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

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

This chapter is a continuation of differentiation of functions which we already study in Class XI. In this chapter, we are going to study new concepts like continuity, differentiability and relations between them and also study about differentiation of inverse trigonometric functions. This chapter contains 8 exercises with 121 questions. The NCERT Solutions for Class 12 Maths Chapter-5 Continuity and Differentiability are solved and prepared by subject experts.

## The mathematical definition of Continuity and Differentiability -

Let f be a real function and c be a point in the domain of f. Then f is continuous at c if$\lim_{x\rightarrow c}f(x)=f(c)$. A function f is differentiable at a point c in its domain if it is continuous at point c. A function is said to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b]. NCERT Class 12 Maths Chapter-5 Continuity and Differentiability is one of the time-consuming chapters because it contains 8 exercises with 121 questions and also 23 questions in the miscellaneous exercise. To understand it in the better way 48 solved examples are given in the NCERT textbook.

## Topics of NCERT Grade 12 Maths Chapter-5 Continuity and Differentiability

5.1 Introduction

5.2 Continuity

5.2.1 Algebra of continuous functions

5.3. Differentiability

5.3.1 Derivatives of composite functions

5.3.2 Derivatives of implicit functions

5.3.3 Derivatives of inverse trigonometric functions

5.4 Exponential and Logarithmic Functions

5.5. Logarithmic Differentiation

5.6 Derivatives of Functions in Parametric Form

Some solved examples-

Que 1. Prove that the function $\dpi{100} f ( x) = 5 x -3$ is continuous at $\dpi{100} x = 0, at\: \: x = - 3$ and at $x = 5$

Solution-

Given function is
$f ( x) = 5 x -3$
$f(x) = 5(0)-3 = -3$
$\lim_{x\rightarrow 0} f(x) = 5(0)-3 = -3$
$\lim_{x\rightarrow 0} f(x) =f(0)$
Hence, function is continous at x = 0

$\dpi{100} f(-3)= 5(-3)-3=-15-3=-18\\\Rightarrow \lim_{x\rightarrow -3} f(x) = 5(-3)-3 = -15-3=-18\\\Rightarrow \lim_{x\rightarrow -3} f(x) = f(-3)$
Hence, function is continous at x = -3

$f(5)= 5(5)-3=25-3=22\\\Rightarrow \lim_{x\rightarrow 5} f(x) = 5(5)-3 = 25-3=-22\\ \Rightarrow \lim_{x\rightarrow 5} f(x) = f(5)$
Hence, fucnction is continous at x = 5

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability- Solved Exercise Questions

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.1

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

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.3

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.4

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.5

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.6

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.8

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous

## NCERT Solutions for class 12- Maths

 Chapter 1 Relations and Functions Chapter 2 Inverse Trigonometric Functions Chapter 3 Matrices Chapter 4 Determinants Chapter 6 Application of Derivatives Chapter 7 Integrals Chapter 8 Application of Integrals Chapter 9 Differential Equations Chapter 10 Vector Algebra Chapter 11 Three Dimensional Geometry Chapter 12 Linear Programming Chapter 13 Probability