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NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 (Applications of Derivatives)

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 contains all the questions with detailed solutions. Students are advised to solve these questions for better understanding of the concepts in exercise 6.2.
authorImageKrati Saraswat31 Jan, 2024
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NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 (Applications of Derivatives)

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 Applications of Derivatives is prepared by the academic team of Physics Wallah. We have prepared NCERT Solutions for all exercise of Chapter 6. Given below is step by step solutions of all questions given in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2.

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 Overview

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 addresses these significant subjects. In order to fully comprehend the concepts presented in the chapter and make effective use of the provided solutions, it is recommended that students go over each topic in great detail. The intention is for students to effortlessly achieve excellent exam scores after reviewing and practicing these responses.

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2

Solve The Following Questions of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2:

NCERT Solutions for Class 12 Maths Chapter 6 Miscellaneous Exercise

Question 1. Show that the function given by f(x) = e 2x is strictly increasing on R. Solution : Given: f'(x) = e 2x Now, x ∈ R Since the value of  e 2x is always positive for any real value of x, e 2x > 0. ⇒2e 2x > 0 ⇒f'(x) > 0 So f(x) is incerasing on R.

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1

Question 2. Show that the function given by f ( x ) = 3 x + 17 is strictly increasing on R. Solution : NCERT Solutions for Class 12 Maths Application of Derivatives/image010.png Question 3. Show that the function given by f ( x ) = sin x is (a) strictly increasing (0,π/2) (b) strictly decreasing in (π/2,π)  (c) neither increasing nor decreasing in (0,π) Solution : Given: The given function is f ( x ) = sin x . chapter 6-Application Of Derivatives Exercise 6.2

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3

Question 4. Find the intervals in which the function  F given by 2x 2 - 3x  is (a) strictly increasing, (b) strictly decreasing. Solution : Given: chapter 6-Application Of Derivatives Exercise 6.2

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4

Question 5. Find the intervals in which the function F given by f ( x ) = 2 x 3 − 3 x 2 − 36 x + 7 is (a) strictly increasing, (b) strictly decreasing. Solution : (a)Given: NCERT Solutions for Class 12 Maths Application of Derivatives/image037.png

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5

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

(a) x 2 + 2 x − 5 (b) 10 − 6 x − 2 x 2

(c) −2 x 3 − 9 x 2 − 12 x + 1 (d) 6 − 9 x x 2

(e) ( x + 1) 3 ( x − 3) 3

Solution : (a) Given: NCERT Solutions for Class 12 Maths Application of Derivatives/image065.png NCERT Solutions for Class 12 Maths Application of Derivatives/image075.png chapter 6-Application Of Derivatives Exercise 6.2 NCERT Solutions for Class 12 Maths Application of Derivatives/image096.png NCERT Solutions for Class 12 Maths Application of Derivatives/image098.png Question 7. Show that NCERT Solutions for Class 12 Maths Application of Derivatives/image121.png is an increasing function of x throughout its domain. Solution : Given: NCERT Solutions for Class 12 Maths Application of Derivatives/image123.png ∴dydx=11+x-(2+x)(2)-2x(1)(2+x)2=11+x-4(2+x)2=x2(1+x)(2+x)2 Now, dydx=0 ⇒x2(1+x)(2+x)2=0⇒x2=0      [(2+x)≠0 as x>-1]⇒x=0 Since x > -1 , point = 0 divides the domain (−1, ∞) in two disjoint intervals i.e., −1 < x  < 0 and x > 0 When −1 < x < 0, we have: x<0⇒x2>0x>-1⇒(2+x)>0⇒(2+x2)>0 ∴ y'=x2(1+x)(2+x)2>0 Also, when x > 0 x>0⇒x2>0, (2+x)2>0 ∴ y'=x2(1+x)(2+x)2>0 Hence, function f  is increasing throughout this domain Question 8. Find the value of x  for which NCERT Solutions for Class 12 Maths Application of Derivatives/image122.png is an increasing function. Solution : NCERT Solutions for Class 12 Maths Application of Derivatives/image136.png Question 9. Prove that NCERT Solutions for Class 12 Maths Application of Derivatives/image156.png is an increasing function of θ  in [0, π/2] Solution : NCERT Solutions for Class 12 Maths Application of Derivatives/image156.png chapter 6-Application Of Derivatives Exercise 6.2 Question 10. Prove that the logarithmic function is strictly increasing on (0,∞) Solution : Given: The given function is f(x) =  logx ∴ f'(x) = 1/x It is clear that for x > 0,  f'(x) = 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) = x 2 - x + 1  is neither strictly increasing nor strictly decreasing on (-1,1) Solution : Given: The given function f (x) = x 2 - x + 1 chapter 6-Application Of Derivatives Exercise 6.2 hence,f is neither strictly increasing nor decreasing on the interval (-1,1) Question 12. Which of the following functions are strictly decreasing on (0,π/2)

