πh³
The rate of change of volume with respect to time (t) is given by
\(\frac{d V}{d t}\) = 12π \(\frac{d}{d h}\) (h³) . \(\frac{d h}{d t}\)

= 12π (3h²) \(\frac{d h}{d t}\)

= 36πh² \(\frac{d h}{d t}\)

It is also given that \(\frac{d V}{d t}\) = 12 cm³/ s
Therefore, when h = 4 cm, we have
12 = 36π(4)² \(\frac{d h}{d t}\)
Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of\(\frac{1}{48 \pi}\) cm/s.

Question 15.
The total cost C(x) in rupees associated with the production of x units of an item is given by
C(x) = 0.007 x³ – 0.003 x² + 15x + 4000.
Find the marginal cost when 17 units are produced.
Solution.
Marginal cost is the rate of change of total cost with respect to output.

Marginal cost (MC) = \(\frac{d C}{d x}\)
= 0.007 (3x²) – 0.003 (2x) + 15
= 0.021 x² – 0.006x + 15
When x = 17, then MC = 0.021 (17²) – 0.006 (17) +15
= 0.021 (289) – 0.006 (17) +15
= 6.069 – 0.102 + 15 = 20.967
Hence, when 17 units are produced, the marginal cost is ₹ 20.967.

Question 16.
The total revenue in rupees received from the sale of x units of a product is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.
Solution.
Marginal revenue is the rate of change of total revenue with respect to the number of units sold.
Marginal Revenue (MR) = \(\frac{d R}{d x}\)
= 13(2x) + 26
= 26x + 26

When x = 7, then MR = 26(7) + 26
= 182 + 26 = 208
Hence, the required marginal revenue is ₹ 208.

Direction (17 – 18):
Choose the correct answer.

Question 17.
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π
(B) 12π
(C) 8π
(D) 11π
Solution.
The area of a circle (A) with radius (r) is given by
A = πr²
Therefore, the rate of change of the area with respect to its radius r is
\(\frac{d A}{d r}\) = \(\frac{d}{d r}\) (πr²) = 2πr
∴ When r = 6 cm
⇒ \(\frac{d A}{d r}\) = 2π × 6 = 12π cm²/s
Hence, the required rate of change of the area of a circle is 12π cm²/s.
The correct answer is (B).

Question 18.
The total revenue in rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue, when x = 15, is
(A) 116
(B) 96
(C) 90
(D) 126
Solution.
Marginal revenue is the rate of change of total revenue with respect to the number of units sold.
∴ Marginal Revenue (MR) = \(\frac{d R}{d x}\)
= 3(2x) + 36 = 6x + 36
∴ When x = 15, then MR = 6(15) + 36 = 90 + 36 = 126
Hence, the required marginal revenue is ₹ 126.
The correct answer is (D).