Worked out from the 140 one-mark questions of the 2020–2026 public exams. The table shows how many board questions came from each chapter; below it are the questions the board asked more than once, with the answer.
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 \(\frac{2}{3}\): \(\frac{1}{3}\cdot\frac{\pi}{2}=\frac{\pi}{6}\).
If \(\rho(A)=\rho([A|B])\), then the system \(AX=B\) of linear equations is :
- A. consistent and has infinitely many solutions
- B. consistent and has a unique solution
- C. inconsistent
- D. consistentCorrect
Explanation. By the Rouché–Capelli theorem, equal ranks mean the system is consistent. Whether the solution is unique or infinite depends on the number of unknowns, which is not given.
If \(x<0\), then \(\tan^{-1}\left(\frac{1}{x}\right)\) is equal to :
- A. \(-\pi+\cot^{-1}(x)\)Correct
- B. \(\tan^{-1}(x)\)
- C. \(-\pi+\tan^{-1}x\)
- D. \(\cot^{-1}(x)\)
Explanation. For \(x<0\), \(\cot^{-1}x\in(\frac{\pi}{2},\pi)\) while \(\tan^{-1}\frac{1}{x}\in(-\frac{\pi}{2},0)\), so \(\tan^{-1}\frac{1}{x}=\cot^{-1}x-\pi\). Check x = −1: \(-\frac{\pi}{4}=\frac{3\pi}{4}-\pi\).
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 \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\).
The value of 'c' satisfied by the Rolle's theorem for the function \(f(x)=x^3-3x^2,\ x\in[0,3]\) 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; the value in (0, 3) is c = 2.
If the function \(f(x)=\frac{1}{12}\) for \(a<x<b\), represents a probability density function of a continuous random variable X, then which of the following cannot be the value of a and b?
- A. 7 and 19
- B. 0 and 12
- C. 16 and 24Correct
- D. 5 and 17
Explanation. Total probability \(\frac{b-a}{12}=1\) needs b − a = 12. For 16 and 24, b − a = 8, so it cannot be a pdf.
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 \(\mathbb{R}, \mathbb{Q}, \mathbb{Z}\).
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).
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 \(\lambda=19\).
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. \(\frac{\partial u}{\partial x}=e^{x^2+y^2}\cdot 2x=2xu\).
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.
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 \(1+i\).
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: \(\frac{\pi}{2}\).
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 \((1, 0)\).
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 \(=\left|\begin{vmatrix}1 & 1 & 0\\ 1 & 2 & 0\\ 1 & 1 & \pi\end{vmatrix}\right|=\pi(2-1)=\pi\).
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 = \(2\int_0^1 2\sqrt{x}\,dx = 4\cdot\frac{2}{3} = \frac{8}{3}\).
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, \(PF_1+PF_2=2a=10\).
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 \(df=\frac{1}{(x+1)^2}dx\).
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: \(|1+i|^2|1+2i|^2\cdots|1+ni|^2 = 2\cdot5\cdot10\cdots(1+n^2) = x^2+y^2\).
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. \(\sum\frac{1}{\alpha}=\frac{\sum\alpha\beta}{\alpha\beta\gamma}=\frac{q}{-r}=-\frac{q}{r}\).
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. \(\sigma=\sqrt{npq}=\sqrt{25\times0.8\times0.2}=\sqrt{4}=2\).
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.
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.
The value of \(\left(\frac{1+i}{\sqrt{2}}\right)^8 + \left(\frac{1-i}{\sqrt{2}}\right)^8\) is :
- A. 8
- B. 4
- C. 2Correct
- D. 6
Explanation. \(\left(\frac{1+i}{\sqrt{2}}\right)^2 = i\) and \(\left(\frac{1-i}{\sqrt{2}}\right)^2 = -i\), so the sum is \(i^4 + (-i)^4 = 1 + 1 = 2\).
If \(|z| = 1\), then the value of \(\frac{1+z}{1+\bar{z}}\) is :
- A. \(\frac{1}{z}\)
- B. \(z\)Correct
- C. 1
- D. \(\bar{z}\)
Explanation. Since \(\bar{z} = \frac{1}{z}\), \(\frac{1+z}{1+\frac{1}{z}} = \frac{z(1+z)}{z+1} = z\).
The Principal value of \(\sin^{-1}\left(\frac{-1}{2}\right)\) is :
- A. \(\frac{-\pi}{6}\)Correct
- B. 0
- C. \(\frac{-\pi}{2}\)
- D. \(\frac{\pi}{2}\)
Explanation. The principal range of \(\sin^{-1}\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and \(\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\).
The solution of \(\frac{dy}{dx}+p(x)y=0\) is :
- A. \(x=ce^{-\int p\,dy}\)
- B. \(y=ce^{\int p\,dx}\)
- C. \(x=ce^{\int p\,dy}\)
- D. \(y=ce^{-\int p\,dx}\)Correct
Explanation. Separating variables: \(\frac{dy}{y}=-p\,dx\Rightarrow \log y=-\int p\,dx+\log c\), so \(y=ce^{-\int p\,dx}\).
The order and degree of the differential equation \(\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^{\frac{1}{3}} + x^{\frac{1}{4}} = 0\) are :
- A. 2, 6
- B. 2, 3Correct
- C. 2, 4
- D. 3, 3
Explanation. The highest derivative is \(\frac{d^2y}{dx^2}\), so order 2. Writing \(\left(\frac{dy}{dx}\right)^{\frac{1}{3}} = -\left(\frac{d^2y}{dx^2} + x^{\frac{1}{4}}\right)\) and cubing gives \(\left(\frac{d^2y}{dx^2} + x^{\frac{1}{4}}\right)^3 = -\frac{dy}{dx}\), so the degree is 3.
The position of a particle 's' moving at any time t is given by \(s(t) = 5t^2 - 2t - 8\). The time at which the particle is at rest, is :
- A. 1
- B. 0
- C. 3
- D. \(\frac{1}{3}\)Correct
Explanation. At rest, \(v = s'(t) = 0\). As printed, \(10t - 2 = 0\) gives \(t = \frac{1}{5}\), which is not an option. The paper has a misprint: the textbook question is \(s(t) = 3t^2 - 2t - 8\), for which \(6t - 2 = 0\) gives \(t = \frac{1}{3}\) (intended answer).
