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# NCERT Solutions Class 12 Maths Chapter 5 Exercise 5.2 - Continuity and Differentiability

Last updated date: 10th Aug 2024
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## NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 - FREE PDF Download

The NCERT Solutions for Class 5 Exercise 5.2 Class 12 Maths Continuity and Differentiability provides complete solutions to the problems in the Exercise. These NCERT Solutions are intended to assist students with the CBSE Class 12 board examination.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 5 Exercise 5.2 Class 12 | Vedantu
3. Topics Covered in Class 12 Maths Chapter 5 Exercise 5.2
4. Access NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability Exercise 5.2
5. Class 12 Maths Chapter 5: Exercises Breakdown
6. CBSE Class 12 Maths Chapter 12 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Students should thoroughly study this NCERT solution in order to solve all types of questions based on Continuity and Differentiability. By completing these practice questions with the NCERT Maths Solutions Chapter 5 Exercise 5.2 Class 12, you will be better prepared to understand all of the different types of questions that may be asked in the Class 12 board exams.

## Glance on NCERT Solutions Maths Chapter 5 Exercise 5.2 Class 12 | Vedantu

• This exercise deals with the core concepts of calculus.

• A function is continuous at a point x=a if the one-sided limits exist and are equal to the two-sided limit.

• A function is differentiable at a point x=a if the limit of the difference quotient exists as h approaches zero.

• Learn how to determine the continuity of various types of functions at specific points.

• Understand the concept of continuous functions over an interval.

• Apply the properties of continuous functions to solve problems.

• Work through problems involving polynomial, rational, trigonometric, exponential, and logarithmic functions to test their continuity.

## Topics Covered in Class 12 Maths Chapter 5 Exercise 5.2

• Differentiability

• Derivatives of composite functions

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability Exercise 5.2

1. Differentiate the function with respect to $x$. $\sin \left( {{x^2} + 5} \right)$

Ans: Let $f(x) = \sin \left( {{x^2} + 5} \right),u(x) = {x^2} + 5$, and $v(t) = \sin t$

Then, $(vou)(x) = v(u(x)) = v\left( {{x^2} + 5} \right)$

$= \tan \left( {{x^2} + 5} \right) = f(x)$

Thus, $f$ is a composite of two functions.

Put $t = u(x) = {x^2} + 5$

Then, we obtain $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos \left( {{x^2} + 5} \right)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2} + 5} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}(5) = 2x + 0 = 2x$

Therefore, by chain rule. $\dfrac{{df}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos \left( {{x^2} + 5} \right) \times 2x$

$= 2x\cos \left( {{x^2} + 5} \right)$

2. Differentiate the functions with respect of ${\text{x}}$. $\cos (\sin x)$

Ans: Let $f(x) = \cos (\sin x),u(x) = \sin x$, and $v(t) = \cos t$

Then, $(\operatorname{vou} )(x) = v(u(x)) = v(\sin x) = \cos (\sin x) = f(x)$

Thus, $f$ is a composite function of two functions.

Put $t = u(x) = \sin x$

$\therefore \dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}[\cos t] = - \sin t = - \sin (\sin x)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sin x) = \cos x$

By chain rule, $\dfrac{{df}}{{dx}},\dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = - \sin (\sin x) \cdot \cos x$

$= - \cos x\sin (\sin x)$

3. Differentiate the functions with respect of ${\text{x}}$.

$\sin (ax + b)$

Ans: Let $f(x) = \sin (ax + b),u(x) = ax + b$, and $v(t) = \sin t$

Then, $(vou)(x) = v(u(x)) = v(ax + b) = \sin (ax + b) = f(x)$

Thus, $f$ is a composite function of two functions $u$ and $v$.

