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 radius of a circle is increasing at \(2\) cm/s. The rate at which its area is increasing when the radius is \(5\) cm is
Why: dA/dt = 2πr dr/dt = 2π × 5 × 2.
\(A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt} = 2\pi(5)(2) = 20\pi\ \text{cm}^2/\text{s}\).
The volume of a cube is increasing at \(12\ \text{cm}^3/\text{s}\). How fast is its edge increasing when the edge is \(2\) cm?
Why: dV/dt = 3a² da/dt, so 12 = 3 × 4 × da/dt.
\(V = a^3 \Rightarrow \dfrac{dV}{dt} = 3a^2\dfrac{da}{dt}\). \(12 = 3(2)^2\dfrac{da}{dt} \Rightarrow \dfrac{da}{dt} = 1\) cm/s.
The function \(f(x) = x^2 - 6x + 5\) is strictly increasing on
Why: f′(x) = 2x − 6 > 0 exactly when x > 3.
\(f'(x) = 2x - 6 \gt 0 \Leftrightarrow x \gt 3\). So \(f\) is strictly increasing on \((3, \infty)\).
The function \(f(x) = e^{2x}\) is
Why: f′(x) = 2e2x > 0 for every x.
\(f'(x) = 2e^{2x} \gt 0\) for all real \(x\), so \(f\) is strictly increasing on \(\mathbb{R}\).
The local maximum value of \(f(x) = -x^2 + 4x + 1\) is
Why: f′(x) = −2x + 4 = 0 at x = 2, and f(2) = 5.
\(f'(x) = -2x + 4 = 0 \Rightarrow x = 2\); \(f''(x) = -2 \lt 0\), so a local maximum. Value \(f(2) = -4 + 8 + 1 = 5\). (\(2\) is where it occurs, not the value.)
The critical points of \(f(x) = 2x^3 - 6x\) are
Why: f′(x) = 6x² − 6 = 0 gives x = ±1.
\(f'(x) = 6x^2 - 6 = 6(x - 1)(x + 1) = 0 \Rightarrow x = \pm 1\). (\(\pm\sqrt3\) and \(0\) are the zeros of \(f\), not of \(f'\).)
The local minimum value of \(f(x) = 2x^3 - 6x\) is
Why: f″(1) = 12 > 0, so x = 1 gives the local minimum f(1) = −4.
Critical points \(x = \pm 1\); \(f''(x) = 12x\). \(f''(1) = 12 \gt 0\): local minimum \(f(1) = 2 - 6 = -4\). (\(f(-1) = 4\) is the local maximum.)
The greatest value of \(f(x) = \sin x + \sqrt3\cos x\) for \(0 \le x \le \dfrac{\pi}{2}\) is
Why: f′(x) = cos x − √3 sin x = 0 at x = π/6, where f = 1/2 + 3/2.
\(f'(x) = \cos x - \sqrt3\sin x = 0 \Rightarrow \tan x = \dfrac{1}{\sqrt3} \Rightarrow x = \dfrac{\pi}{6}\). \(f\left(\tfrac{\pi}{6}\right) = \tfrac12 + \tfrac32 = 2\); \(f(0) = \sqrt3\), \(f\left(\tfrac{\pi}{2}\right) = 1\). Greatest value \(= 2\).
The absolute maximum value of \(f(x) = x^2 - 4x\) on \([0, 5]\) is
Why: Compare f at the critical point x = 2 and at the end points 0 and 5.
\(f'(x) = 2x - 4 = 0 \Rightarrow x = 2\). \(f(0) = 0\), \(f(2) = -4\), \(f(5) = 5\). Absolute maximum \(= 5\) (at \(x = 5\)); absolute minimum \(= -4\).
Two positive numbers have sum \(16\). The greatest possible value of their product is
Why: P = x(16 − x) is greatest at x = 8.
\(P(x) = x(16 - x)\), \(P'(x) = 16 - 2x = 0 \Rightarrow x = 8\); \(P''(x) = -2 \lt 0\). Maximum \(P = 8 \times 8 = 64\).
The minimum value of \(f(x) = xe^{x}\) is
Why: f′(x) = ex(1 + x) = 0 at x = −1, and f″(−1) > 0.
\(f'(x) = e^x(1 + x) = 0 \Rightarrow x = -1\); \(f''(x) = e^x(2 + x)\), \(f''(-1) = e^{-1} \gt 0\). Minimum \(= f(-1) = -\dfrac{1}{e}\).
The total cost of producing \(x\) units of an item is \(C(x) = 2x^2 + 10x + 100\) rupees. The marginal cost when \(5\) units are produced is
Why: Marginal cost = C′(x) = 4x + 10.
\(C'(x) = 4x + 10\), so \(C'(5) = 30\): about ₹\(30\) for the next unit. (\(C(5) = 200\) is the total cost.)
The function \(f(x) = x^3 - 6x^2 + 9x\) is strictly decreasing on
Why: f′(x) = 3(x − 1)(x − 3) is negative between 1 and 3.
\(f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3) \lt 0\) for \(1 \lt x \lt 3\). So \(f\) is strictly decreasing on \((1, 3)\).
For \(f(x) = x^4 - 2x^2\), which statement about \(x = 0\) is true?
Why: f′(0) = 0 and f″(0) = −4 < 0.
\(f'(x) = 4x^3 - 4x\), \(f'(0) = 0\); \(f''(x) = 12x^2 - 4\), \(f''(0) = -4 \lt 0\). By the second derivative test, \(x = 0\) is a point of local maximum.
A rectangle has perimeter \(36\) cm. Its greatest possible area is
Why: A = x(18 − x) is greatest at x = 9 (a square).
Sides \(x\) and \(18 - x\): \(A = 18x - x^2\), \(A' = 18 - 2x = 0 \Rightarrow x = 9\), \(A'' = -2 \lt 0\). Maximum area \(= 81\ \text{cm}^2\).
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.
Anil has \(40\) m of fencing to enclose a rectangular kitchen garden next to the long straight wall of his house. The wall forms one side, so only the other three sides need fencing. Let each side perpendicular to the wall be \(x\) m.
For a school fete, a spherical balloon is pumped up with a small electric pump so that its radius increases at a steady \(0.5\) cm/s. (Volume \(V = \dfrac43\pi r^3\), surface area \(S = 4\pi r^2\).)
This free set is separate from the chapter's question bank. On the Application of Derivatives 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.