12th Standard Mathematics: Previous Year Mixed Test

Explanation. \(BB^T=A^{-1}A^T\,A\,(A^{-1})^T=A^{-1}(AA^T)(A^T)^{-1}=(A^{-1}A)(A^T(A^T)^{-1})=I_3\), using \(A^TA=AA^T\).

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.

Explanation. Let \(z=re^{i\theta}\). The imaginary part of \(z+\frac{1}{z}\) is \((r-\frac{1}{r})\sin\theta=0\). Since z is not real, \(\sin\theta\neq0\), so \(r=1\).

Explanation. \((\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})^3=\cos\pi+i\sin\pi=-1\), so the four values are the roots of \(w^4+1=0\). Their product is the constant term 1.

Explanation. The leading term of f is \(ax^m\); substituting g (leading term \(bx^n\)) gives leading term \(ab^m x^{mn}\), so the degree is mn.

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: \(-\frac{\pi}{4}=\frac{3\pi}{4}-\pi\).

Explanation. A circle is an ellipse with a = b, so \(e=\sqrt{1-\frac{b^2}{a^2}}=0\).

Explanation. Three coplanar vectors have scalar triple product zero, so \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\).

Explanation. The foot of the perpendicular is the midpoint of AA': \(\left(\frac{1+3}{2},\frac{2+6}{2},\frac{3+11}{2}\right)=(2,4,7)\). Check: 2 + 8 + 28 = 38.

Explanation. \(D^2=(x-6)^2+y^2=(x-6)^2+x^2-4\). \(\frac{d}{dx}D^2=2(x-6)+2x=0\) gives x = 3, so \(y^2=5\); the point is \((3,\sqrt{5})\).

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.

Explanation. If \(y=x^{1/5}\), then \(\frac{dy}{y}=\frac{1}{5}\frac{dx}{x}\), so the percentage error of the fifth root is \(\frac{1}{5}\) times that of x.

Explanation. A harmonic function satisfies Laplace's equation \(u_{xx}+u_{yy}=0\) at every point of A. (The paper prints \(u:A\to\mathbb{R}^2\); the textbook has \(\mathbb{R}\), which does not change the answer.)

Explanation. By the fundamental theorem of calculus, \(\frac{d}{dx}\int_0^x t\cos t\,dt=x\cos x\).

Explanation. Put \(u=x^2\), \(du=2x\,dx\), so \(\frac{dx}{x}=\frac{du}{2u}\); limits 1 to 9. The integral becomes \(\frac{1}{2}\int_1^9\frac{e^{\sin u}}{u}du=\frac{1}{2}[f(9)-f(1)]\), so a = 9.

Explanation. Separate variables: \(\frac{2\,dy}{y+3}=\frac{dx}{x}\), so \(2\log(y+3)=\log x+\log C\), i.e. \((y+3)^2=Cx\), a family of parabolas.

Explanation. The amount decreases: \(\frac{dP}{dt}=-kP\) (k > 0), so \(\frac{dP}{P}=-k\,dt\) and \(P=Ce^{-kt}\).

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.

Explanation. X is uniform on (0, l): \(E(X)=\frac{l}{2}\), \(E(X^2)=\frac{l^2}{3}\), so \(Var(X)=\frac{l^2}{3}-\frac{l^2}{4}=\frac{l^2}{12}\).

Explanation. The dual interchanges \(\vee\) and \(\wedge\) and keeps \(\neg\): \(\neg(p\wedge q)\wedge[p\wedge(p\vee\neg r)]\).

Explanation. In \(\mathbb{N}\), \(2-3=-1\notin\mathbb{N}\), so subtraction is not closed there; it is closed in \(\mathbb{R}, \mathbb{Q}, \mathbb{Z}\).

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).

Explanation. \(|A^{-1}|=\frac{1}{|A|}=\frac{1}{5}\).

Explanation. At maximum height the velocity is zero: \(\frac{dx}{dt}=80-32t=0\), so \(t=2.5\) seconds.

Explanation. Only the first derivative occurs, so the order is 1. Removing the radical: \(\frac{dy}{dx}=\left(4\frac{dy}{dx}+7x\right)^2\), whose highest power of \(\frac{dy}{dx}\) is 2, so the degree is 2.

Explanation. Multiplying by A: \(\lambda I=A^2\). \(A^2=\begin{bmatrix}19 & 0\\ 0 & 19\end{bmatrix}=19I\), so \(\lambda=19\).

Explanation. Integrating, \(y=x^3+C\). At \((-1,1)\): \(1=-1+C\), so \(C=2\) and \(y=x^3+2\).

Explanation. We need \(0\le\sqrt{x-1}\le 1\), i.e. \(0\le x-1\le 1\), so \(x\in[1, 2]\).