Put $t = u(x) = ax + b$

Therefore, $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos (ax + b)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(ax + b) = \dfrac{d}{{dx}}(ax) + \dfrac{d}{{dx}}(b) = a + 0 = a$

Hence, by chain rule, we obtain

$\dfrac{{df}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos (ax + b) \cdot a$

$= a\cos (ax + b)$

4. Differentiate the functions with respect of ${\text{x}}$.

$\sec (\tan (\sqrt x ))$

Ans: Let $f(x) = \sec (\tan (\sqrt x )),u(x) = \sqrt x ,v(t) = \tan t$, and $w(s) = \sec s$

Then, $(vou)(x) = w[v(u(x))] = w[v(\sqrt x )] = w(\tan \sqrt x ) = \sec (\tan \sqrt x ) = f(x)$

Thus, $f$ is a composite function of three functions, $u, v$ and $w$. Put $s = v(t) = \tan t$ and $t = u(x) = \sqrt x$

$[s = \tan t]$

Then, $\dfrac{{dw}}{{ds}} = \dfrac{d}{{ds}}(\sec s) = \sec s\tan s$

$= \sec (\tan t) \cdot \tan (\tan t)$

$= \sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x )\quad [t = \sqrt x ]$

$\dfrac{{ds}}{{dt}} = \dfrac{d}{{dt}}(\tan t) = {\sec ^2}t = {\sec ^2}\sqrt x$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sqrt x ) = \dfrac{d}{{dx}}\left( {{x^{\frac{1}{2}}}} \right)$

$= \dfrac{1}{2} \cdot {x^{\frac{1}{2} - 1}} = \dfrac{1}{{2\sqrt x }}$

Hence, by chain rule, we obtain $\dfrac{{dt}}{{dx}} = \dfrac{{dw}}{{ds}} \cdot \dfrac{{ds}}{{dt}} \cdot \dfrac{{dt}}{{dx}}$

$= \sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x ) \times {\sec ^2}\sqrt x \times \dfrac{1}{{2\sqrt x }}$

$= \dfrac{1}{{2\sqrt x }}{\sec ^2}\sqrt x (\tan \sqrt x )\tan (\tan \sqrt x )$

$= \dfrac{{{{\sec }^2}\sqrt x \sec (\tan \sqrt x )\tan (\tan \sqrt x )}}{{2\sqrt x }}$

5. Differentiate the functions with respect of ${\text{X}}$. $\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}$

Ans: The given function is $f(x) = \dfrac{{\sin (ax + b)}}{{\cos (cx + d)}} = \dfrac{{g(x)}}{{h(x)}},$where $g(x) = \sin (ax + b)$ and

$h(x) = \cos (cx + d)$

$\therefore f = \dfrac{{{g^\prime }h - g{h^\prime }}}{{{h^2}}}$

Consider $g(x) = \sin (ax + b)$

Let $u(x) = ax + b,v(t) = \sin t$

Then $(vou)(x) = v(u(x)) = v(ax + b) = \sin (ax + b) = g(x)$

$\therefore g$ is a composite function of two functions, $u$ and $v$.

Put $t = u(x) = ax + b$

$\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos (ax + b)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(ax + b) = \dfrac{d}{{dx}}(ax) + \dfrac{d}{{dx}}(b) = a + 0 = a$

Therefore, by chain rule, we obtain ${g^\prime } = \dfrac{{dg}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos (ax + b) \cdot a = a\cos (ax + b)$

Consider $h(x) = \cos (cx + d)$

Let $p(x) = cx + d,q(y) = \cos y$

Then, $(qop)(x) = q(p(x)) = q(cx + d) = \cos (cx + d) = h(x)$

$\therefore h$ is a composite function of two functions, $p$ and $q$.

