12th Standard Mathematics Supplementary 2021: Online Practice Test

Explanation. \(|A| = 6-5 = 1\), so \(A^{-1} = \frac{1}{1}\begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}\).

Explanation. Comparing with \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the centre is \((h, k) = (1, -1)\).

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.

Explanation. \(X^{-1}(7) = \{(3,4), (4,3), (5,2), (6,1)\}\), which has 4 elements.

Explanation. Since \(z\bar{z} = |z|^2 = 1\), \(\bar{z} = \frac{1}{z}\). Then \(\frac{1+z}{1+\frac{1}{z}} = \frac{z(1+z)}{z+1} = z\).

Explanation. Put \(t = \sin x\): the integral is \(\left[\frac{\sin^3 x}{3}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}\).

Explanation. \(f'(x) = 2x \ge 0\) on \([0, \infty)\) (zero only at \(x = 0\)), so \(f\) is increasing there.

Explanation. Volume \(= \left|\begin{vmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & \pi \end{vmatrix}\right| = \pi(2-1) = \pi\).

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.

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).

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.

Explanation. \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) is an ellipse with \(a = 5\). For any point on it, \(PF_1 + PF_2 = 2a = 10\).

Explanation. Normals must be parallel: \(\frac{2}{4} = \frac{-\lambda}{1} = \frac{1}{-\mu}\), so \(\lambda = -\frac{1}{2}\) and \(\mu = -2\).

Explanation. \((-4)^3 + 64 = -64 + 64 = 0\), so \(-4\) is a zero.

Explanation. Separating variables, \(\frac{dy}{y} = -P\,dx\), so \(\log y = -\int P\,dx + \log c\), i.e. \(y = ce^{-\int P\,dx}\).

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 \(\frac{16}{35}\).)

Explanation. \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\) for \(|x| \le 1\); here \(\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}\).

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.

Explanation. \(i^{25} = i^{24} \cdot i = i\), so \((i^{25})^3 = i^3 = -i\).

Explanation. \(V = x^3\), so \(dV = 3x^2\,dx = 3(16)(0.1) = 4.8\) cubic cm.