Explanation. \(\frac{\partial u}{\partial x}=e^{x^2+y^2}\cdot 2x=2xu\).

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.

Explanation. \(\left(\frac{1+i}{\sqrt{2}}\right)^2=\frac{1+2i-1}{2}=i\), so the square roots are \(\pm\frac{1}{\sqrt{2}}(1+i)\).

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 \(1+i\).

Explanation. \(9\binom{6}{4}p^4q^2=\binom{6}{2}p^2q^4\) gives \(9p^2=q^2\), so \(q=3p\). With \(p+q=1\), \(p=0.25\).

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: \(\frac{\pi}{2}\).

Explanation. \(y''=6(x-1)\) is zero at \(x=1\) and changes sign there; \(y(1)=0\), so the point is \((1, 0)\).

Explanation. \(\int\frac{dx}{\sqrt{4-9x^2}}=\frac{1}{3}\sin^{-1}\frac{3x}{2}\). From 0 to \(\frac{2}{3}\): \(\frac{1}{3}\cdot\frac{\pi}{2}=\frac{\pi}{6}\).

Explanation. \(V=\pi\int_0^a(ax-x^2)\,dx=\pi\left(\frac{a^3}{2}-\frac{a^3}{3}\right)=\frac{\pi a^3}{6}\).

Explanation. Angle FBF' = 90° means angle FBO = 45°, so OF = OB, i.e. \(ae=b\). Then \(a^2e^2=a^2(1-e^2)\), so \(e^2=\frac{1}{2}\) and \(e=\frac{1}{\sqrt{2}}\).

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

Explanation. \(dg=g'(x)\,dx=\frac{f'(x)}{f(x)}\,dx\) by the chain rule.

Explanation. Latus rectum is x = 1. Area = \(2\int_0^1 2\sqrt{x}\,dx = 4\cdot\frac{2}{3} = \frac{8}{3}\).

Explanation. Put \(x=a\sin\theta\): integral = \(a^4\int_0^{\pi/2}\cos^4\theta\,d\theta = a^4\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3\pi a^4}{16}\).

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

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 \(|z_1+z_2+z_3|=2\).

Explanation. There are 3 simple statements p, q, r, so the truth table has \(2^3=8\) rows.

Explanation. Three coplanar vectors have scalar triple product zero, so \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\).

Explanation. \(y'=Ae^x-Be^{-x}\) and \(y''=Ae^x+Be^{-x}=y\), so \(\frac{d^2y}{dx^2}-y=0\).

Explanation. As \(x\to\pm\infty\), \(\frac{1}{x}\to 0\), so the horizontal asymptote is \(y=0\).

Explanation. \(f'(x)=\frac{(x+1)-x}{(x+1)^2}=\frac{1}{(x+1)^2}\), so \(df=\frac{1}{(x+1)^2}dx\).

Explanation. Taking modulus squared on both sides: \(|1+i|^2|1+2i|^2\cdots|1+ni|^2 = 2\cdot5\cdot10\cdots(1+n^2) = x^2+y^2\).

Explanation. \(f(0)=f(3)=0\). \(f'(c)=3c^2-6c=0\) gives c = 0 or 2; c = 2 lies in (0, 3).

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.

Explanation. \((A^T)^{-1}=(A^{-1})^T=\begin{bmatrix}5 & -2\\ 3 & -1\end{bmatrix}\).

Explanation. Direction (2, 1, −2), normal (1, 1, 0): \(\sin\theta=\frac{|2+1+0|}{3\cdot\sqrt{2}}=\frac{1}{\sqrt{2}}\), so \(\theta=45^\circ\).

Explanation. \(\sin^{-1}x=\frac{\pi}{2}-\cot^{-1}\frac{1}{2}=\tan^{-1}\frac{1}{2}\), so \(x=\sin\left(\tan^{-1}\frac{1}{2}\right)=\frac{1}{\sqrt{5}}\).

Explanation. \(\sum\frac{1}{\alpha}=\frac{\sum\alpha\beta}{\alpha\beta\gamma}=\frac{q}{-r}=-\frac{q}{r}\).

Explanation. The variance of a constant is zero, so Var(3) = 0.

Explanation. \(\sigma=\sqrt{npq}=\sqrt{25\times0.8\times0.2}=\sqrt{4}=2\).

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.

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

Explanation. A square matrix of order n is invertible exactly when it is non-singular, i.e. when its rank equals n.

Explanation. Distance = \(\frac{|7|}{\sqrt{9+36+4}} = \frac{7}{7} = 1\).

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 \(\frac{1}{2} \le x \le 1\).

Explanation. Separating variables, \(\frac{dy}{y} = \frac{dx}{x}\) gives \(\log y = \log x + \log k\), so \(y = kx\).

