NCERT Solutions Class 12 maths Chapter-6 (Applications of Derivatives)Exercise 6.2

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-6 (Applications of Derivatives)Exercise 6.2 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 6.2 

Set 1

Question 1. Show that the function given by f (x) = 3x + 17 is increasing on R. 

Solution:

If for a function f(x), f'(x) > 0 for all x, then the function is a strictly increasing function.  (vice-versa is not true)

Given: f(x) = 3x + 17

f'(x) = 3 > 0        -(Always greater than zero)

Hence, 3x + 17 is strictly increasing on R.

Question 2. Show that the function is given by f (x) = e^2x is increasing on R. 

Solution:

If for a function f(x), f'(x) > 0 for all x, then the function is a strictly increasing function.  (vice-versa is not true)

Given: f(x) = e2x

f’(x) = 2e2x > 0

Hence, f(x) = e2x is strictly increasing on ∞

Question 3. Show that the function given by f (x) = sin x is

(i) increasing in(0, π/2)            

(ii) decreeing in(π/2, π)      

(iii) neither increasing nor decreasing in (0, π)

Solution:

Given: f(x) = sin x

So, f’(x) = d/dx(sin x) = cos x

(i) Now in (0, π/2), f’(x) = cos x > 0 (positive in first quadrant)

Hence, f(x) = sin x is strictly increasing in (0, π/2).

(ii) In (π/2, π), f’(x) = cos x < 0           -(negative in second quadrant)

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

(iii) As we know that f’(x) = cos x is positive in interval(0, π/2)

and f’(x) = cos x is negative in interval (π/2, π)

So, it is neither increasing nor decreasing.

Question 4. Find the intervals in which the function f given by  f(x) = 2x² – 3x is 

(i) increasing  

(ii) decreasing

Solution:

Given: f(x) = 2x2 – 3x

f'(x) =  = 4x – 3           -(1)

= x = 3/4

So the intervals are (-∞, 3/4) and (3/4, ∞)

(i) Interval (3/4, ∞) let take x = 1

So, from eq(1) f'(x) > 0

Hence, f is strictly increasing in interval (3/4, ∞)

(ii) Interval (-∞, 3/4) let take x = 0.5

So, from eq(1) f'(x) < 0

Hence, f is strictly decreasing in interval (-∞, 3/4)

Question 5. Find the intervals in which the function f given by f(x) = 2x³ – 3x² – 36x + 7 is

(i) increasing 

(ii) decreasing 

Solution:

Given: f(x) = 2x3 – 3x2 – 36x + 7

f'(x) =  = 6x2 – 6x – 36         -(1)

f'(x) = 6(x2 – x – 6)

On putting f'(x) = 0, we get

6(x2 – x – 6) = 0

(x2 – x – 6) = 0

x = -2, x = 3

So, the intervals are (-∞, -2), (-2, 3), and (3, ∞)

For (-∞, -2) interval, take x = -3

From eq(1), we get

f'(x) = (+)(-)(-) = (+) > 0

So, f is strictly increasing in interval (-∞, -2)

For (-2, 3) interval, take x = 2

From eq(1), we get

f'(x) = (+)(+)(-) = (-) < 0

So, f is strictly decreasing in interval (-2, 3)

For (3, ∞)interval, take x = 4

From eq(1), we get

f'(x) = (+)(+)(+) = (+) > 0

So, f is strictly increasing in interval (3, ∞)

(i) f is strictly increasing in interval (-∞, -2) and (3, ∞)

(ii) f is strictly decreasing in interval (-2, 3)  

Question 6. Find the intervals in which the following functions are strictly increasing or decreasing: 

(i) x² + 2x – 5 

(ii) 10 – 6x – 2x² 

(iii) -2x³ – 9x² – 12x + 1 

(iv) 6 – 9x – x² 

(v) (x + 1)³ (x – 3)³

Solution:

(i) f(x) = x² + 2x – 5 

f'(x) = 2x + 2         -(1)

On putting f'(x) = 0, we get

2x + 2 = 0

x = -1

So, the intervals are (-∞, -1) and (-1, ∞)

For (-∞, -1) interval take x = -2 

From eq(1), f'(x) = (-) < 0

So, f is strictly decreasing 

For (-1, ∞) interval take x = 0

From eq(1), f'(x) = (+) > 0

So, f is strictly increasing

(ii) f(x) = 10 – 6x – 2x² 

f'(x) = -6 – 4x

On putting f'(x) = 0, we get

-6 – 4x = 0

x = -3/2

So, the intervals are (-∞, -3/2) and (-3/2, ∞)

