NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

 

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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 f ( x) = 5 x -3 is continuous at 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

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

NCERT Solutions for Class 12

 

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