TN Online TestSamacheer Kalvi practice

12th Standard Mathematics Supplementary 2021: MCQs with Answers

20 MCQs 90 marks 180 minutes Paper code 4712
Share this chapter: Telegram

The 20 one-mark questions from the Supplementary 2021 12th Standard Mathematics public exam, in paper order, with the correct option marked and a short explanation.

Answer key at a glance

Take online test →
Q1
The inverse of \(\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}\) is :
  • A. \(\begin{bmatrix} 3 & -1 \\ -5 & -3 \end{bmatrix}\)
  • B. \(\begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}\)Correct
  • C. \(\begin{bmatrix} -3 & 5 \\ 1 & -2 \end{bmatrix}\)
  • D. \(\begin{bmatrix} -2 & 5 \\ 1 & -3 \end{bmatrix}\)
Explanation. \(|A| = 6-5 = 1\), so \(A^{-1} = \frac{1}{1}\begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}\).
Q2
The centre of the hyperbola \(\frac{(x-1)^2}{16} - \frac{(y+1)^2}{25} = 1\) is :
  • A. \(\left(\frac{1}{2}, -\frac{1}{2}\right)\)
  • B. \((-1, 1)\)
  • C. \((1, -1)\)Correct
  • D. \((0, 0)\)
Explanation. Comparing with \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the centre is \((h, k) = (1, -1)\).
Q3
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.
Q4
A pair of dice numbered 1, 2, 3, 4, 5, 6 of a six sided die and 1, 2, 3, 4 of a four sided die is rolled and the sum is determined. If the random variable X denote the sum, then the number of elements in the inverse image of 7 is :
  • A. 3
  • B. 1
  • C. 4Correct
  • D. 2
Explanation. \(X^{-1}(7) = \{(3,4), (4,3), (5,2), (6,1)\}\), which has 4 elements.
Q5
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 \(z\bar{z} = |z|^2 = 1\), \(\bar{z} = \frac{1}{z}\). Then \(\frac{1+z}{1+\frac{1}{z}} = \frac{z(1+z)}{z+1} = z\).
Chapter: Complex Numbers
Q6
The value of \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos x \, dx\) is :
  • A. 0
  • B. \(\frac{3}{2}\)
  • C. \(\frac{2}{3}\)Correct
  • D. \(\frac{1}{2}\)
Explanation. Put \(t = \sin x\): the integral is \(\left[\frac{\sin^3 x}{3}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}\).
Q7
The function \(f(x) = x^2\), in the interval \([0, \infty)\) is :
  • A. cannot be determined
  • B. increasing functionCorrect
  • C. increasing and decreasing function
  • D. decreasing function
Explanation. \(f'(x) = 2x \ge 0\) on \([0, \infty)\) (zero only at \(x = 0\)), so \(f\) is increasing there.
Q8
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 \(= \left|\begin{vmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & \pi \end{vmatrix}\right| = \pi(2-1) = \pi\).
Q9
In the set \(\mathbb{R}\) of real numbers '\(*\)' is defined as follows. Which one of the following is not a binary operation on \(\mathbb{R}\) ?
  • A. \(a * b = a\)
  • B. \(a * b = \min(a, b)\)
  • C. \(a * b = a^b\)Correct
  • D. \(a * b = \max(a, b)\)
Explanation. \(a^b\) need not be a real number, e.g. \((-1)^{\frac{1}{2}}\) is not real, so \(a * b = a^b\) is not closed on \(\mathbb{R}\). The other three always give real numbers.
Q10
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).
Q11
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 values of a and b ?
  • A. 7 and 19
  • B. 0 and 12
  • C. 16 and 24Correct
  • D. 5 and 17
Explanation. \(\int_a^b \frac{1}{12}\,dx = 1\) gives \(b - a = 12\). For 16 and 24, \(b - a = 8\), so these cannot be the values.
Q12
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. \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) is an ellipse with \(a = 5\). For any point on it, \(PF_1 + PF_2 = 2a = 10\).
Q13
If the planes \(\vec{r} \cdot (2\hat{i} - \lambda\hat{j} + \hat{k}) = 3\) and \(\vec{r} \cdot (4\hat{i} + \hat{j} - \mu\hat{k}) = 5\) are parallel, then the values of \(\lambda\) and \(\mu\) are respectively :
  • A. \(-\frac{1}{2}, -2\)Correct
  • B. \(\frac{1}{2}, -2\)
  • C. \(\frac{1}{2}, 2\)
  • D. \(-\frac{1}{2}, 2\)
Explanation. Normals must be parallel: \(\frac{2}{4} = \frac{-\lambda}{1} = \frac{1}{-\mu}\), so \(\lambda = -\frac{1}{2}\) and \(\mu = -2\).
Q14
A zero of \(x^3 + 64\) is :
  • A. \(4i\)
  • B. 0
  • C. \(-4\)Correct
  • D. 4
Explanation. \((-4)^3 + 64 = -64 + 64 = 0\), so \(-4\) is a zero.
Q15
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, \(\frac{dy}{y} = -P\,dx\), so \(\log y = -\int P\,dx + \log c\), i.e. \(y = ce^{-\int P\,dx}\).
Q16
\(\int_0^{\frac{\pi}{2}} \sin^7 x \, dx =\)
  • A. \(\frac{\pi}{2}\)
  • B. \(\int_0^{\frac{\pi}{2}} \cos^7 x \, dx\)Correct
  • C. 0
  • D. 1
Explanation. Using \(\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx\) with \(a = \frac{\pi}{2}\), \(\sin^7 x\) becomes \(\cos^7 x\). (Both equal \(\frac{16}{35}\).)
Q17
The value of \(\sin^{-1}\left(\frac{1}{2}\right) + \cos^{-1}\left(\frac{1}{2}\right)\) is :
  • A. 0
  • B. \(\frac{\pi}{2}\)Correct
  • C. \(\frac{\pi}{3}\)
  • D. \(\pi\)
Explanation. \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\) for \(|x| \le 1\); here \(\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}\).
Q18
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 \(\text{adj}(AB) = (\text{adj } B)(\text{adj } A)\) (reversal law), so option (4) is not true; the other three are standard results.
Q19
The value of the complex number \((i^{25})^3\) is equal to :
  • A. 1
  • B. \(i\)
  • C. \(-i\)Correct
  • D. \(-1\)
Explanation. \(i^{25} = i^{24} \cdot i = i\), so \((i^{25})^3 = i^3 = -i\).
Chapter: Complex Numbers
Q20
If we measure the side of a cube to be 4 cm with an error of 0.1 cm, then the error in calculation of the volume is (in cubic cm) :
  • A. 2
  • B. 0.4
  • C. 4.8Correct
  • D. 0.45
Explanation. \(V = x^3\), so \(dV = 3x^2\,dx = 3(16)(0.1) = 4.8\) cubic cm.
Take online test → All previous year papers

About this paper

These are the Part I one-mark multiple-choice questions from the Supplementary 2021 12th Standard Mathematics public examination conducted by the Tamil Nadu Directorate of Government Examinations. The answers and explanations are prepared by TN Online Test for revision. Each question links to the textbook chapter it comes from, so you can go back to that chapter's notes and MCQs.

Other years

2026 · March 2026 2025 · March 2025 2024 · March 2024 2023 · March 2023 2022 · May 2022 2020 · March 2020