Class 12 Mathematics Chapter 5 Continuity and Differentiability – Extra Questions with Answers

Extra practice questions for Class 12 Maths Chapter 5 (Continuity and Differentiability), beyond the textbook. These Class 12 Mathematics Chapter 5 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. State the condition for continuity of f at x=a.
Ans: lim_x→af(x)=f(a).

Q2. Is |x| differentiable at x=0?
Ans: No, the left and right derivatives differ (−1 and 1).

Q3. What is d/dx(e^x)?
Ans: e^x.

Q4. What is d/dx(log x)?
Ans: 1/x.

Q5. State the chain rule for y=f(g(x)).
Ans: dy/dx=f'(g(x))·g'(x).

Short Answer Questions (2–3 marks)

Q6. Differentiate y=sin(x²) with respect to x.
Ans: Using chain rule, dy/dx=cos(x²)·2x=2x cos(x²).

Q7. Find dy/dx if x²+y²=25 (implicit differentiation).
Ans: Differentiating both sides: 2x+2y(dy/dx)=0, so dy/dx=−x/y.

Q8. Differentiate y=x^x using logarithmic differentiation.
Ans: Take log: log y=x log x. Differentiate: (1/y)(dy/dx)=log x+1, so dy/dx=y(log x+1)=x^x(log x+1).

Higher-Order Thinking / Application Questions

Q9. Examine the continuity and differentiability of f(x)=|x−1| at x=1, showing all reasoning using left-hand and right-hand limits and derivatives.
Ans: For continuity at x=1: LHL=lim_x→1⁻|x−1|=lim_h→0|(1−h)−1|=lim_h→0h=0. RHL=lim_x→1⁺|x−1|=lim_h→0|(1+h)−1|=lim_h→0h=0. f(1)=|1−1|=0. Since LHL=RHL=f(1)=0, f is continuous at x=1. For differentiability: LHD=lim_h→0[f(1−h)−f(1)]/(−h)=lim_h→0[h−0]/(−h)=−1. RHD=lim_h→0[f(1+h)−f(1)]/h=lim_h→0[h−0]/h=1. Since LHD=−1≠RHD=1, f is NOT differentiable at x=1, despite being continuous there — this is a classic example demonstrating that continuity does not guarantee differentiability, because the graph has a sharp corner (kink) at x=1 even though there is no break or jump.

Q10. State Rolle’s Theorem and verify it for f(x)=x²−4x+3 on the interval [1,3], finding the value of c explicitly.
Ans: Rolle’s Theorem states that if f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then there exists at least one c∈(a,b) such that f'(c)=0. Here f(x)=x²−4x+3 is a polynomial, so it is continuous on [1,3] and differentiable on (1,3), satisfying the first two conditions. Check f(1)=1−4+3=0 and f(3)=9−12+3=0, so f(1)=f(3)=0, satisfying the third condition. By Rolle’s Theorem, there must exist c∈(1,3) with f'(c)=0. Now f'(x)=2x−4, so setting f'(c)=0 gives 2c−4=0, i.e., c=2. Since 2∈(1,3), this confirms Rolle’s Theorem holds, and geometrically it means the tangent to the parabola is horizontal at x=2, which is the vertex of the parabola exactly midway between the two roots x=1 and x=3.

Written by Satish

NCERTBooks.org is an independent educational resource run by a small team focused on making official NCERT textbooks easy to find, read, and download for students, parents, and teachers across India. We are not affiliated with NCERT or the Ministry of Education -- we organise publicly available NCERT content by class and subject, verify links against official sources, and build tools (like our in-browser reader) that make studying more convenient. Every guide we publish is written and reviewed by our team based on the actual NCERT curriculum and syllabus.