The 20 one-mark questions from the March 2024 12th Standard Mathematics public exam, in paper order, with the correct option marked and a short explanation.
Q1
The area between \(y^2=4x\) and its latus rectum is :
- A. \(\frac{8}{3}\)Correct
- B. \(\frac{2}{3}\)
- C. \(\frac{5}{3}\)
- D. \(\frac{4}{3}\)
Explanation. Latus rectum is x = 1. Area = \(2\int_0^1 2\sqrt{x}\,dx = 4\cdot\frac{2}{3} = \frac{8}{3}\).
Q2
The value of \(\int_0^a \left(\sqrt{a^2-x^2}\right)^3 dx\) is :
- A. \(\frac{3\pi a^2}{8}\)
- B. \(\frac{\pi a^3}{16}\)
- C. \(\frac{3\pi a^4}{8}\)
- D. \(\frac{3\pi a^4}{16}\)Correct
Explanation. Put \(x=a\sin\theta\): integral = \(a^4\int_0^{\pi/2}\cos^4\theta\,d\theta = a^4\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3\pi a^4}{16}\).
Q3
If \(P(x, y)\) be any point on \(16x^2+25y^2=400\) with foci \(F_1(3, 0)\) and \(F_2(-3, 0)\), then \(PF_1+PF_2\) is :
- A. 10Correct
- B. 8
- C. 12
- D. 6
Explanation. The ellipse is \(\frac{x^2}{25}+\frac{y^2}{16}=1\), so a = 5. For any point on an ellipse, \(PF_1+PF_2=2a=10\).
Q4
If \(|z_1|=1, |z_2|=2, |z_3|=3\) and \(|9z_1z_2+4z_1z_3+z_2z_3|=12\) then the value of \(|z_1+z_2+z_3|\) is :
- A. 3
- B. 1
- C. 4
- D. 2Correct
Explanation. Since \(9=z_3\bar{z}_3,\ 4=z_2\bar{z}_2,\ 1=z_1\bar{z}_1\), the expression is \(|z_1z_2z_3|\,|\bar{z}_1+\bar{z}_2+\bar{z}_3| = 6|z_1+z_2+z_3| = 12\), so \(|z_1+z_2+z_3|=2\).
Q5
The number of rows in the truth table of \((p\vee q)\wedge(p\vee r)\) is :
- A. 6
- B. 9
- C. 3
- D. 8Correct
Explanation. There are 3 simple statements p, q, r, so the truth table has \(2^3=8\) rows.
Q6
If a vector \(\vec{\alpha}\) lies in the plane of \(\vec{\beta}\) and \(\vec{\gamma}\), then :
- A. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\)Correct
- B. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=1\)
- C. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=2\)
- D. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=-1\)
Explanation. Three coplanar vectors have scalar triple product zero, so \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\).
Q7
The differential equation of the family of curves \(y=Ae^x+Be^{-x}\), where A and B are arbitrary constants is :
- A. \(\frac{dy}{dx}+y=0\)
- B. \(\frac{d^2y}{dx^2}+y=0\)
- C. \(\frac{dy}{dx}-y=0\)
- D. \(\frac{d^2y}{dx^2}-y=0\)Correct
Explanation. \(y'=Ae^x-Be^{-x}\) and \(y''=Ae^x+Be^{-x}=y\), so \(\frac{d^2y}{dx^2}-y=0\).
Q8
The horizontal asymptote of \(f(x)=\frac{1}{x}\) is :
- A. \(x=c\)
- B. \(y=0\)Correct
- C. \(y=c\)
- D. \(x=0\)
Explanation. As \(x\to\pm\infty\), \(\frac{1}{x}\to 0\), so the horizontal asymptote is \(y=0\).
Q9
If \(f(x)=\frac{x}{x+1}\), then its differential is given by :
- A. \(\frac{1}{x+1}dx\)
- B. \(\frac{-1}{(x+1)^2}dx\)
- C. \(\frac{-1}{x+1}dx\)
- D. \(\frac{1}{(x+1)^2}dx\)Correct
Explanation. \(f'(x)=\frac{(x+1)-x}{(x+1)^2}=\frac{1}{(x+1)^2}\), so \(df=\frac{1}{(x+1)^2}dx\).
Q10
If \((1+i)(1+2i)(1+3i)\cdots(1+ni)=x+iy\) then \(2\cdot5\cdot10\cdots(1+n^2)\) is :
- A. \(x^2+y^2\)Correct
- B. 1
- C. \(1+n^2\)
- D. \(i\)
Explanation. Taking modulus squared on both sides: \(|1+i|^2|1+2i|^2\cdots|1+ni|^2 = 2\cdot5\cdot10\cdots(1+n^2) = x^2+y^2\).
