CBSE Class 12 Mathematics NCERT Solutions for Continuity and Differentiability
This section provides detailed NCERT Solutions for Class 12 Mathematics, focusing on Chapter 5: Continuity and Differentiability. It covers the fundamental concepts of continuity for various types of functions, including polynomial, rational, and absolute value functions. The solutions meticulously explain how to check for continuity at specific points and over intervals by evaluating the function's value and its limit. Key principles such as the limit from the left, limit from the right, and the function's value at a point are applied. These solutions are designed to help students understand the theoretical underpinnings and practical application of continuity, aiding in thorough exam preparation and concept reinforcement.
Quick info
| Board | CBSE |
|---|---|
| Class | Class 12 |
| Subject | Mathematics |
| Session | 2026 |
| Language | English |
| Type | NCERT Solutions |
| Chapter | Continuity and Differentiability |
Chapter summary
This chapter's NCERT Solutions for Class 12 Mathematics delve into the concept of continuity and differentiability. It focuses on understanding continuity at a point and over an interval, using the definition involving limits and function values. The exercises cover proving continuity for simple functions and examining continuity for more complex ones, including rational and absolute value functions. The solutions provide step-by-step derivations, ensuring students grasp the application of limit properties in determining continuity.
Learning outcomes
- Understand the definition of continuity of a function at a point.
- Evaluate the continuity of polynomial functions.
- Examine the continuity of rational functions at specified points.
- Determine the continuity of functions involving absolute values.
- Apply the limit definition to prove continuity.
- Verify continuity at multiple points for a given function.
Topics covered
Paper topics
- Continuity at a point
- Continuity over an interval
- Limit of a function
- Function value
- Polynomial functions
- Rational functions
- Absolute value functions
- Proving continuity
- Examining continuity
Important topics
- Definition of continuity at a point
- Continuity of polynomial functions
- Continuity of rational functions
- Continuity of absolute value functions
- Checking continuity at specific points
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Questions and Solutions
Class XII
Chapter 5 - Continuity and Differentiability
Maths
Exercise 5.1
Question 1:
Prove that the function <math>f(x) = 5x - 3</math> is continuous at <math>x = 0</math>, at <math>x = -3</math> and at <math>x = 5</math>. Answer
The given function is <math>f(x) = 5x - 3</math>
At <math>x = 0</math>, <math>f(0) = 5 \times 0 - 3 = 3</math>
<math display="block">\lim_{x\to 0} f(x) = \lim_{x\to 0} (5x-3) = 5 \times 0 - 3 = -3</math>
<math display="block">\therefore \lim_{x \to 0} f(x) = f(0)</math>
Therefore, f is continuous at <math>x = 0</math>
At <math>x = -3</math>, <math>f(-3) = 5 \times (-3) - 3 = -18</math>
<math display="block">\lim_{x \to -3} f(x) = \lim_{x \to -3} (5x - 3) = 5 \times (-3) - 3 = -18</math>
<math display="block">\therefore \lim_{x \to -3} f(x) = f(-3)</math>
Therefore, f is continuous at <math>x = -3</math>
At <math>x = 5</math>, <math>f(x) = f(5) = 5 \times 5 - 3 = 25 - 3 = 22</math>
<math display="block">\lim_{x \to 5} f(x) = \lim_{x \to 5} (5x - 3) = 5 \times 5 - 3 = 22</math>
<math display="block">\therefore \lim_{x \to 5} f(x) = f(5)</math>
Therefore, f is continuous at <math>x = 5</math>
Question 2:
Examine the continuity of the function <math>f(x) = 2x^2 - 1</math> at <math>x = 3</math>
Answer
The given function is <math>f(x) = 2x^2 - 1</math>
At <math>x = 3</math>, <math>f(x) = f(3) = 2 \times 3^2 - 1 = 17</math>
<math display="block">\lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 - 1) = 2 \times 3^2 - 1 = 17</math>
<math display="block">\therefore \lim_{x \to 3} f(x) = f(3)</math>
Thus, f is continuous at <math>x = 3</math>
Class XII
Chapter 5 - Continuity and Differentiability
Maths
Question 3:
Examine the following functions for continuity.
