The 20 one-mark questions from the March 2025 12th Standard Mathematics public exam, in paper order, with the correct option marked and a short explanation.
Q1
Subtraction is not a binary operation in :
- A. \(\mathbb{N}\)Correct
- B. \(\mathbb{R}\)
- C. \(\mathbb{Q}\)
- D. \(\mathbb{Z}\)
Explanation. In \(\mathbb{N}\), \(2-3=-1\notin\mathbb{N}\), so subtraction is not closed there; it is closed in \(\mathbb{R}, \mathbb{Q}, \mathbb{Z}\).
Q2
Suppose that X takes on one of the values 0, 1 and 2. If for some constant k, \(P(X=i)=k\,P(X=i-1)\) for \(i=1, 2\) and \(P(X=0)=\frac{1}{7}\), then the value of k is :
- A. 3
- B. 1
- C. 4
- D. 2Correct
Explanation. \(P(0)=\frac{1}{7}, P(1)=\frac{k}{7}, P(2)=\frac{k^2}{7}\). Total probability 1 gives \(1+k+k^2=7\), so \((k+3)(k-2)=0\) and k = 2 (k must be positive).
Q3
If A is a non-singular matrix of order \(3\times 3\) and \(|A|=5\) then \(|A^{-1}|\) is :
- A. \(5^2\)
- B. 5
- C. \(\frac{1}{5^2}\)
- D. \(\frac{1}{5}\)Correct
Explanation. \(|A^{-1}|=\frac{1}{|A|}=\frac{1}{5}\).
Q4
A stone is thrown up vertically. The height it reaches at time t seconds is given by \(x=80t-16t^2\). The stone reaches the maximum height in time t seconds is given by :
- A. 3
- B. 2
- C. 3.5
- D. 2.5Correct
Explanation. At maximum height the velocity is zero: \(\frac{dx}{dt}=80-32t=0\), so \(t=2.5\) seconds.
Q5
The order and degree of the differential equation \(\sqrt{\frac{dy}{dx}}-4\frac{dy}{dx}-7x=0\) are respectively :
- A. 1, 2Correct
- B. 2, 1
- C. 2, 2
- D. 1, 1
Explanation. Only the first derivative occurs, so the order is 1. Removing the radical: \(\frac{dy}{dx}=\left(4\frac{dy}{dx}+7x\right)^2\), whose highest power of \(\frac{dy}{dx}\) is 2, so the degree is 2.
Q6
If \(A=\begin{bmatrix}2 & 3\\ 5 & -2\end{bmatrix}\) be such that \(\lambda A^{-1}=A\), then \(\lambda\) is :
- A. 19Correct
- B. 17
- C. 21
- D. 14
Explanation. Multiplying by A: \(\lambda I=A^2\). \(A^2=\begin{bmatrix}19 & 0\\ 0 & 19\end{bmatrix}=19I\), so \(\lambda=19\).
Q7
The slope at any point of a curve \(y=f(x)\) is given by \(\frac{dy}{dx}=3x^2\) and it passes through \((-1, 1)\). Then the equation of the curve is :
- A. \(y=3x^3+4\)
- B. \(y=x^3+2\)Correct
- C. \(y=x^3+5\)
- D. \(y=3x^2+4\)
Explanation. Integrating, \(y=x^3+C\). At \((-1,1)\): \(1=-1+C\), so \(C=2\) and \(y=x^3+2\).
Q8
The domain of the function defined by \(f(x)=\sin^{-1}\sqrt{x-1}\) is :
- A. \([0, 1]\)
- B. \([1, 2]\)Correct
- C. \([-1, 0]\)
- D. \([-1, 1]\)
Explanation. We need \(0\le\sqrt{x-1}\le 1\), i.e. \(0\le x-1\le 1\), so \(x\in[1, 2]\).
Q9
If \(u(x, y)=e^{x^2+y^2}\), then \(\frac{\partial u}{\partial x}\) is equal to :
- A. \(x^2u\)
- B. \(e^{x^2+y^2}\)
- C. \(y^2u\)
- D. \(2xu\)Correct
Explanation. \(\frac{\partial u}{\partial x}=e^{x^2+y^2}\cdot 2x=2xu\).
Q10
The number of real numbers in \([0, 2\pi]\) satisfying \(\sin^4x-2\sin^2x+1\) is :
- A. 1
- B. 2Correct
- C. \(\infty\)
- D. 4
Explanation. The paper omits "= 0"; the intended equation is \(\sin^4x-2\sin^2x+1=0\). Then \((\sin^2x-1)^2=0\), so \(\sin x=\pm 1\), giving \(x=\frac{\pi}{2}, \frac{3\pi}{2}\): 2 values.
Q11
The square root of i are :
- A. \(\pm\frac{1}{2}(1+i)\)
- B. \(\pm\frac{1}{\sqrt{2}}(1+i)\)Correct
- C. \(\pm\frac{1}{2}(1-i)\)
- D. \(\pm\frac{1}{\sqrt{2}}(1-i)\)
Explanation. \(\left(\frac{1+i}{\sqrt{2}}\right)^2=\frac{1+2i-1}{2}=i\), so the square roots are \(\pm\frac{1}{\sqrt{2}}(1+i)\).
