Mathematics

(a) 2, 3

Equal matrices have equal entries: $2x=4\Rightarrow x=2$ and $x+y=5\Rightarrow y=3$.

(b) parallel

Slopes are $\frac73$ and $\frac{14}{6}=\frac73$, equal. The constant terms do not follow the same ratio ($7x-3y=5$ becomes $14x-6y=10\ne3$), so the lines are parallel and not coincident.

(b) $2a^2$

$|\vec a\times\hat i|^2=a^2-a_x^2$, and similarly for $\hat j,\hat k$. Sum $=3a^2-(a_x^2+a_y^2+a_z^2)=3a^2-a^2=2a^2$.

(d) $-\frac32\leq x\leq\frac32$

$(9-4x^2)^{1/2}=3\left(1-\frac{4x^2}{9}\right)^{1/2}$, valid when $\left|\frac{4x^2}{9}\right|<1$, i.e. $|x|<\frac32$.

(d) $\frac{n\pi}{4}$

$\sin9\theta-\sin\theta=2\cos5\theta\sin4\theta=0$. From $\sin4\theta=0$, $\theta=\frac{n\pi}{4}$ (the other family $\cos5\theta=0$ gives $\theta=\frac{(2n+1)\pi}{10}$), so the option that matches is $\frac{n\pi}{4}$.

(c) 3

$a^4=81=3^4$, so $a=3$ (the base must be positive).

(b) $-2:3$

The yz-plane is $x=0$. If the ratio is $k:1$, then $\frac{6k+4}{k+1}=0\Rightarrow k=-\frac23$, so the ratio is $-2:3$ (external division).

(a) 0

$\sin^3x\cos^2x$ is an odd function, and the limits are symmetric, so the integral is $0$.

(a) 1

Put $t=\frac\pi2-x$: $\cos x=\sin t$, so the limit is $\lim_{t\to0}\frac{\sin t}{t}=1$.

(b) $k=6$

One root is $0$, so substituting $x=0$ gives $6-k=0$, i.e. $k=6$.

(a) 9

For $y^2=12x$, $a=3$. A normal with slope $m$ is $y=mx-2am-am^3$. The line $x+y=k$ has $m=-1$, so $k = 2(3)+3 = 9.$

(c) $\frac{y(x\log y-y)}{x(y\log x-x)}$

Take logs: $y\log x=x\log y$. Differentiating, $\frac{dy}{dx}\log x+\frac yx=\log y+\frac xy\frac{dy}{dx}$ $\frac{dy}{dx}\left(\log x-\frac xy\right)=\log y-\frac yx$ $\frac{dy}{dx}=\frac{y(x\log y-y)}{x(y\log x-x)}.$

(b) $-1$

The bisectors of $x^2-2pxy-y^2=0$ satisfy $\frac{x^2-y^2}{2}=\frac{xy}{-p}$, i.e. $x^2-2\cdot\left(-\frac1p\right)xy\cdot(-1)\ldots$ which simplifies to $x^2+\frac{2}{p}xy-y^2=0$. For this to be the other pair, $-2q=\frac2p$, so $pq=-1$.

(c) an isosceles

Given $\frac{\cos A}{\cos B}=\frac ab=\frac{\sin A}{\sin B}$, so $\tan A=\tan B$, hence $A=B$ and the triangle is isosceles.

(d) 0, 2, 3

The denominator is intended to be $x^3-5x^2+6x=x(x-2)(x-3)$, which is zero at $x=0,2,3$. The function is discontinuous at these points.

(b) $\pm\frac23$

Let $\theta=\sin^{-1}x$. Then $\cos2\theta=1-2\sin^2\theta=1-2x^2=\frac19$, so $x^2=\frac49$ and $x=\pm\frac23$.

(a) 729

$2+5\omega+2\omega^2=2(1+\omega+\omega^2)+3\omega=3\omega$. Therefore $(3\omega)^6=729\,\omega^6=729.$

(c) $\frac x\alpha+\frac y\beta+\frac z\gamma=3$

Let the plane be $\frac xa+\frac yb+\frac zc=1$ with $A(a,0,0)$, $B(0,b,0)$, $C(0,0,c)$. The centroid is $\left(\frac a3,\frac b3,\frac c3\right)=(\alpha,\beta,\gamma)$, so $a=3\alpha$, etc. The plane becomes $\frac{x}{3\alpha}+\frac{y}{3\beta}+\frac{z}{3\gamma}=1$, i.e. $\frac x\alpha+\frac y\beta+\frac z\gamma=3$.

(c) $8h^2=9ab$

Let the slopes be $m$ and $2m$. Then $3m=-\frac{2h}{b}$ and $2m^2=\frac ab$. From the first, $m^2=\frac{4h^2}{9b^2}$, so $\frac{8h^2}{9b^2}=\frac ab$, giving $8h^2=9ab$.

(c) 4

Intercepts: $x=\frac{12}{k}$, $y=4$. Area $=\frac12\cdot\frac{12}{k}\cdot4=\frac{24}{k}=6$, so $k=4$.

(c) $\frac{e^x}{(x+1)^2}+c$

Write $\frac{x-1}{(x+1)^3}=\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}$. Let $f=\frac1{(x+1)^2}$; then $\frac{df}{dx}=-\frac{2}{(x+1)^3}$, so the integrand is $e^x\left(f+\frac{df}{dx}\right)$ and the integral is $e^xf$. The result is $\frac{e^x}{(x+1)^2}+c$.

(b) 1

Put $u=\tan^{-1}x$, $du=\frac{dx}{1+x^2}$. Then $\int_0^{\pi/2}\sin u\,du=\left[-\cos u\right]_0^{\pi/2}=1.$

(d) $\frac74$

$y=\frac{x+2}{2}$, so $A=\int_1^2\frac{x+2}{2}dx=\frac12\left[\frac{x^2}{2}+2x\right]_1^2=\frac12\left(6-\frac52\right)=\frac74.$

(a) 900

"ARRANGE" has 7 letters with A twice and R twice. Total arrangements $=\frac{7!}{2!\,2!}=1260$. With the two Rs together (treated as one): $\frac{6!}{2!}=360$. Required $=1260-360=900$.

(b) $\frac32$

Let the terms be $\frac ar, a, ar$. Product $a^3=216$, so $a=6$. Sum: $\frac6r+6+6r=19$, so $6r^2-13r+6=0$, giving $r=\frac32$ or $\frac23$. Among the options, $r=\frac32$.