CBSE Class 12 Mathematics Chapter 5 Continuity and Differentiability NCERT Solutions

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This resource provides comprehensive NCERT Solutions for CBSE Class 12 Mathematics, focusing on Chapter 5: Continuity and Differentiability. It meticulously explains the concept of continuity for various functions, including polynomial, rational, and absolute value functions, at specific points and over their domains. The solutions demonstrate how to verify continuity by comparing the limit of the function as x approaches a point with the function's value at that point. Key topics covered include proving continuity at given points and examining continuity for different types of functions. These solutions are designed to help students grasp the fundamental principles of continuity, build problem-solving skills, and prepare effectively for their board examinations by offering clear, step-by-step explanations.

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Quick info

BoardCBSE
ClassClass 12
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 5

Chapter summary

Chapter 5 of the CBSE Class 12 Mathematics syllabus, Continuity and Differentiability, introduces the foundational concepts of continuity. This section focuses on understanding and verifying the continuity of functions at specific points and across their entire domains. The NCERT Solutions provided here break down the process of checking continuity using the limit definition, ensuring students can apply these principles to polynomial, rational, and absolute value functions.

Learning outcomes

  • Understand the definition of continuity of a function at a point.
  • Prove the continuity of polynomial functions at specified points.
  • Examine the continuity of rational and absolute value functions.
  • Apply the limit definition to verify continuity.
  • Identify points of discontinuity for given functions.

Topics covered

Paper topics

  • Continuity of a function at a point
  • Continuity of a function in an interval
  • Limit of a function
  • Function value at a point
  • Polynomial functions
  • Rational functions
  • Absolute value functions
  • Verifying continuity
  • Points of discontinuity

Important topics

  • Definition of continuity
  • Continuity of polynomial functions
  • Continuity of rational functions
  • Continuity of absolute value functions
  • Verifying continuity using limits

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Questions and Solutions

Question 1

Prove that the function f(x) = 5x - 3 is continuous at x = 0, x = -3, and at x = 5.
Solution:

The given function is f(x) = 5x - 3. To prove continuity at a point, we need to show that the limit of the function as x approaches that point is equal to the function's value at that point.

At x = 0:

First, find the value of the function at x = 0:

f(0) = 5(0) - 3 = -3

Next, find the limit of the function as x approaches 0:

\lim_{x \to 0} f(x) = \lim_{x \to 0} (5x - 3) = 5(0) - 3 = -3

Since \lim_{x \to 0} f(x) = f(0) (both are -3), the function is continuous at x = 0.

At x = -3:

Find the value of the function at x = -3:

f(-3) = 5(-3) - 3 = -15 - 3 = -18

Find the limit of the function as x approaches -3:

\lim_{x \to -3} f(x) = \lim_{x \to -3} (5x - 3) = 5(-3) - 3 = -15 - 3 = -18

Since \lim_{x \to -3} f(x) = f(-3) (both are -18), the function is continuous at x = -3.

At x = 5:

Find the value of the function at x = 5:

f(5) = 5(5) - 3 = 25 - 3 = 22

Find the limit of the function as x approaches 5:

\lim_{x \to 5} f(x) = \lim_{x \to 5} (5x - 3) = 5(5) - 3 = 25 - 3 = 22

Since \lim_{x \to 5} f(x) = f(5) (both are 22), the function is continuous at x = 5.

Conclusion: The function f(x) = 5x - 3 is continuous at x = 0, x = -3, and x = 5.

Question 2

Examine the continuity of the function f(x) = 2x^2 - 1 at x = 3.
Solution:

The given function is f(x) = 2x^2 - 1. To examine its continuity at x = 3, we need to check if the limit of the function as x approaches 3 equals the function's value at x = 3.

First, calculate the value of the function at x = 3:

f(3) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17

Next, calculate the limit of the function as x approaches 3:

\lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 - 1) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17

We observe that \lim_{x \to 3} f(x) = f(3), as both values are equal to 17.

Conclusion: Therefore, the function f(x) = 2x^2 - 1 is continuous at x = 3.

Question 3

Examine the following functions for continuity.

(i) f(x)=x-5

(ii) f(x) = \frac{1}{x-5}, x \neq 5

(iii) f(x) = \frac{x^2 - 25}{x+5}, x \neq -5

(iv) f(x) = |x-5|, x \neq 5

Solution:

We will examine the continuity of each function.

(i) f(x) = x - 5

This is a polynomial function. Polynomial functions are continuous for all real numbers. Let's verify this for an arbitrary real number k.

The value of the function at k is f(k) = k - 5.

