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

### Question 11. Prove that the function f given by f(x) = x2 – x + 1 is neither strictly increasing nor decreasing on (– 1, 1).

Solution:

Given: f(x) = x2 – x + 1

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) = x2 – x + 1 is neither strictly increasing nor decreasing.

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

Solution:

(A) f(x) = cos x

f'(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 2x

f'(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 3x

f'(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 x

f'(x) = sec2x

Now in x ∈ (0, π/2), sec2x > 0

Hence, f(x) is strictly increasing on(0, π/2).

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

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

Solution:

f(x) = x100 + sin x – 1

f'(x) = 100x99 + cos x

(A) In (0, 1) interval, x > 0, so 100x99 > 0

and 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 100x99: x ∈ (π/2, π) = (11/7, 22/7) = (1.5, 3.1) > 1

So, x99 > 1. Hence 100x99 > 100

For 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 100x99 > 0

So 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) = x2 + ax + 1 is increasing on (1, 2)?

Solution:

Given: f(x) = x2 + ax + 1

f'(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 > 0

2 + 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) = x3 – 3x2 + 3x – 100 is increasing in R.

Solution:

f(x) = x3 – 3x2 + 3x – 100

f'(x) = 3x2 – 6x + 3

f'(x) = 3(x2 – 2x + 1)

f'(x) = 3(x – 1)2 ≥ 0 ∀ x in R

So f(x) is strictly increasing in R.

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

Solution:

Given, f(x) = x2e-x

f'(x) = x2(-e-x) + e-x.2x

f'(x) = e-x(2x – x2)

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) ≥ 0

x.(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.

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