CBSE Class 12 Mathematics Continuity and Differentiability MCQs Set 03

Practice CBSE Class 12 Mathematics Continuity and Differentiability MCQs Set 03 provided below. The MCQ Questions for Class 12 Chapter 5 Continuity and Differentiability Mathematics with answers and follow the latest CBSE/ NCERT and KVS patterns. Refer to more Chapter-wise MCQs for CBSE Class 12 Mathematics and also download more latest study material for all subjects

MCQ for Class 12 Mathematics Chapter 5 Continuity and Differentiability

Class 12 Mathematics students should review the 50 questions and answers to strengthen understanding of core concepts in Chapter 5 Continuity and Differentiability

Chapter 5 Continuity and Differentiability MCQ Questions Class 12 Mathematics with Answers

Question. The function \( f(x) = \begin{cases} \frac{\sin x}{x} + \cos x, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \) is continuous at \( x = 0 \), then the value of \( k \) is
(a) 3
(b) 2
(c) 1
(d) 1.5
Answer: (b) 2

 

Question. The function \( f(x) = [x] \), where \( [x] \) denotes the greatest integer function, is continuous at
(a) 4
(b) -2
(c) 1
(d) 1.5
Answer: (d) 1.5

 

Question. The number of points at which the function \( f(x) = \frac{1}{x - [x]} \) is not continuous is
(a) 1
(b) 2
(c) 3
(d) None of the options
Answer: (d) None of the options

 

Question. The value of \( k \) which makes the function defined by \( f(x) = \begin{cases} \sin \frac{1}{x}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \), continuous at \( x = 0 \) is
(a) 8
(b) 1
(c) -1
(d) None of the options
Answer: (d) None of the options

 

Question. The value of \( c \) in Rolle's Theorem for the function \( f(x) = e^x \sin x \), \( \in [0, \pi] \) is 
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{3\pi}{4} \)
Answer: (d) \( \frac{3\pi}{4} \)

 

Question. The value of \( c \) in Mean Value Theorem for the function \( f(x) = x(x - 2) \), \( x \in [1, 2] \) is
(a) \( \frac{3}{2} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{7}{4} \)
Answer: (a) \( \frac{3}{2} \)

 

Question. If \( f(x) = 2x \) and \( g(x) = \frac{x^2}{2} + 1 \), then which of the following can be a discontinuous function?
(a) \( f(x) + g(x) \)
(b) \( f(x) - g(x) \)
(c) \( f(x) \cdot g(x) \)
(d) \( \frac{g(x)}{f(x)} \)
Answer: (d) \( \frac{g(x)}{f(x)} \)
 

Question. The set of points where the function \( f \) given by \( f(x) = |2x - 1|\sin x \) is differentiable is
(a) \( R \)
(b) \( R - \left\{ \frac{1}{2} \right\} \)
(c) \( (0, \infty) \)
(d) None of the options
Answer: (b) \( R - \left\{ \frac{1}{2} \right\} \)

 

Question. The function \( f(x) = \cot x \) is discontinuous on the set
(a) \( \{x = n\pi : n \in Z\} \)
(b) \( \{x = 2n\pi : n \in Z\} \)
(c) \( \left\{ x = (2n + 1)\frac{\pi}{2} ; n \in Z \right\} \)
(d) \( \left\{ x = \frac{n\pi}{2} ; n \in Z \right\} \)
Answer: (a) \( \{x = n\pi : n \in Z\} \)

 

Question. The function \( f(x) = e^{|x|} \) is
(a) continuous everywhere but not differentiable at \( x = 0 \)
(b) continuous and differentiable everywhere
(c) not continuous at \( x = 0 \)
(d) None of the options
Answer: (a) continuous everywhere but not differentiable at \( x = 0 \)

 

Question. If \( f(x) = x^2 \sin \frac{1}{x} \), where \( x \neq 0 \), then the value of the function \( f \) at \( x = 0 \), so that the function is continuous at \( x = 0 \), is
(a) 0
(b) -1
(c) 1
(d) None of the options
Answer: (a) 0

 

