The 20 one-mark questions from the March 2023 12th Standard Mathematics public exam, in paper order, with the correct option marked and a short explanation.
Q1
A square matrix A of order n has inverse if and only if :
- A. \(\rho(A) > n\)
- B. \(\rho(A) = n\)Correct
- C. \(\rho(A) \neq n\)
- D. \(\rho(A) < n\)
Explanation. A square matrix of order n is invertible exactly when it is non-singular, i.e. when its rank equals n.
Q2
Distance from the origin to the plane \(3x-6y+2z+7=0\) is :
- A. 2
- B. 0
- C. 3
- D. 1Correct
Explanation. Distance = \(\frac{|7|}{\sqrt{9+36+4}} = \frac{7}{7} = 1\).
Q3
If \(3\cos^{-1}x = \cos^{-1}(4x^3-3x)\),
- A. \(x \in \left(\frac{1}{2}, 1\right)\)
- B. \(x \in \left[\frac{1}{2}, 1\right]\)Correct
- C. \(x \in (-\infty, 1]\)
- D. \(x \in \left[\frac{1}{2}, \infty\right)\)
Explanation. With \(x=\cos\theta\), \(\cos^{-1}(\cos 3\theta)=3\theta\) needs \(0 \le 3\theta \le \pi\), i.e. \(0 \le \theta \le \frac{\pi}{3}\), so \(\frac{1}{2} \le x \le 1\).
Q4
The general solution of the differential equation \(\frac{dy}{dx} = \frac{y}{x}\) is :
- A. \(y = kx\)Correct
- B. \(xy = k\)
- C. \(\log y = kx\)
- D. \(y = k\log x\)
Explanation. Separating variables, \(\frac{dy}{y} = \frac{dx}{x}\) gives \(\log y = \log x + \log k\), so \(y = kx\).
Q5
The number of normals that can be drawn from a point to the parabola \(y^2 = 4ax\) is :
- A. 3Correct
- B. 2
- C. 0
- D. 1
Explanation. The normal \(y = mx - 2am - am^3\) is a cubic in m, so at most three normals can be drawn from a point.
Q6
If \(\vec{a}\) and \(\vec{b}\) are parallel vectors then \([\vec{a}, \vec{c}, \vec{b}]\) is equal to :
- A. 1
- B. 2
- C. 0Correct
- D. -1
Explanation. Since \(\vec{b} = \lambda\vec{a}\), the scalar triple product contains two parallel vectors, so it is 0.
Q7
The number of real numbers in \([0, 2\pi]\) satisfying \(\sin^4 x - 2\sin^2 x + 1\) is :
- A. 1
- B. 2Correct
- C. \(\infty\)
- D. 4
Explanation. The paper omits "= 0"; the intended equation is \(\sin^4 x - 2\sin^2 x + 1 = 0\), i.e. \((\sin^2 x - 1)^2 = 0\), so \(\sin^2 x = 1\) giving \(x = \frac{\pi}{2}, \frac{3\pi}{2}\): 2 values.
Q8
Suppose that X takes on one of the values 0, 1, 2. If for some constant k, \(P(X=i) = kP(X=i-1)\) 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. \(P(X=1)=\frac{k}{7}, P(X=2)=\frac{k^2}{7}\); total \(\frac{1+k+k^2}{7}=1\) gives \(k^2+k-6=0\), so \(k=2\) (k must be positive).
Q9
The maximum value of the function \(x^2 e^{-2x}, x > 0\) is :
- A. \(\frac{1}{e^2}\)Correct
- B. \(\frac{1}{e}\)
- C. \(\frac{4}{e^4}\)
- D. \(\frac{1}{2e}\)
Explanation. \(f'(x) = 2xe^{-2x}(1-x) = 0\) gives \(x=1\) (a maximum), so the maximum value is \(f(1) = e^{-2}\).
Q10
The operation * defined by \(a * b = \frac{ab}{7}\) is not a binary operation on :
- A. R
- B. \(Q^+\)
- C. C
- D. ZCorrect
Explanation. For integers, \(\frac{ab}{7}\) need not be an integer (e.g. \(1 * 1 = \frac{1}{7}\)), so * is not closed on Z.
Q11
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 \(= 2\int_0^1 2\sqrt{x}\,dx = 4 \cdot \frac{2}{3} = \frac{8}{3}\).
Q12
Angle between the curves \(y^2 = x\) and \(x^2 = y\) at the origin is :
- A. \(\frac{\pi}{2}\)Correct
- B. \(\tan^{-1}\left(\frac{3}{4}\right)\)
- C. \(\frac{\pi}{4}\)
- D. \(\tan^{-1}\left(\frac{4}{3}\right)\)
Explanation. At the origin, \(y^2 = x\) has the y-axis as tangent and \(x^2 = y\) has the x-axis as tangent, so the angle is \(\frac{\pi}{2}\).
Q13
\(|\text{adj}(\text{adj}A)| = |A|^{16}\), then the order of the square matrix A is :
- A. 2
- B. 3
- C. 5Correct
- D. 4
Explanation. \(|\text{adj}(\text{adj}A)| = |A|^{(n-1)^2}\), so \((n-1)^2 = 16\) and \(n = 5\).
Q14
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 \(i^4 + (-i)^4 = 1 + 1 = 2\).
Q15
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 \(\bar{z} = \frac{1}{z}\), \(\frac{1+z}{1+\frac{1}{z}} = \frac{z(1+z)}{z+1} = z\).
Q16
The abscissa of the point on the curve \(f(x) = \sqrt{8-2x}\) at which the slope of the tangent is \(-0.25\) ?
- A. -2
- B. -8
- C. 0
- D. -4Correct
Explanation. \(f'(x) = \frac{-1}{\sqrt{8-2x}} = -\frac{1}{4}\) gives \(8-2x = 16\), so \(x = -4\).
Q17
The value of \(\int_0^{\pi/3} \tan x\,dx\) is :
- A. \(-\log 2\)
- B. \(\log 2\)Correct
- C. \(-\log 3\)
- D. \(\log 3\)
Explanation. \(\int_0^{\pi/3} \tan x\,dx = [\log \sec x]_0^{\pi/3} = \log 2 - \log 1 = \log 2\).
Q18
The number of positive zeros of the polynomial \(\sum_{r=0}^{n} {}^nC_r (-1)^r x^r\) is :
- A. \(< n\)
- B. 0
- C. r
- D. nCorrect
Explanation. The polynomial is \((1-x)^n\), whose only zero \(x = 1\) is positive with multiplicity n (the coefficients also change sign n times); so there are n positive zeros.
Q19
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 \(\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\).
Q20
Area of the greatest rectangle inscribed in the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is :
- A. \(\sqrt{ab}\)
- B. 2abCorrect
- C. \(\frac{a}{b}\)
- D. ab
Explanation. With vertex \((a\cos\theta, b\sin\theta)\), area \(= 4ab\sin\theta\cos\theta = 2ab\sin 2\theta\), which is greatest at \(\theta = \frac{\pi}{4}\): \(2ab\).