Explanation. The normal \(y = mx - 2am - am^3\) is a cubic in m, so at most three normals can be drawn from a point.

Explanation. Since \(\vec{b} = \lambda\vec{a}\), the scalar triple product contains two parallel vectors, so it is 0.

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.

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).

Explanation. \(f'(x) = 2xe^{-2x}(1-x) = 0\) gives \(x=1\) (a maximum), so the maximum value is \(f(1) = e^{-2}\).

Explanation. For integers, \(\frac{ab}{7}\) need not be an integer (e.g. \(1 * 1 = \frac{1}{7}\)), so * is not closed on Z.

Explanation. Latus rectum is \(x=1\). Area \(= 2\int_0^1 2\sqrt{x}\,dx = 4 \cdot \frac{2}{3} = \frac{8}{3}\).

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 \(\frac{\pi}{2}\).

Explanation. \(|\text{adj}(\text{adj}A)| = |A|^{(n-1)^2}\), so \((n-1)^2 = 16\) and \(n = 5\).

Explanation. \(\left(\frac{1+i}{\sqrt{2}}\right)^2 = i\) and \(\left(\frac{1-i}{\sqrt{2}}\right)^2 = -i\), so the sum is \(i^4 + (-i)^4 = 1 + 1 = 2\).

Explanation. Since \(\bar{z} = \frac{1}{z}\), \(\frac{1+z}{1+\frac{1}{z}} = \frac{z(1+z)}{z+1} = z\).

Explanation. \(f'(x) = \frac{-1}{\sqrt{8-2x}} = -\frac{1}{4}\) gives \(8-2x = 16\), so \(x = -4\).

Explanation. \(\int_0^{\pi/3} \tan x\,dx = [\log \sec x]_0^{\pi/3} = \log 2 - \log 1 = \log 2\).

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.

Explanation. The principal range of \(\sin^{-1}\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and \(\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\).

Explanation. With vertex \((a\cos\theta, b\sin\theta)\), area \(= 4ab\sin\theta\cos\theta = 2ab\sin 2\theta\), which is greatest at \(\theta = \frac{\pi}{4}\): \(2ab\).

Explanation. Total probability must be 1: \(\int_0^a 2x\,dx=a^2=1\), so \(a=1\) (a > 0).

Explanation. A cumulative distribution function F(x) is non-decreasing, not decreasing; the other three properties are true.

Explanation. \(f'(x)=\frac{(x+1)-x}{(x+1)^2}=\frac{1}{(x+1)^2}\), so \(df=\frac{1}{(x+1)^2}\,dx\).

Explanation. Using \(\int_0^a f(x)dx=\int_0^a f(a-x)dx\): \(\int_0^1(1-x)x^{99}dx=\frac{1}{100}-\frac{1}{101}=\frac{1}{10100}\).

Explanation. \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\) and \(\frac{\pi}{6}\in[0,\pi]\), so the principal value is \(\frac{\pi}{6}\).

Explanation. \(\lambda A^{-1}=A\Rightarrow A^2=\lambda I\). Here \(A^2=\begin{bmatrix}19&0\\0&19\end{bmatrix}=19I\), so \(\lambda=19\).

Explanation. \(\sum\frac{1}{\alpha}=\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\frac{q}{-r}=-\frac{q}{r}\).

Explanation. Taking modulus squared on both sides: \(|1+i|^2|1+2i|^2\cdots|1+ni|^2=2\cdot5\cdot10\cdots(1+n^2)=x^2+y^2\).

Explanation. \(|3-x|\ge 0\) with equality at \(x=3\), so the minimum value is \(0+9=9\).

Explanation. Any four consecutive powers of \(i\) add to 0 \((i-1-i+1=0)\); 12 terms form three such groups, so the sum is 0.

Explanation. Coplanar means the scalar triple product is 0: \(\begin{vmatrix}2&-1&3\\3&2&1\\1&m&4\end{vmatrix}=2(8-m)+11+3(3m-2)=21+7m=0\), so \(m=-3\).

Explanation. \((x+3)^2+(y+4)^2=9\) gives \(x^2+y^2+6x+8y+16=0\).

Explanation. Separating variables: \(\frac{dy}{y}=-p\,dx\Rightarrow \log y=-\int p\,dx+\log c\), so \(y=ce^{-\int p\,dx}\).

Explanation. Using \(\int_0^\infty e^{-ax}x^n\,dx=\frac{n!}{a^{n+1}}\): \(\frac{2!}{3^3}=\frac{2}{27}\).

Explanation. \(y''=6(x-1)\) is 0 at \(x=1\) and changes sign there; \(y(1)=0\), so the point is \((1,0)\).

Explanation. Direction ratios \((2,1,-2)\) and \((4,-4,2)\): dot product \(8-4-4=0\), so the lines are perpendicular.

