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

The chapter "Application of Derivatives" in Class 12 Mathematics is a critical part of the NCERT curriculum. It focuses on using derivatives to the solve real-world problems including rate of change, maxima and minima, tangents and normals. Exercise 6.2 specifically deals with the problems related to finding maxima and minima of the functions which is a vital concept for the students preparing for the board exams and competitive exams like JEE. Understanding these applications is essential as they lay the foundation for the more advanced topics in calculus and its applications in the various fields such as the physics, engineering and economics.

Application of Derivatives

In Class 12 Mathematics, Chapter "Application of Derivatives" is a crucial segment focusing on using the derivatives to solve practical problems. Exercise 6.2 in this chapter deals with the finding maxima and minima of functions essential for the optimizing real-world scenarios. The Mastery of this topic is important for the understanding how to analyze and model various applications in the fields like economics, engineering and physics. This exercise helps students apply theoretical concepts to solve complex problems reinforcing their understanding and problem-solving skills.

Content of this article has been merged with Chapter 5: Continuity and Differentiability - Exercise 6.2 as per the revised syllabus of NCERT.

Class 12 NCERT Solutions- Mathematics Part I - Exercise 6.2

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 f(x)=x+\frac{1}{x}  is increasing on I.

Solution:

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

Now, f(x) = x+\frac{x}{1}

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

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) = \frac{1}{\sin x}.\cos x=\cot 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. 

Read more :

• Application of Derivatives
• Derivatives | First and Second Order Derivatives, Formulas and Examples
• Class 12 NCERT Solutions- Mathematics Part I – Application of Derivatives – Exercise 6.1
• Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives -Exercise 6.3