# Class 12 NCERT Solutions- Mathematics Part I – Application of Derivatives – Exercise 6.2| Set 2

### Chapter 6 Application of Derivatives – Exercise 6.2| Set 1

### Question 11. Prove that the function f given by f(x) = x^{2} – x + 1 is neither strictly increasing nor decreasing on (– 1, 1).

**Solution:**

Given: f(x) = x

^{2 }– x + 1Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

Demo Class for First Step to Coding Course,specificallydesigned for students of class 8 to 12.The students will get to learn more about the world of programming in these

free classeswhich will definitely help them in making a wise career choice in the future.f'(x) = 2x – 1

For strictly increasing, f'(x) > 0

2x – 1 > 0

x > 1/2

So, f(x) function is increasing for x > 1/2 in the interval (1/2, 1) -(Given interval is (-1, 1)

Similarly, for decreasing f'(x) < 0

2x – 1 < 0

x < 1/2

So, f(x) function is increasing for x < 1/2 in the interval (-1, 1/2) -(Given interval is (-1, 1)

Hence, the function f(x) = x

^{2 }– x + 1 is neither strictly increasing nor decreasing.

### Question 12. Which of the following functions are decreasing on (0, π/2).

### (A) cos x (B) cos 2x (C) cos 3x (D) tan x

**Solution:**

(A)f(x) = cos xf'(x) = -sin x

Now in (0, π/2) interval, sin x is positive(because it is second quadrant)

So, -sin x < 0

∴ f'(x) < 0

f(x) = cos x is strictly decreasing on(0, π/2).

(B)f(x) = cos 2xf'(x) = -2 sin 2x

Now in (0, π/2) interval, sin x is positive(because it is second quadrant)

-sin 2x < 0

∴ f'(x) < 0,

f(x) = cos 2x is strictly decreasing on(0, π/2).

(C)f(x) = cos 3xf'(x) = -3sin 3x

Let 3x = t

So in sin 3x = sin t

When t ∈(0, π), sin t + >0 or 3x ∈ (0, π)

But when π/3 < x < π/2

π < 3x < 3π/2

Here sin 3x < 0

So, in x ∈ (0, π/3),

f'(x) = -3sin 3x < 0 & in x∈(π/3, π/2), f'(x) = -3sin 3x > 0

f'(x) is changing signs, hence f(x) is not strictly decreasing.

(D)f(x) = tan xf'(x) = sec

^{2}xNow in x ∈ (0, π/2), sec

^{2}x > 0Hence, f(x) is strictly increasing on(0, π/2).

So, option (A) and (B) are decreasing on (0, π/2).

### Question 13. On which of the following intervals is the function f given by f(x) = x^{100} + sin x – 1 decreasing ?

### (A) (0, 1) (B) π/2, π (C) 0, π/2 (D) None of these

**Solution:**

f(x) = x

^{100 }+ sin x – 1f'(x) = 100x

^{99 }+ cos x

(A)In(0, 1) interval, x > 0, so 100x^{99 }> 0and for cos x: (0, 1°) = (0, 0.57°) > 0

Hence, f(x)is strictly increasing in interval(0, 1)

(B)In(π/2, π) interval,For 100x

^{99}: x ∈ (π/2, π) = (11/7, 22/7) = (1.5, 3.1) > 1So, x

^{99 }> 1. Hence 100x^{99 }> 100For Cos x: (π/2, π) in second quadrant and in second quadrant cos x is negative, so the value is in be -1 and 0.

Hence, f(x)is strictly increasing in interval (π/2, π)

(C)In (0, π/2) interval, both cos x > 0 and 100x^{99 }> 0So f'(x) > 0

Hence, f(x)is strictly increasing in interval (0, π/2)

So, the correct option is (D).

### Question 14. For what values of a the function f given by f(x) = x^{2} + ax + 1 is increasing on (1, 2)?

**Solution:**

Given: f(x) = x

^{2 }+ ax + 1f'(x) = 2x + a

Now, x ∈ (1, 2), 2x ∈ (2, 4)

2x + a ∈ (2 + a, 4 + a)

For f(x) to be strictly increasing, f'(x) > 0

If the minimum value of f'(x) > 0 then

f'(x) on its entire domain will be > 0.

f'(x)

_{min }> 02 + a > 0

a > -2

### Question 15. Let I be any interval disjoint from [–1, 1]. Prove that the function f given by is increasing on I.

**Solution:**

Clearly the maximum interval I is R-(-1,1)

Now, f(x) =

f'(x) =

It is given that I be any interval disjoint from [–1, 1]

So, for every x ∈ I either x < -1 or x > 1

So, for x < -1, f'(x) is positive.

So, for x < 1, f'(x) is positive.

Hence, f'(x) > 0 ∀ x ∈ I, so, f(x) is strictly increasing on I.

### Question 16. Prove that the function f given by f(x) = log sin x is increasing on (0, π/2) and decreasing on (π/2, π).

**Solution:**

f(x) = log sin x

f'(x) =

Interval (0, π/2), it is first quadrant, here cot x is positive.

So, f'(x) = cot x is positive (i.e., cot x > 0)

Hence, f(x) is strictly increasing in interval (0, π/2)

Interval (π/2, π), it is second quadrant, here cot x is negative.

So, f'(x) = cot x is negative (i.e., cot x < 0)

Hence, f(x) is strictly decreasing in interval (π/2, π)

### Question 17. Prove that the function f given by f(x) = log|cos x| is decreasing on (0, π/2) and increasing on (π/2, π).

**Solution:**

f(x) = log cos x

f'(x) = 1/cos x (-sin x) = -tan x

Interval (0, π/2), it is first quadrant, here tan x is positive.

So, f'(x) = -tan x is negative(i.e., tan x < 0)

Hence, f(x) is strictly decreasing in interval (0, π/2)

Interval (π/2, π), it is second quadrant, here tan x is negative.

So, f'(x) = -tan x is positive (i.e., tan x > 0)

Hence, f(x) is strictly increasing in interval (π/2, π)

### Question 18. Prove that the function given by f(x) = x^{3} – 3x^{2} + 3x – 100 is increasing in R.

**Solution:**

f(x) = x

^{3 }– 3x^{2 }+ 3x – 100f'(x) = 3x

^{2 }– 6x + 3f'(x) = 3(x

^{2 }– 2x + 1)f'(x) = 3(x – 1)

^{2}≥ 0 ∀ x in RSo f(x) is strictly increasing in R.

### Question 19. The interval in which y = x^{2} e^{-x} is increasing is

### (A) (– ∞, ∞) (B) (– 2, 0) (C) (2, ∞) (D) (0, 2)

**Solution:**

Given, f(x) = x

^{2}e^{-x}f'(x) = x

^{2}(-e^{-x}) + e^{-x}.2xf'(x) = e

^{-x}(2x – x^{2})f'(x) = e

^{-x}.x(2 – x)For f(x) to be increasing, f'(x) ≥ 0

So, f'(x) ≥ 0

e

^{-x}.x.(2 – x) ≥ 0x.(2 – x) ≥ 0

x(x – 2) ≥ 0

x ∈ [0, 2]

So, the f(x) is strictly increasing in interval (0, 2). Correct option in D.