Explanation. The product of two natural numbers is a natural number, but \(2-3\) and \(2\div3\) are not in N.

Explanation. Each cofactor of \(\lambda A\) is an \((n-1)\)-order minor, so \(\text{Adj}(\lambda A)=\lambda^{n-1}\,\text{Adj}\,A\), not \(\lambda^n\).

Explanation. \(e^{\int P\,dx}=\sin x\Rightarrow\int P\,dx=\log\sin x\Rightarrow P=\cot x\).

Explanation. Comparing with \(x^2=4ay\), \(4a=24\), so the latus rectum length is \(4a=24\).

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.

Explanation. Treating y as constant, \(\frac{\partial u}{\partial x}=e^{x^2+y^2}\cdot 2x=2xu\).

Explanation. In \(\mathbb{N}\), for example \(2-5=-3\notin\mathbb{N}\), so subtraction is not closed; it is closed in \(\mathbb{Z},\mathbb{Q},\mathbb{R}\).

Explanation. \(\int_0^{\pi}\sin^4x\,dx=2\int_0^{\pi/2}\sin^4x\,dx=2\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}=\frac{3\pi}{8}\).

Explanation. By the fundamental theorem of algebra, a polynomial equation of degree n has exactly n roots (counting multiplicity); they need not be distinct or real.

Explanation. By the Rouche-Capelli theorem, equal ranks mean the system is consistent; whether the solution is unique depends on comparing the rank with the number of unknowns, which is not given.

Explanation. \(x^2=8\left(y-\frac{1}{8}\right)\), so the vertex is \(\left(0,\frac{1}{8}\right)\).

Explanation. \(\cos^{-1}x+\cos^{-1}y=\left(\frac{\pi}{2}-\sin^{-1}x\right)+\left(\frac{\pi}{2}-\sin^{-1}y\right)=\pi-\frac{2\pi}{3}=\frac{\pi}{3}\).

Explanation. Any 4 consecutive powers of \(i\) add to 0. \(\sum_{n=1}^{13}i^n=i^{13}=i\) and \(\sum_{n=1}^{13}i^{n-1}=i^{12}=1\), so the sum is \(1+i\). (The paper prints the index as \(i=1\); it means \(n=1\).)

Explanation. Every point is \((s,t,0)\) with z = 0 and s, t free, so the equation represents the xoy plane.

Explanation. \((x-h)^2+(y-k)^2=a^2\) has three arbitrary constants, so eliminating them gives a differential equation of order 3.

Explanation. The complex number 0 has modulus 0 and no direction, so its argument is not defined.

Explanation. \(\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}=\tan^{-1}\frac{17/36}{17/18}=\tan^{-1}\frac{1}{2}\). Since \(\cos\left(2\tan^{-1}\frac{1}{2}\right)=\frac{1-1/4}{1+1/4}=\frac{3}{5}\), option (2) is also equal to it; the paper has two correct options.

Explanation. Velocity \(v=s'(t)=6t-2=0\) gives \(t=\frac{1}{3}\).

Explanation. \(P=2\left(x+\frac{100}{x}\right)\); \(P'=0\) gives \(x=10\), a square of side 10 m, so \(P=40\) m.

Explanation. \(\sigma=\sqrt{npq}=\sqrt{25\times0.8\times0.2}=\sqrt{4}=2\).

Explanation. For a circle the coefficients of \(x^2\) and \(y^2\) are equal, so \(b=3\): \(x^2+y^2+4x-6y+3=0\). Radius \(=\sqrt{4+9-3}=\sqrt{10}\).

Explanation. Write the second plane as \(x+2y+3z+\frac{7}{2}=0\). Distance \(=\frac{|7-7/2|}{\sqrt{14}}=\frac{7}{2\sqrt{14}}=\frac{\sqrt{7}}{2\sqrt{2}}\).

Explanation. \((AB)^{-1}=B^{-1}A^{-1}\), so \(B^{-1}=(AB)^{-1}A\). Here \(A=(A^{-1})^{-1}=\begin{bmatrix}3 & 1\\ 2 & 1\end{bmatrix}\), giving \(B^{-1}=\begin{bmatrix}2 & -5\\ -3 & 8\end{bmatrix}\).

Explanation. \(\int\frac{dx}{\sqrt{4-9x^2}}=\frac{1}{3}\sin^{-1}\frac{3x}{2}\); from 0 to \(\frac{2}{3}\) this is \(\frac{1}{3}\cdot\frac{\pi}{2}=\frac{\pi}{6}\).

Explanation. Multiplying by \(\frac{dy}{dx}\) gives \(\left(\frac{dy}{dx}\right)^2+1=0\): the highest derivative is of first order and its power is 2, so order 1, degree 2.