Put $y = p(x) = cx + d$

$\dfrac{{dq}}{{dy}} = \dfrac{d}{{dy}}(\cos y) = - \sin y = - \sin (cx + d)$

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(cx + d) = \dfrac{d}{{dx}}(cx) + \dfrac{d}{{dx}}(d) = c$

Therefore, by chain rule, we obtain ${h^\prime } = \dfrac{{dh}}{{dx}} = \dfrac{{dq}}{{dy}} \cdot \dfrac{{dy}}{{dx}} = - \sin (cx + d){\text{xc}} = - c\sin (cx + d)$

$\therefore {f^\prime } = \dfrac{{a\cos (ax + b) \cdot \cos (cx + d) - \sin (ax + b)\{ - c\sin cx + d\} }}{{{{[\cos (cx + d)]}^2}}}$

$= \dfrac{{a\cos (ax + b)}}{{\cos (cx + d)}} + c\sin (ax + b) \cdot \dfrac{{\sin (cx + d)}}{{\cos (cx + d)}} \times \dfrac{1}{{\cos (cx + d)}}$

$= a\cos (ax + b)\sec (cx + d) + c\sin (ax + b)\tan (cx + d)\sec (cx + d)$

6. Differentiate the function with respect to ${\text{x}}$.

$\cos {x^3} \cdot {\sin ^2}\left( {{x^5}} \right)$

Ans: $\cos {x^3} \cdot {\sin ^2}\left( {{x^5}} \right)$

$\dfrac{d}{{dx}}\left[ {\cos {x^3} \cdot {{\sin }^2}\left( {{x^5}} \right)} \right] = {\sin ^2}\left( {{x^5}} \right){\text{x}}\dfrac{d}{{dx}}\left( {\cos {x^3}} \right) + \cos {x^3}{\text{x}}\dfrac{d}{{dx}}\left[ {{{\sin }^2}\left( {{x^5}} \right)} \right]$

$= {\sin ^2}\left( {{x^5}} \right){\text{x}}\left( { - \sin {x^3}} \right){\text{x}}\dfrac{d}{{dx}}\left( {{x^3}} \right) + \cos {x^3} + 2\sin \left( {{x^5}} \right) \cdot \dfrac{d}{{dx}}\left[ {\sin {x^5}} \right]$

The given function is

$= \sin {x^3}{\sin ^2}\left( {{x^5}} \right) \times 3{x^2} + 2\sin {x^5}\cos {x^3} \cdot \cos {x^5} \times \dfrac{d}{{dx}}\left( {{x^5}} \right)$

$= 3{x^2}\sin {x^3} \cdot {\sin ^3}\left( {{x^5}} \right) + 2\sin {x^5}\cos {x^5}\cos {x^3} \cdot x5{x^4}$

$= 10{x^4}\sin {x^5}\cos {x^5}\cos {x^3} - 3{x^2}\sin {x^3}{\sin ^2}\left( {{x^5}} \right)$

7. Differentiate the functions with respect to ${\text{x}}$. $2\sqrt {\cot \left( {{x^2}} \right)}$

Ans: $\dfrac{d}{{dx}}\left[ {2\sqrt {\cot \left( {{x^2}} \right)} } \right]$

$= 2 \cdot \dfrac{1}{{2\sqrt {\cot \left( {{x^2}} \right)} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {\cot \left( {{x^2}} \right)} \right]$

$= \sqrt {\dfrac{{\sin \left( {{x^2}} \right)}}{{\cos \left( {{x^2}} \right)}}} {\text{x}} - {\operatorname{cosec} ^2}\left( {{x^2}} \right) \times \dfrac{d}{{dx}}\left( {{x^2}} \right)$

$= \sqrt {\dfrac{{\sin \left( {{x^2}} \right)}}{{\cos \left( {{x^2}} \right)}}} \times \dfrac{1}{{{{\sin }^2}\left( {{x^2}} \right)}}x(2x)$

$= \dfrac{{ - 2x}}{{\sqrt {\cos {x^2}\sqrt {\sin {x^2}\sin {x^2}} } }}$

$= \dfrac{{ - 2\sqrt 2 x}}{{\sqrt {2\sin {x^2}\cos {x^2}} \sin {x^2}}}$

$= \dfrac{{ - 2\sqrt 2 x}}{{\sin {x^2}\sqrt {\sin 2{x^2}} }}$

8. Differentiate the functions with respect to $x$ $\cos (\sqrt x )$

Ans: Let $f(x) = \cos (\sqrt x )$

Also, let $u(x) = \sqrt x$

And, $v(t) = \cos t$

Then, $(vou)(x) = v(u(x))$

$= v(\sqrt x )$

$= \cos \sqrt x$

$= f(x)$

Clearly, $f$ is a composite function of two functions, $u$ and $v$, such that $t = u(x) = \sqrt x$