Question. Let \( f(x) = |\sin x| \). Then
(a) \( f \) is everywhere differentiable
(b) \( f \) is everywhere continuous but not differentiable at \( x = n\pi : n \in Z \)
(c) \( f \) is everywhere continuous but not differentiable at \( x = (2n + 1)\frac{\pi}{2}, n \in Z \)
(d) None of the options
Answer: (b) \( f \) is everywhere continuous but not differentiable at \( x = n\pi : n \in Z \)

 

Question. The value of \( c \) in Rolle's theorem for the function \( f(x) = x^3 - 3x \) in the interval \( [0, \sqrt{3}] \) is
(a) 1
(b) -1
(c) \( \frac{3}{2} \)
(d) \( \frac{1}{3} \)
Answer: (a) 1

 

Question. The set of points where the functions \( f \) given by \( f(x) = |x - 3| \cos x \) is differentiable is
(a) \( R \)
(b) \( R - \{3\} \)
(c) \( (0, \infty) \)
(d) None of the options
Answer: (b) \( R - \{3\} \)

 

Question. Let \( f(x) = |\cos x| \). Then,
(a) \( f \) is everywhere differentiable.
(b) \( f \) is everywhere continuous but not differentiable at \( n = n\pi, n \in Z \).
(c) \( f \) is everywhere continuous but not differentiable at \( x = (2n + 1)\frac{\pi}{2}, n \in Z \).
(d) None of the options.
Answer: (c) \( f \) is everywhere continuous but not differentiable at \( x = (2n + 1)\frac{\pi}{2}, n \in Z \).

 

Question. The function \( f(x) = |x| + |x - 1| \) is
(a) continuous at \( x = 0 \) as well as at \( x = 1 \).
(b) continuous at \( x = 1 \) but not at \( x = 0 \).
(c) discontinuous at \( x = 0 \) as well as at \( x = 1 \).
(d) continuous at \( x = 0 \) but not at \( x = 1 \).
Answer: (a) continuous at \( x = 0 \) as well as at \( x = 1 \).

 

Question. Differential coefficient of \( \sec (\tan^{-1} x) \) w.r.t. \( x \) is
(a) \( \frac{x}{\sqrt{1 + x^2}} \)
(b) \( \frac{x}{1 + x^2} \)
(c) \( x \sqrt{1 + x^2} \)
(d) \( \frac{1}{\sqrt{1 + x^2}} \)
Answer: (a) \( \frac{x}{\sqrt{1 + x^2}} \)

 

Question. If \( u = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \) and \( v = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \), then \( \frac{du}{dv} \) is
(a) \( \frac{1}{2} \)
(b) \( x \)
(c) \( \frac{1 - x^2}{1 + x^2} \{4, -4\}, \phi \)
(d) 1
Answer: (d) 1

 

Question. The function \( f(x) = \frac{4 - x^2}{4x - x^3} \) is
(a) discontinuous at only one point
(b) discontinuous at exactly two points
(c) discontinuous at exactly three points
(d) None of the options
Answer: (c) discontinuous at exactly three points

 

Question. If \( y = \sqrt{\sin x + y} \), then \( \frac{dy}{dx} \) is equal to
(a) \( \frac{\cos x}{2y - 1} \)
(b) \( \frac{\cos x}{1 - 2y} \)
(c) \( \frac{\sin x}{1 - 2y} \)
(d) \( \frac{\sin x}{2y - 1} \)
Answer: (a) \( \frac{\cos x}{2y - 1} \)
 

Assertion-Reason Questions

The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.

 

Question. Assertion (A) : If \( f(x).g(x) \) is continuous at \( x = a \). then \( f(x) \) and \( g(x) \) are separately continuous at \( x = a \).
Reason (R) : Any function \( f(x) \) is said to be continuous at \( x = a \), if \( \lim_{h \to 0} f(a + h) = f(a) \).
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.
Solution: Let \( f(x) = x \) and \( g(x) = \sin \left(\frac{1}{x}\right) \) then \( F(x) = f(x)g(x) = x \sin \left(\frac{1}{x}\right) \).
Then \( \lim_{h \to 0} F(0 + h) = F(0) \). Hence \( F(x) = f(x) g(x) \) is continuous at \( x = 0 \), But \( g(x) = \sin \left(\frac{1}{x}\right) \) is not continuous at \( x = 0 \). Assertion (A) is false and Reason (R) is true.
Hence, (d) is the correct option.