Q12
The value of \(\sum_{n=1}^{13}(i^n+i^{n-1})\) is :
- A. 1
- B. \(1+i\)Correct
- C. 0
- D. i
Explanation. Any four consecutive powers of i add to 0. So \(\sum_{n=1}^{13}i^n=i^{13}=i\) and \(\sum_{n=1}^{13}i^{n-1}=i^{12}=1\); the total is \(1+i\).
Q13
If in 6 trials, X is a binomial variable which follows the relation \(9P(X=4)=P(X=2)\), then the probability of success is :
- A. 0.375
- B. 0.125
- C. 0.75
- D. 0.25Correct
Explanation. \(9\binom{6}{4}p^4q^2=\binom{6}{2}p^2q^4\) gives \(9p^2=q^2\), so \(q=3p\). With \(p+q=1\), \(p=0.25\).
Q14
The angle between the lines \(\frac{x-2}{3}=\frac{y+1}{-2}, z=2\) and \(\frac{x-1}{1}=\frac{2y+3}{3}=\frac{z+5}{2}\) is :
- A. \(\frac{\pi}{3}\)
- B. \(\frac{\pi}{6}\)
- C. \(\frac{\pi}{2}\)Correct
- D. \(\frac{\pi}{4}\)
Explanation. Direction ratios are \((3, -2, 0)\) and \(\left(1, \frac{3}{2}, 2\right)\). Dot product \(=3-3+0=0\), so the lines are perpendicular: \(\frac{\pi}{2}\).
Q15
The point of inflection of the curve \(y=(x-1)^3\) is :
- A. \((1, 0)\)Correct
- B. \((0, 0)\)
- C. \((1, 1)\)
- D. \((0, 1)\)
Explanation. \(y''=6(x-1)\) is zero at \(x=1\) and changes sign there; \(y(1)=0\), so the point is \((1, 0)\).
Q16
The value of \(\int_0^{\frac{2}{3}}\frac{dx}{\sqrt{4-9x^2}}\) is :
- A. \(\frac{\pi}{4}\)
- B. \(\frac{\pi}{6}\)Correct
- C. \(\pi\)
- D. \(\frac{\pi}{2}\)
Explanation. \(\int\frac{dx}{\sqrt{4-9x^2}}=\frac{1}{3}\sin^{-1}\frac{3x}{2}\). From 0 to \(\frac{2}{3}\): \(\frac{1}{3}\cdot\frac{\pi}{2}=\frac{\pi}{6}\).
Q17
The volume of solid of revolution of the region bounded by \(y^2=x(a-x)\) about x-axis is :
- A. \(\frac{\pi a^3}{5}\)
- B. \(\pi a^3\)
- C. \(\frac{\pi a^3}{6}\)Correct
- D. \(\frac{\pi a^3}{4}\)
Explanation. \(V=\pi\int_0^a(ax-x^2)\,dx=\pi\left(\frac{a^3}{2}-\frac{a^3}{3}\right)=\frac{\pi a^3}{6}\).
Q18
An ellipse has OB as semi minor axes, F and F' its foci and the angle FBF' is a right angle. Then the eccentricity of the ellipse is :
- A. \(\frac{1}{4}\)
- B. \(\frac{1}{\sqrt{2}}\)Correct
- C. \(\frac{1}{\sqrt{3}}\)
- D. \(\frac{1}{2}\)
Explanation. Angle FBF' = 90° means angle FBO = 45°, so OF = OB, i.e. \(ae=b\). Then \(a^2e^2=a^2(1-e^2)\), so \(e^2=\frac{1}{2}\) and \(e=\frac{1}{\sqrt{2}}\).
Q19
The volume of the parallelepiped with its edges represented by the vectors \(\hat{i}+\hat{j}, \hat{i}+2\hat{j}, \hat{i}+\hat{j}+\pi\hat{k}\) is :
- A. \(\pi\)Correct
- B. \(\frac{\pi}{2}\)
- C. \(\frac{\pi}{4}\)
- D. \(\frac{\pi}{3}\)
Explanation. Volume \(=\left|\begin{vmatrix}1 & 1 & 0\\ 1 & 2 & 0\\ 1 & 1 & \pi\end{vmatrix}\right|=\pi(2-1)=\pi\).
Q20
If \(f(x)>0\) for all x and \(g(x)=\log(f(x))\), then dg is :
- A. \(\frac{1}{f(x)}\,dx\)
- B. \(\frac{1}{f(x)}f'(x)\,dx\)Correct
- C. \(\frac{1}{x}\,dx\)
- D. \(\frac{1}{x}f(x)\,dx\)
Explanation. \(dg=g'(x)\,dx=\frac{f'(x)}{f(x)}\,dx\) by the chain rule.