Then,

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sqrt x ) = \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right)$

$\dfrac{1}{2}{x^{ - \dfrac{1}{2}}} = \dfrac{1}{{2\sqrt x }}$

And, $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\cos t) = - \sin t = - \sin \sqrt x$

By using chain rule, we obtain $\dfrac{{dt}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}}$

$= - \sin (\sqrt x ) \cdot \dfrac{1}{{2\sqrt x }}$

$= - \dfrac{1}{{2\sqrt x }}\sin (\sqrt x )$

$= - \dfrac{{\sin (\sqrt x )}}{{2\sqrt x }}$

9. Prove that the function $f$ given by $f(x) = |x - 1|,x \in {\mathbf{R}}$ is not differentiable at $x = 1$.

Ans: The given function is $f(x) = |x - 1|,x \in {\mathbf{R}}$

It is known that a function $f$ is differentiable at a point $x = c$ in its domain if both $\mathop {\lim }\limits_{k \to {0^ - }} \dfrac{{f(c + h) - f(c)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(c + h) - f(c)}}{h}$ are finite and equal.

To check the differentiability of the given function at $x = 1$, Consider the left hand limit of $f$ at $x = 1$ $\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f|I + h - 1||1 - 1|}}{h}$

$= \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{|h| - 0}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - h}}{h}\quad (h < 0 \Rightarrow |h| = - h)$

$= - 1$

Consider the right hand limit of $f$ at $x = 1$

$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f|I + h - 1| - |1 - 1|}}{h}$

$= \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{|h| - 0}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{h}{h}\quad (h > 0 \Rightarrow |h| = h)$

$= 1$

Since the left and right hand limits of $f$ at $x = 1$ are not equal, $f$ is not differentiable at $x = 1$

10. Prove that the greatest integer function defined by $f = (x) = [x],0 < x < 3$ is not differentiable at $x = 1$ and $x = 2$.

Ans: The given function $f$ is $f = (x) = [x],0 < x < 3$

It is known that a function $f$ is differentiable at a point $x = c$ in its domain if both $\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(c + h) - f(c)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(c + h) - f(c)}}{h}$ are finite and equal.

To check the differentiable of the given function at $x = 1$, consider the left hand limit of $f$ at

$x = 1$

$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{[1 + h] - [1]}}{h}$

$= \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{0 - 1}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} = \dfrac{{ - 1}}{h} = \infty$

Consider the right hand limit of $f$ at $x = 1$ $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{[1 + h][1]}}{h}$

$= \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{1 - 1}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$

Since the left and right limits of $f$ at $x = 1$ are not equal, $f$ is not differentiable at $x = 1$

To check the differentiable of the given function at $x = 2$, consider the left hand limit of $f$ at

$x = 2$

$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(2 + h) - f(2)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{[2 + h] - [2]}}{h}$

$= \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{1 - 2}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - 1}}{h} = \infty$

Consider the right hand limit of $f$ at $x = 1$

$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(2 + h) - f(2)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{[2 + h] - [2]}}{h}$

$= \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{1 - 2}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$

Since the left and right hand limits of $f$ at $x = 2$ are not equal, $f$ is not differentiable at $x = 2$

## Conclusion

Continuity and Differentiability are crucial for a solid foundation in Math. Referring to these NCERT Solutions by Vedantu students can significantly enhance your understanding of continuity and differentiability. Class 12 Ex 5.2 focuses on the continuity of functions, a crucial topic that forms the basis for understanding differentiability and further calculus concepts. Pay attention to the step-by-step solutions provided in Class 12 Maths Ex 5.2, grasp the underlying principles, and ensure clarity on the concepts before moving forward.