 

Question. Assertion (A) : If \( f(x) \) and \( g(x) \) are two continuous functions such that \( f(0) = 3 \), \( g(0) = 2 \), then \( \lim_{x \to 0} \{f(x) + g(x)\} = 5 \).
Reason (R) : If \( f(x) \) and \( g(x) \) are two continuous functions at \( x = a \) then \( \lim_{x \to a} \{f(x) + g(x)\} = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \).
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation for A.
Solution: \( \lim_{x \to 0} \{f(x) + g(x)\} = f(0) + g(0) = 3 + 2 = 5 \)
Hence, (a) is the correct option.

 

Question. Assertion (A) : \( |\sin x| \) is a continuous function.
Reason (R) : If \( f(x) \) and \( g(x) \) both are continuous functions, then \( gof(x) \) is also a continuous function.
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation for A.
Solution: We have \( f(x) = \sin x \) and \( g(x) = |x| \)
\( \therefore gof(x) = g(f(x)) = g(\sin x) = |\sin x| \)
Here, \( \sin x \) and \( |x| \) both are continuous functions.
\( \therefore |\sin x| \) is also continuous function.
Clearly, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Hence, (a) is the correct option.

 

Question. Assertion (A) : If \( y = \sin x \), then \( \frac{d^3 y}{dx^3} = -1 \) at \( x = 0 \).
Reason (R) : If \( y = f(x) \cdot g(x) \), then \( \frac{dy}{dx} = f(x) \cdot \frac{d}{dx}g(x) + g(x) \frac{d}{dx}f(x) \).
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (b) Both A and R are true and R is not the correct explanation for A.
Solution: We have, \( y = \sin x \)
\( \implies \) \( \frac{dy}{dx} = \cos x \)
\( \implies \) \( \frac{d^2 y}{dx^2} = -\sin x \)
\( \therefore \quad \frac{d^3 y}{dx^3} = -\cos x = -1 \), at \( x = 0 \)
Clearly, both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Hence, (b) is the correct option.

 

Question. Assertion (A) : If \( f(x) = \sin^{-1} x + \cos^{-1} x + 2 \) then \( f'(1) = 0 \).
Reason (R) : \( \frac{d}{dx} \sin x = \cos x \)
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (b) Both A and R are true and R is not the correct explanation for A.
Solution: We have, \( f(x) = \sin^{-1} x + \cos^{-1} x + 2 \)
\( f(x) = \frac{\pi}{2} + 2 \quad \left[ \because \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \right] \)
\( \therefore \quad f'(x) = 0 \)
\( \implies \) \( f'(1) = 0 \)
Clearly, both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Hence, (b) is the correct option.

MCQs for Chapter 5 Continuity and Differentiability Mathematics Class 12

Students can use these MCQs for Chapter 5 Continuity and Differentiability to quickly test their knowledge of the chapter. These multiple-choice questions have been designed as per the latest syllabus for Class 12 Mathematics released by CBSE. Our expert teachers suggest that you should practice daily and solving these objective questions of Chapter 5 Continuity and Differentiability to understand the important concepts and better marks in your school tests.

Chapter 5 Continuity and Differentiability NCERT Based Objective Questions

Our expert teachers have designed these Mathematics MCQs based on the official NCERT book for Class 12. We have identified all questions from the most important topics that are always asked in exams. After solving these, please compare your choices with our provided answers. For better understanding of Chapter 5 Continuity and Differentiability, you should also refer to our NCERT solutions for Class 12 Mathematics created by our team.

Online Practice and Revision for Chapter 5 Continuity and Differentiability Mathematics

To prepare for your exams you should also take the Class 12 Mathematics MCQ Test for this chapter on our website. This will help you improve your speed and accuracy and its also free for you. Regular revision of these Mathematics topics will make you an expert in all important chapters of your course.

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Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics Class 12 material?

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