(A) cos x

(B) cos 2 x

(C) cos 3 x

(D) tan x

Solution : chapter 6-Application Of Derivatives Exercise 6.2 NCERT Solutions for Class 12 Maths Application of Derivatives/image189.png chapter 6-Application Of Derivatives Exercise 6.2 Question 13. On which of the following intervals is the function f given by f(x) = x 100 + sinx - 1 is strictly decreasing: (A) (0, 1) (B) (π/2,π) (C) (0,π/2) (D) None of these Solution : Given: chapter 6-Application Of Derivatives Exercise 6.2 Question 14. Find the least value of a  such that the function f given by f(x) = x 2 + ax + 1 strictly increasing on (1, 2). Solution : chapter 6-Application Of Derivatives Exercise 6.2 Question 15. Let I be any interval disjoint from (-1,1)  Prove that the function f given by f(x) = x + 1/x  is strictly increasing on I. Solution : Given: chapter 6-Application Of Derivatives Exercise 6.2 ∴ f is strictly increasing on (-∞, 1) and (1, ∞) Hence, function f  is strictly increasing in interval  I  disjoint from (−1, 1). Hence, the given result is proved. Question 16. Prove that the function f given by f(x) = log sin x is strictly increasing on (0,π/2) and strictly decreasing on (π/2,π) Solution : Given: chapter 6-Application Of Derivatives Exercise 6.2 Question 17. Prove that the function f given by f(x) = log cos x is strictly decreasing on (0,π/2) and strictly decreasing on (π/2,π) Solution : Given: chapter 6-Application Of Derivatives Exercise 6.2 On the Question 18. Prove that the function given by f(x) = x 3 - 3x 2 + 3x - 100 is increasing in R. Solution : Given: chapter 6-Application Of Derivatives Exercise 6.2 Question 19. The interval in which y = x 2 e -x is increasing in: (A) (-∞, ∞) (B)(-2,0) (C) (2, ∞) (D) (0, 2) Solution : Given: chapter 6-Application Of Derivatives Exercise 6.2

NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.2 FAQs

How many exercise are there in application of derivatives?

Class 12 Maths Chapter 6 Application of Derivatives has 102 questions in 6 exercises along with 24 more provided in a miscellaneous exercise.

What are the applications of derivatives?

To calculate the profit and loss in business using graphs.

Who invented application of derivatives?

Calculus was discovered by Isaac Newton and Gottfried Leibniz in 17th Century. But it was not possible without the early developments of Isaac Barrow about the derivatives in 16th century.

What is sin differentiated?

For example, the derivative of the trigonometric function sin x is denoted as sin' (x) = cos x, it is the rate of change of the function sin x at a specific angle x is stated by the cosine of that particular angle. (i.e) The derivative of sin x is cos x.

What is dt in math?

Similarly, "dt" represents an infinitesimally small change in the independent variable t.
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