## Class 12 Maths Chapter 5: Exercises Breakdown

 Exercise Number of Questions Exercise 5.1 34 Questions & Solutions Exercise 5.3 15 Questions and Answers Exercise 5.4 10 Questions and Answers Exercise 5.6 11 Questions and Answers Exercise 5.7 17 Questions and Answers

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 12 Maths Chapter 5 Exercise 5.2 - Continuity and Differentiability

1. What is Continuity, and what are the criteria to prove the continuity of a function?

Continuity is a unique feature or characteristic of a function under which the function is represented in a graphical manner. The graph of continuity always shows a continuous wave which indicates that a function is continuous. Continuity can be found in many aspects of nature and our surroundings.

The three criteria that prove a function to be continuous are:

If the value of f(a) is finite.

Lim x→a is available.

Lim x→a f(x) = f(a).

2. How are the NCERT Solutions of Class 12 Maths Chapter 5 exercise 5.2 Helpful?

The solutions of class 12 maths chapter 5 are very much helpful to the students. They guide the students and lead them to the path of success. These solutions are structured in simple language for the betterment of the students. The solutions are prepared by our experts and guarantee error-free solutions.

3. How many questions are there in NCERT Solutions Class 12 Maths exercise 5.2?

NCERT Solutions Class 12 Maths Exercise 5.2 consists of ten questions. All the ten questions in Exercise 5.2 are based on the application of Product Rule, Quotient Rule, and Chain Rule. They focus on finding Left Hand Derivatives (LHD) and Right Hand Derivatives (RHD). Only a few of them are difficult, but the vast majority are simple. You must practice them regularly to understand the concepts properly. Finally, this is a useful exercise for honing your skills in revising functions' derivatives.

4. Is Class 12 Maths Chapter 5 important for board exams?

In NCERT Class 12 Maths Chapter 5 you will learn about continuity and differentiability, as well as its more advanced theorems and logarithms. You will be taught a more advanced version of what you learned in your last class. You'll learn about the ideas of continuity, differentiability, and their connections. You'll also learn about inverse trigonometric function differentiation. It belongs to the ‘Calculus’ unit which holds 35 marks in the board exams. You can also get questions worth two-four marks from this chapter.

Students can easily get access to the NCERT Solutions for Class 12 Maths Chapter 5 both in online and offline mode. You can access NCERT Solutions by visiting the website of Vedantu or download the Vedantu app and go through the solutions online. And if you want to access this later offline, then you can download the solution in PDFs free of cost. Once downloaded from Vedantu, you can access them on your device anytime and anywhere. All these steps can be done free of cost.

6. What are the main topics explained in Class 12 Maths Chapter 5?

Class 12 Maths Chapter 5 ‘Continuity and Differentiability’ is a very important chapter for a 12th grader. This chapter includes some very crucial concepts to understand such as introduction to continuity and differentiability, exponential and logarithmic functions, logarithmic differentiation, derivation of functions in parametric forms, second order derivatives and mean value theorem. To get well-versed with the entire chapter, you should visit Vedantu. The study material provided by Vedantu will help you clear all your concepts. You can go through the NCERT Solutions to revise your basics.

7. What are the benefits of choosing Vedantu for Class 12 Maths Chapter 5?

When it comes to Class 12 Maths Chapter 5, the advantages of using Vedantu are virtually limitless. Vedantu has a team of educators who work together to create high-quality study resources for students. You can benefit from the free study materials provided by Vedantu to its students. You will receive many more benefits if you join the Vedantu family. Live meetings with the teachers can assist you to clear up any confusion you may have. All Vedantu’s NCERT Solutions are prepared by the best teachers in India and are based on the latest syllabus and exam patterns.

8. How many exercises are there in Continuity and Differentiability?

Class 12 Maths Exercise 5.2 has 10 Questions but the overall number of questions is 97 in total of 7 Exercises.