15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
The value of \(k\) for which \(f(x) = \begin{cases} \dfrac{x^2 - 9}{x - 3}, & x \ne 3 \\ k, & x = 3 \end{cases}\) is continuous at \(x = 3\) is
Why: For x ≠ 3, (x² − 9)/(x − 3) = x + 3, which tends to 6.
\(\lim_{x \to 3}\dfrac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3}(x + 3) = 6\). Continuity needs \(k = 6\).
The function \(f(x) = |x - 2|\) at \(x = 2\) is
Why: The graph has a corner at x = 2: left derivative −1, right derivative 1.
\(\lim_{x \to 2}|x - 2| = 0 = f(2)\), so \(f\) is continuous. Left derivative \(= -1\), right derivative \(= 1\); they differ, so \(f\) is not differentiable at \(2\).
\(\dfrac{d}{dx}\left[\cos(x^2)\right]\) equals
Why: Chain rule: derivative of cos u is −sin u × u′, with u = x².
\(\dfrac{d}{dx}\cos(x^2) = -\sin(x^2) \cdot 2x = -2x\sin(x^2)\).
\(\dfrac{d}{dx}\left[e^{3x}\sin x\right]\) equals
Why: Product rule: 3e3x sin x + e3x cos x.
\((e^{3x})'\sin x + e^{3x}(\sin x)' = 3e^{3x}\sin x + e^{3x}\cos x = e^{3x}(3\sin x + \cos x)\).
\(\dfrac{d}{dx}\left[\tan^{-1}(2x)\right]\) equals
Why: d/dx tan⁻¹u = u′/(1 + u²) with u = 2x.
\(\dfrac{1}{1 + (2x)^2} \cdot 2 = \dfrac{2}{1 + 4x^2}\).
For \(0 \lt x \lt \pi\), \(\dfrac{d}{dx}\left[\log(\sin x)\right]\) equals (Here \(\log\) means \(\log_e\).)
Why: Derivative of log u is u′/u = cos x/sin x.
\(\dfrac{1}{\sin x} \cdot \cos x = \cot x\).
For \(-2 \lt x \lt 2\), \(\dfrac{d}{dx}\left[\sin^{-1}\dfrac{x}{2}\right]\) equals
Why: (1/√(1 − x²/4)) × 1/2 = 1/√(4 − x²).
\(\dfrac{1}{\sqrt{1 - \frac{x^2}{4}}} \cdot \dfrac12 = \dfrac{1}{2} \cdot \dfrac{2}{\sqrt{4 - x^2}} = \dfrac{1}{\sqrt{4 - x^2}}\).
If \(x^2 + y^2 = 25\), then \(\dfrac{dy}{dx}\) at the point \((3, 4)\) is
Why: 2x + 2y y′ = 0 gives y′ = −x/y.
Differentiating: \(2x + 2y\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{x}{y} = -\dfrac34\) at \((3, 4)\).
If \(x = 2t^2\) and \(y = 4t\), then \(\dfrac{dy}{dx}\) at \(t = 2\) is
Why: dy/dx = (dy/dt)/(dx/dt) = 4/(4t) = 1/t.
\(\dfrac{dx}{dt} = 4t\), \(\dfrac{dy}{dt} = 4\), so \(\dfrac{dy}{dx} = \dfrac{1}{t} = \dfrac12\) at \(t = 2\).
For \(x \gt 0\), if \(y = x^x\), then \(\dfrac{dy}{dx}\) equals (Here \(\log\) means \(\log_e\).)
Why: Take logs: log y = x log x, so y′/y = log x + 1.
\(\log y = x\log x \Rightarrow \dfrac{1}{y}\dfrac{dy}{dx} = \log x + 1 \Rightarrow \dfrac{dy}{dx} = x^x(1 + \log x)\).
If \(y = e^{2x} + x^3\), then \(\dfrac{d^2y}{dx^2}\) equals
Why: Differentiate twice: 2e2x + 3x², then 4e2x + 6x.
\(y' = 2e^{2x} + 3x^2\), \(y'' = 4e^{2x} + 6x\).
If \(y = 3\cos 2x\), then \(\dfrac{d^2y}{dx^2} + 4y\) equals
Why: y″ = −12 cos 2x = −4y.
\(y' = -6\sin 2x\), \(y'' = -12\cos 2x = -4y\). So \(y'' + 4y = 0\).
The function \(f(x) = \begin{cases} kx^2, & x \le 2 \\ 3, & x \gt 2 \end{cases}\) is continuous at \(x = 2\) when \(k\) equals
Why: Match the two pieces at x = 2: 4k = 3.
\(f(2) = 4k\) and \(\lim_{x \to 2^+} 3 = 3\). So \(4k = 3 \Rightarrow k = \dfrac34\).
For a constant \(a \gt 0\), \(\dfrac{d}{dx}\left(a^x\right)\) equals (Here \(\log\) means \(\log_e\).)
Why: ax = ex log a; differentiate by the chain rule.
\(a^x = e^{x\log a}\), so \(\dfrac{d}{dx}a^x = e^{x\log a} \cdot \log a = a^x\log a\). (\(x\,a^{x-1}\) is the power rule, which does not apply: the variable is in the exponent.)
The function \(f(x) = |x - 1| + |x + 2|\) is not differentiable at
Why: Each modulus term has a corner where its inside is zero.
\(|x - 1|\) has a corner at \(x = 1\) and \(|x + 2|\) at \(x = -2\); the other term is differentiable there, so the sum has corners at both points. (E.g. at \(x = 1\): left derivative \(0\), right derivative \(2\).)
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
A city taxi charges ₹\(F(d)\) for a trip of \(d\) km, where \(F(d) = 50 + 12d\) for \(0 \le d \le 5\) and \(F(d) = k + 10d\) for \(d \gt 5\). The transport authority insists that the fare must not jump at \(5\) km.
Priya pours a cup of tea and leaves it on the kitchen table while she answers the door. The room is at \(25\,^\circ\)C, and the tea cools towards room temperature. Its temperature (in °C) \(t\) minutes after pouring is \(T(t) = 25 + 60e^{-t/10}\).
This free set is separate from the chapter's question bank. On the Continuity and Differentiability chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 40 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.