25 original CUET-style maths questions to sit in one go, then mark. Half the length of a full CUET maths paper, so you can fit it into a study session.
On \(\{1, 2, 3\}\), the relation \(R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}\) is
Every \((a, a)\) is there (reflexive); \((1, 2)\) and \((2, 1)\) are both there (symmetric); the only chains \((1, 2), (2, 1)\) give \((1, 1)\), which is there (transitive).
The function \(f : \mathbb{R} \to \mathbb{R}\), \(f(x) = x^5 + 2x\), is
\(f'(x) = 5x^4 + 2 > 0\), so \(f\) is strictly increasing and hence one-one. It is continuous and goes to \(\pm\infty\) as \(x \to \pm\infty\), so it takes every real value (onto).
\(\sin^{-1}\!\left(-\dfrac12\right) + \cos^{-1}\!\left(-\dfrac{\sqrt3}{2}\right)\) equals
\(\sin^{-1}\left(-\tfrac12\right) = -\dfrac{\pi}{6}\); \(\cos^{-1}\left(-\tfrac{\sqrt3}{2}\right) = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}\). Sum \(\dfrac{4\pi}{6} = \dfrac{2\pi}{3}\).
\(\cos\left(\sin^{-1}\dfrac{5}{13}\right)\) equals
If \(\sin\theta = \tfrac{5}{13}\) with \(\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]\), then \(\cos\theta = \sqrt{1 - \tfrac{25}{169}} = \tfrac{12}{13}\) (positive in this range).
If \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), then \(A + A^T\) is
\(A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\), so \(A + A^T = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix}\) (always symmetric).
\(A\) is a \(2 \times 3\) matrix and \(B\) is a \(3 \times 4\) matrix. Then \(AB\) is
The inner sizes match (3 and 3), and the product takes the outer sizes: \(2 \times 4\).
\(\begin{vmatrix} 2 & -1 \\ 3 & 4 \end{vmatrix}\) equals
\(2 \times 4 - (-1) \times 3 = 8 + 3 = 11\).
\(A\) is a \(3 \times 3\) matrix with \(|A| = 3\). Then \(|3A|\) equals
\(|kA| = k^3|A|\) for a \(3 \times 3\) matrix: \(27 \times 3 = 81\).
Trap: \(|3A| = 3|A| = 9\) multiplies only one row by 3.
\(f(x) = \dfrac{\sin x}{x}\) for \(x \ne 0\) and \(f(0) = k\). If \(f\) is continuous at 0, then \(k\) is
\(\displaystyle\lim_{x \to 0}\frac{\sin x}{x} = 1\), so \(k = 1\).
\(\dfrac{d}{dx}\left(e^{\sin x}\right)\) equals
Chain rule: \(e^{\sin x} \cdot \cos x\).
If \(y = \log(\log x)\), \(x > 1\), then \(\dfrac{dy}{dx}\) is
\(\dfrac{1}{\log x} \cdot \dfrac1x\).
The slope of the normal to \(y = x^3\) at \((1, 1)\) is
Slope of the tangent: \(3x^2 = 3\). The normal is perpendicular: slope \(-\dfrac13\).
\(f(x) = 2x^3 - 3x^2 - 12x + 5\) is strictly decreasing on
\(f'(x) = 6x^2 - 6x - 12 = 6(x - 2)(x + 1) < 0\) for \(-1 < x < 2\).
The maximum value of \(f(x) = -2x^2 + 8x - 3\) is
\(f'(x) = -4x + 8 = 0\) at \(x = 2\); \(f(2) = -8 + 16 - 3 = 5\).
\(\displaystyle\int \frac{\sec^2 x}{\tan x}\,dx\) equals
With \(t = \tan x\), \(dt = \sec^2 x\,dx\): \(\displaystyle\int\frac{dt}{t} = \log|\tan x| + C\).
\(\displaystyle\int_1^2 (3x^2 - 2x)\,dx\) equals
\(\left[x^3 - x^2\right]_1^2 = (8 - 4) - (1 - 1) = 4\).
\(\displaystyle\int_{-1}^{1} x^3\cos x\,dx\) equals
\(x^3\cos x\) is odd (odd × even), so its integral over \([-1, 1]\) is 0.
The area between \(y = 3x\), the \(x\)-axis and the line \(x = 4\) is
\(\displaystyle\int_0^4 3x\,dx = \frac{3 \cdot 16}{2} = 24\) (a triangle with base 4 and height 12).
The order of \(\dfrac{d^3y}{dx^3} + 2\dfrac{dy}{dx} = y\) is
The highest derivative present is the third, so the order is 3.
Solve \(\dfrac{dy}{dx} = e^{2x - y}\). The general solution is
\(e^{y}\,dy = e^{2x}\,dx\). Integrate: \(e^y = \tfrac12 e^{2x} + c\), i.e. \(2e^y = e^{2x} + C\).
Trap: \(\int e^{2x}\,dx = \tfrac12 e^{2x}\): dropping the \(\tfrac12\) gives the wrong family.
The magnitude of \(2\hat i - 3\hat j + 6\hat k\) is
\(\sqrt{4 + 9 + 36} = \sqrt{49} = 7\).
\((\hat i + 2\hat j + 3\hat k)\cdot(3\hat i - 2\hat j + \hat k)\) equals
\(3 - 4 + 3 = 2\).
The plane through \((1, 0, 0)\), \((0, 2, 0)\) and \((0, 0, 3)\) is
Intercept form: \(\dfrac{x}{1} + \dfrac{y}{2} + \dfrac{z}{3} = 1\). Multiply by 6: \(6x + 3y + 2z = 6\).
The corner points of a feasible region are \((0, 0)\), \((4, 0)\), \((3, 2)\) and \((0, 3)\). The maximum of \(Z = 5x + 3y\) is
\(Z\): 0, 20, 21, 9. Maximum 21 at \((3, 2)\).
For two events, \(P(B) = 0.8\) and the probability of \(A\) given \(B\) is \(0.35\). The probability that both happen is
\(P(A \cap B) = P(B)\,P(A \mid B) = 0.8 \times 0.35 = 0.28\).
All questions are original, written by CBSE Math Revision in the style of each exam; none are taken from NTA, IIT, CBSE or NCERT papers. Every answer was checked twice: by a computer re-solve and by hand. No calculator needed. Spotted a mistake? Tell us.
CBSE Math Revision is independent and not affiliated with NTA, the IITs or CBSE. Exam details marked to confirm are from the exam bodies' published documents but have not yet been re-checked against the current edition: always follow the official information bulletin.