For (-∞, -3/2) interval take x = -2 

From eq(1), f'(x) = (-)(-) = (+) > 0

So, f is strictly increasing

For (-3/2, ∞) interval take x = -1

From eq(1), f'(x) = (-)(+) = (-) < 0

So, f is strictly decreasing 

(iii) f(x) = -2x³ – 9x² – 12x + 1 

f'(x) = -6x² – 8x – 12

On putting f'(x) = 0, we get

-6x² – 8x – 12 = 0

-6(x + 1)(x + 2) = 0

x = -1, x = -2

So, the intervals are (-∞, -2), (-2, -1), and (-1, ∞)

For (-∞, -2) interval take x = -3 

From eq(1), f'(x) = (-)(-)(-) = (-) < 0

So, f is strictly decreasing

For (-2, -1) interval take x = -1.5

From eq(1), f'(x) = (-)(-)(+) = (+) > 0

So, f is strictly increasing

For (-1, ∞) interval take x = 0

From eq(1), f'(x) = (-)(+)(+) = (-) < 0

So, f is strictly decreasing

(iv) f(x) = 6 – 9x – x² 

f'(x) = -9 – 2x

On putting f'(x) = 0, we get

-9 – 2x = 0

x = -9/2

So, the intervals are (-∞, -9/2) and (-9/2, ∞)

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

– 9 – 2x > 0

x > -9/2

So f is strictly increasing in interval (-∞, -9/2)

For f to be strictly decreasing, f'(x) < 0

-9 – 2x < 0

x < -9/2

So f is strictly decreasing in interval (-9/2, ∞)

(v) f(x) = (x + 1)³ (x – 3)³

f'(x) = (x + 3)³.3(x – 3)³ + (x – 3)³.3(x + 1)²

f'(x) = 6(x – 3)²(x + 1)²(x – 1)

Now, the factor of (x – 3)² and (x + 1)² are non-negative for all x

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

(x – 1) > 0

x > 1

So, f is strictly increasing in interval (1, ∞)

For f to be strictly decreasing, f'(x) < 0

(x – 1) < 0

x < 1

So, f is strictly decreasing in interval (-∞, 1)

Question 7. Show that y = log(1 + x) – , , is an increasing function of x throughout its domain.

Solution:

f(x) = log(1+x)

        

f'(x)=

So, the domain of the given function is x > -1

Now, x2 > 0, (x + 2)2 ≥ 0, x + 1 > 0

From the above equation f'(x) ≥ 0 ∀ x in the domain(x > -1) and f is an increasing function.

Question 8. Find the values of x for which y = [x(x – 2)]² is an increasing function.

Solution:

Given: y = f(x) = [x(x – 2)]2 = x2(x – 2x)2

= x4 – 4x3 + 4x2

f'(x) = 4x3 – 12x2 + 8x

f'(x) = 4x(x – 2)(x – 1)

x = 0, x = 1, x = 2

So, (∞, 0], [0, 1], [1, 2], [2,∞)

For (∞, 0], let x = -1

So, f'(x) = (-)(-)(-) = (-) ≤ 0

f(x) is decreasing

For [0, 1], let x = 1/2

So, f'(x) = (+)(-)(-) = (+) ≥ 0

f(x) is increasing

Similarly, for [1, 2], f(x) is decreasing

For [2,∞), f(x) is increasing

So, f(x) is increasing in interval [0, 1] and [2,∞)

Question 9. Prove that y = is an increasing function of θ in[0, π/2].

Solution:

y = f(θ) = 

     

Now 0 ≤ θ ≤ π/2, and we have 0 ≤ cosθ ≤ 1, 

So, 4 – cosθ > 0

Therefore f'(θ) ≥ 0 for 0 ≤ θ ≤ π/2

Hence, f'(x) =  is a strictly increasing in the interval (θ, π/2).

Question 10. Prove that the logarithmic function is increasing on (0, ∞).

Solution:

Given: f(x) = log(x)         -(logarithmic function)

f'(x) = 1/x ∀ x in (0, ∞) 

Therefore,  x > 0, so, 1/x > 0

Hence, the logarithmic function is strictly increasing in interval (0, ∞) 

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. 

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 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).

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

Question 19. The interval in which y = x² e^-x is increasing is 
(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.

Chapter-6 (Applications of Derivatives)

• Exercise 6.1
• Exercise 6.3
• Exercise 6.4
• Exercise 6.5