The limit of the function as x approaches k is:

\lim_{x \to k} f(x) = \lim_{x \to k} (x - 5) = k - 5

Since \lim_{x \to k} f(x) = f(k), the function f(x) = x - 5 is continuous at every real number k. Therefore, it is a continuous function everywhere.

(ii) f(x) = \frac{1}{x-5}, x \neq 5

This is a rational function. A rational function is continuous everywhere except where its denominator is zero. In this case, the denominator is zero when x = 5.

For any real number k \neq 5, the value of the function is f(k) = \frac{1}{k-5}.

The limit of the function as x approaches k is:

\lim_{x \to k} f(x) = \lim_{x \to k} \frac{1}{x - 5} = \frac{1}{k - 5}

Since \lim_{x \to k} f(x) = f(k) for all k \neq 5, the function is continuous at every point in its domain.

(iii) f(x) = \frac{x^2 - 25}{x+5}, x \neq -5

This is a rational function. It is defined for all x \neq -5. We need to check continuity for c \neq -5.

For any real number c \neq -5, the value of the function is f(c) = \frac{c^2 - 25}{c+5}.

Let's find the limit as x approaches c:

\lim_{x \to c} f(x) = \lim_{x \to c} \frac{x^2 - 25}{x + 5} = \lim_{x \to c} \frac{(x - 5)(x + 5)}{x + 5}

Since x \neq -5, we can cancel out the (x+5) term:

= \lim_{x \to c} (x - 5) = c - 5

Now, let's evaluate the function at c (where c \neq -5):

f(c) = \frac{c^2 - 25}{c+5} = \frac{(c-5)(c+5)}{c+5} = c - 5

Since \lim_{x \to c} f(x) = f(c) for all c \neq -5, the function is continuous for all x \neq -5.

(iv) f(x) = |x-5|, x \neq 5

The absolute value function f(x) = |x-5| can be written as:

f(x) = \begin{cases} x-5 & \text{if } x \geq 5 \\ -(x-5) & \text{if } x < 5 \end{cases}

Let's examine continuity at x = 5.

The value of the function at x = 5 is f(5) = |5-5| = 0.

Now, we find the left-hand limit (LHL) and the right-hand limit (RHL) as x approaches 5.

LHL: \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} -(x-5) = -(5-5) = 0

RHL: \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (x-5) = (5-5) = 0

Since LHL = RHL = f(5) (all are 0), the function f(x) = |x-5| is continuous at x = 5.

Conclusion: All four functions examined are continuous on their respective domains.

Common mistakes

  • Incorrectly calculating limits.
  • Confusing the limit value with the function's value at a point.
  • Errors in algebraic simplification when evaluating limits.
  • Not considering the domain of the function when examining continuity.

Revision tips

  • Review the definition of continuity thoroughly before attempting problems.
  • Practice evaluating limits for various types of functions.
  • Pay close attention to the points where the function might be undefined.
  • Work through each example step-by-step to reinforce understanding.

Practice MCQs

Q1. For a function f(x) to be continuous at x = c, which condition must be met?

Q2. Which type of function is always continuous everywhere?

Q3. What is the value of f(x) = 5x - 3 at x = 0?

Q4. The function f(x) = \frac{1}{x-5} is continuous for which values of x?

Q5. For f(x) = \frac{x^2 - 25}{x + 5}, what is the limit as x approaches -5?

Frequently asked questions

What is the main concept covered in Chapter 5 of CBSE Class 12 Maths NCERT Solutions?

Chapter 5 focuses on the concept of Continuity and Differentiability of functions. The provided solutions specifically address the continuity of functions at various points and over intervals.

How do these NCERT Solutions help in understanding continuity?

The solutions offer clear, step-by-step explanations for proving and examining the continuity of different types of functions (polynomial, rational, absolute value) using the limit definition, making the concept easier to grasp.

Are the questions in these solutions the same as the NCERT textbook?

Yes, the questions are kept exactly the same as in the NCERT textbook, ensuring that students are practicing with the official material. The wording of the questions has been expanded for clarity.

What is the condition for a function to be continuous at a point 'c'?

A function f(x) is continuous at a point 'c' if three conditions are met: 1. f(c) is defined. 2. The limit of f(x) as x approaches 'c' exists. 3. The limit of f(x) as x approaches 'c' is equal to f(c).

How can I use these solutions for exam revision?

You can use these solutions to review the definition of continuity, practice applying limit concepts to different functions, and check your understanding of continuity checks for various function types. Working through them helps reinforce problem-solving techniques.

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