- <math>f(x) = x - 5</math> (b) <math>f(x) = \frac{1}{x - 5}, x \neq 5</math>
- <math>f(x) = \frac{x^2 - 25}{x + 5}, x \neq -5</math>
<math>f(x) = |x-5|</math> Answer
(a) The given function is <math>f(x) = x - 5</math>
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, <math>\lim_{x \to k} f(x) = \lim_{x \to k} (x-5) = k-5 = f(k)</math>
<math display="block">\therefore \lim_{x \to k} f(x) = f(k)</math>
Hence, f is continuous at every real number and therefore, it is a continuous function.
<math display="block">f(x) = \frac{1}{x-5}, x \neq 5</math>
- The given function is
For any real number <math>k \neq 5</math>, we obtain
<math display="block">\lim_{x \to k} f(x) = \lim_{x \to k} \frac{1}{x - 5} = \frac{1}{k - 5}</math>
Also, <math>f(k) = \frac{1}{k-5}</math> (As <math>k \neq 5</math>)
<math display="block">\therefore \lim_{x \to k} f(x) = f(k)</math>
Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function.
<math>f(x) = \frac{x^2 - 25}{x + 5}, x \neq -5</math>
- The given function is
For any real number <math>c \neq -5</math>, we obtain
Common mistakes
- Incorrectly calculating the limit of a function.
- Confusing the limit of a function with its value at a point.
- Errors in algebraic manipulation when simplifying expressions for limits.
- Not considering the domain of the function when checking for continuity.
Revision tips
- Master the definition of continuity: \(\lim_{x\to c} f(x) = f(c)\).
- Practice evaluating limits for various function types.
- Pay close attention to the points where the function definition might change.
- Work through each example and exercise step-by-step to solidify understanding.
Practice MCQs
Q1. For a function \(f(x)\) to be continuous at \(\), which condition must be met?
Explanation: The fundamental definition of continuity at a point requires that the limit of the function as x approaches that point must exist and be equal to the function's value at that point.
Q2. The function \(f(x) = 5x - 3\) is continuous at which of the following points?
Explanation: Polynomial functions are continuous everywhere. For \(f(x) = 5x - 3\), the limit as x approaches any real number c equals \(5c - 3\), which is also \(f(c)\). Thus, it's continuous at \(\), \(\), and \(\).
Q3. What is the value of \(f(3)\) for the function \(f(x) = 2 - 1\)?
Explanation: Substituting \(\) into the function gives \(f(3) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17\).
Q4. The function \(f(x) = x - 5\) is continuous:
Explanation: Since \(f(x) = x - 5\) is a polynomial function, it is continuous for all real numbers. The limit as \(x k\) is \(k - 5\), which equals \(f(k)\) for any real number \(k\).
Q5. For the function \(f(x) = \), where is it discontinuous?
Explanation: The function is undefined at \(\) because division by zero is not allowed. Therefore, it is discontinuous at \(\). For any other real number \(k 5\), the function is continuous.
Frequently asked questions
What is the main concept covered in the NCERT Solutions for Class 12 Maths Chapter 5?
The main concept is the continuity of functions, both at a specific point and over an interval. It involves understanding and applying the definition of continuity, which relates the limit of a function at a point to its value at that point.
How do these solutions help in understanding continuity?
The solutions provide clear, step-by-step explanations for proving and examining the continuity of various functions. They demonstrate how to correctly calculate limits and compare them with function values, reinforcing the theoretical concepts.
Are these solutions suitable for exam preparation?
Yes, these NCERT Solutions are specifically designed for exam preparation. They cover the types of problems commonly found in board exams related to continuity and differentiability, helping students build confidence and accuracy.
What types of functions are covered in these exercises?
The exercises cover basic polynomial functions, rational functions (where denominators can be zero), and functions involving absolute values, illustrating continuity checks in different scenarios.
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 the limit of \(f(x)\) as \(x\) approaches \(c\) exists, \(f(c)\) is defined, and the limit is equal to \(f(c)\). Mathematically, \(\lim_{x\to c} f(x) = f(c)\).
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