Q11
The number given by the Rolle's theorem for the function \(x^3-3x^2, x\in[0, 3]\) is :
- A. \(\frac{3}{2}\)
- B. 1
- C. 2Correct
- D. \(\sqrt{2}\)
Explanation. \(f(0)=f(3)=0\). \(f'(c)=3c^2-6c=0\) gives c = 0 or 2; c = 2 lies in (0, 3).
Q12
The type of conic section for \(x^2-3=5x+3y\) is :
- A. hyperbola
- B. ellipse
- C. circle
- D. parabolaCorrect
Explanation. Only \(x^2\) appears (no \(y^2\) or xy term), so A = 1, B = 0, C = 0 and \(B^2-4AC=0\): a parabola.
Q13
If A is a non-singular matrix such that \(A^{-1}=\begin{bmatrix}5 & 3\\ -2 & -1\end{bmatrix}\), then \((A^T)^{-1}=\)
- A. \(\begin{bmatrix}-1 & -3\\ 2 & 5\end{bmatrix}\)
- B. \(\begin{bmatrix}-5 & 3\\ 2 & 1\end{bmatrix}\)
- C. \(\begin{bmatrix}5 & -2\\ 3 & -1\end{bmatrix}\)Correct
- D. \(\begin{bmatrix}5 & 3\\ -2 & -1\end{bmatrix}\)
Explanation. \((A^T)^{-1}=(A^{-1})^T=\begin{bmatrix}5 & -2\\ 3 & -1\end{bmatrix}\).
Q14
The angle between the line \(\vec{r}=(\vec{i}+2\vec{j}-3\vec{k})+t(2\vec{i}+\vec{j}-2\vec{k})\) and the plane \(\vec{r}\cdot(\vec{i}+\vec{j})+4=0\) is :
- A. \(45^\circ\)Correct
- B. \(0^\circ\)
- C. \(90^\circ\)
- D. \(30^\circ\)
Explanation. Direction (2, 1, −2), normal (1, 1, 0): \(\sin\theta=\frac{|2+1+0|}{3\cdot\sqrt{2}}=\frac{1}{\sqrt{2}}\), so \(\theta=45^\circ\).
Q15
If \(\sin^{-1}x+\cot^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}\), then x is equal to :
- A. \(\frac{2}{\sqrt{5}}\)
- B. \(\frac{1}{2}\)
- C. \(\frac{\sqrt{3}}{2}\)
- D. \(\frac{1}{\sqrt{5}}\)Correct
Explanation. \(\sin^{-1}x=\frac{\pi}{2}-\cot^{-1}\frac{1}{2}=\tan^{-1}\frac{1}{2}\), so \(x=\sin\left(\tan^{-1}\frac{1}{2}\right)=\frac{1}{\sqrt{5}}\).
Q16
If \(\alpha, \beta\) and \(\gamma\) are zeros of \(x^3+px^2+qx+r\) then \(\sum\frac{1}{\alpha}\) is :
- A. \(\frac{q}{r}\)
- B. \(-\frac{q}{r}\)Correct
- C. \(-\frac{q}{p}\)
- D. \(-\frac{p}{r}\)
Explanation. \(\sum\frac{1}{\alpha}=\frac{\sum\alpha\beta}{\alpha\beta\gamma}=\frac{q}{-r}=-\frac{q}{r}\).
Q17
The value of Var(3) is :
- A. 0Correct
- B. 3
- C. Var(3)
- D. 9
Explanation. The variance of a constant is zero, so Var(3) = 0.
Q18
The random variable X has binomial distribution with n = 25 and p = 0.8 then standard deviation of X is :
- A. 3
- B. 6
- C. 2Correct
- D. 4
Explanation. \(\sigma=\sqrt{npq}=\sqrt{25\times0.8\times0.2}=\sqrt{4}=2\).
Q19
If A, B and C are invertible matrices of some order, then which one of the following is not true ?
- A. \(\det A^{-1}=(\det A)^{-1}\)
- B. \(\text{adj } A=|A|A^{-1}\)
- C. \((ABC)^{-1}=C^{-1}B^{-1}A^{-1}\)
- D. \(\text{adj }(AB)=(\text{adj } A)(\text{adj } B)\)Correct
Explanation. The correct rule is \(\text{adj}(AB)=(\text{adj } B)(\text{adj } A)\), so option (d) is not true; the others are standard results.
Q20
A zero of \(x^3+64\) is :
- A. \(4i\)
- B. 0
- C. −4Correct
- D. 4
Explanation. \((-4)^3+64=-64+64=0\), so −4 is a zero.
