# Continuity And Differentiability Chapter 5 Class 12 Maths NCERT Textbook PDF

NCERT Solutions for Class 12 Maths Chapter 5‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 12 Maths Chapter 5 Exercise Solution’ using the download button.

### Chapter 5: Continuity and Differentiability

#### 5.1 Introduction

This chapter is essentially a continuation of our study of differentiation of functions in Class XI.

We had learned to differentiate certain functions like polynomial functions and trigonometric functions.

In this chapter, we introduce the very important concepts of continuity, differentiability, and relations between them. We will also learn the differentiation of inverse trigonometric functions.

Further, we introduce a new class of functions called exponential and logarithmic functions. These functions lead to powerful techniques of differentiation.

We illustrate certain geometrically obvious conditions through differential calculus. In the process, we will learn some fundamental theorems in this area.

#### 5.3.2 Derivatives of implicit functions

Until now we have been differentiating various functions given in the form y = f (x). But it is not necessary that functions are always expressed in this form.

For example, consider one of the following relationships between x and y:
x – y – π = 0
x + sin XY – y = 0

In the first case, we can solve for y and rewrite the relationship as y = x – π. In the second case, it does not seem that there is an easy way to solve for y.

Nevertheless, there is no doubt about the dependence of y on x in either of the cases.

When a relationship between x and y is expressed in a way that it is easy to solve for y and write y = f (x), we say that y is given as an explicit function of x.

In the latter case, it is implicit that y is a function of x and we say that the relationship of the second type, above, gives function implicitly. In this subsection, we learn to differentiate implicitly functions.

### NCERT Solutions Class 12 Maths Chapter 5 Continuity and Differentiability

Question 1:

Prove that the function is continuous at The given function is f(x)=5x−3At x=0, f(0)=5×0−3=−3limx→0f(x)=limx→0(5x−3)=5×0−3=−3∴limx→0f(x)=f(0)The given function is fx=5x-3At x=0, f0=5×0-3=-3limx→0fx=limx→05x-3=5×0-3=-3∴limx→0fx=f0

Therefore, f is continuous at x = 0

Therefore, is continuous at x = −3

Therefore, f is continuous at x = 5

Question 2:

Examine the continuity of the function .

Thus, f is continuous at x = 3

Question 3:

Examine the following functions for continuity.

(a) (b) (c) (d) (a) The given function is It is evident that f is defined at every real number k and its value at k is k − 5.

It is also observed that, Hence, f is continuous at every real number, and therefore, it is a continuous function.

(b) The given function is For any real number k ≠ 5, we obtain

Hence, f is continuous at every point in the domain of f, and therefore, it is a continuous function.

(c) The given function is For any real number c ≠ −5, we obtain

Hence, f is continuous at every point in the domain of f, and therefore, it is a continuous function.

(d) The given function is This function f is defined at all points of the real line.

Let c be a point on a real line. Then, c < 5 or c = 5 or c > 5

Case I: c < 5

Then, (c) = 5 − c

Therefore, f is continuous at all real numbers less than 5.

Case II : c = 5

Then, Therefore, is continuous at x = 5

Case III: c > 5

Therefore, f is continuous at all real numbers greater than 5.

Hence, f is continuous at every real number, and therefore, it is a continuous function.

Question 4:

Prove that the function is continuous at x = n, where n is a positive integer.

The given function is f (x) = xn

It is evident that f is defined at all positive integers, n, and its value at n is nn.

Therefore, is continuous at n, where n is a positive integer.

Question 5:

Is the function f defined by

continuous at x = 0? At x = 1? At x = 2?

The given function f is At x = 0,

It is evident that f is defined at 0 and its value at 0 is 0.

Therefore, f is continuous at x = 0

At x = 1,

is defined at 1 and its value at 1 is 1.

The left-hand limit of f at x = 1 is,

The right-hand limit of at x = 1 is,

Therefore, f is not continuous at x = 1

At = 2,

is defined at 2 and its value at 2 is 5.

Therefore, f is continuous at = 2