Mathematics

(c) 2

Substituting $x=\frac{5y}{2}$ into $x^2+y^2=5$ gives $\frac{25y^2}{4}+y^2=5$, so $y^2=\frac{20}{29}$. This has two values of $y$, each giving one point, so the line meets the circle at 2 points.

(a) $\{-1,1\}$

For $x>2$, $|x-2|=x-2$ and the value is $1$. For $x<2$, $|x-2|=-(x-2)$ and the value is $-1$. So the range is $\{-1,1\}$.

(d) 0

The point's distance from the centre satisfies $1^2+(\sqrt2)^2=3<4=r^2$, so the point lies inside the circle. No tangent can be drawn from an interior point.

(c) 0

$\sin^7x$ is an odd function, and the limits $-1$ to $1$ are symmetric, so the integral is $0$.

(a) 0

$|\sin x|\le1$, so $\left|\frac{\sin x}{x}\right|\le\frac1{|x|}\to0$ as $x\to\infty$ (squeeze theorem).

(a) $\log_45$

$\lim_{n\to0}\frac{5^n-1}{4^n-1}=\lim\frac{(5^n-1)/n}{(4^n-1)/n}=\frac{\ln5}{\ln4}=\log_45.$

(d) imaginary lines

For $ax^2+2hxy+by^2=0$ here $a=1,\ 2h=1,\ b=1$, so $h^2-ab=\frac14-1<0$. The lines are imaginary (they meet only at the origin).

(c) $2\sin\alpha/2$

$|\vec a-\vec b|^2=1+1-2\cos\alpha=2(1-\cos\alpha)=4\sin^2\frac\alpha2$, so $|\vec a-\vec b|=2\sin\frac\alpha2$.

(b) 210

Total ways to distribute 3 distinct prizes among 6 people is $6^3=216$. Ways in which one person gets all three is $6$. Required number $=216-6=210$.

(b) $e^{-\pi/2}$

Since $i=e^{i\pi/2}$, $i^i=e^{i\cdot i\pi/2}=e^{-\pi/2}$ (principal value).

(b) $\frac12(1+2m)$ $\log_{49}28=\frac{\log_7 28}{\log_7 49}=\frac{\log_7 4+\log_7 7}{2}=\frac{2\log_7 2+1}{2}=\frac12(1+2m)$

(a) $x^2-x-6=0$ Sum $=\alpha+\beta=1$ and $\alpha^2+\beta^2=13$. Then $\alpha\beta=\dfrac{(\alpha+\beta)^2-(\alpha^2+\beta^2)}{2}=\dfrac{1-13}{2}=-6$. The equation is $x^2-(1)x+(-6)=0$, i.e. $x^2-x-6=0$.

(d) $\frac\pi6(2n+1)$ $\tan\theta\tan2\theta=1\Rightarrow \sin\theta\sin2\theta=\cos\theta\cos2\theta\Rightarrow\cos2\theta\cos\theta-\sin2\theta\sin\theta=0\Rightarrow\cos3\theta=0$. So $3\theta=(2n+1)\dfrac\pi2$, giving $\theta=\dfrac\pi6(2n+1)$.

(c) $75^\circ$ By the sine rule $\sin B=\dfrac{b\sin A}{a}=\dfrac{\sqrt6\cdot\frac{1}{\sqrt2}}{2}=\dfrac{\sqrt3}{2}$, so $B=60^\circ$ (the option-consistent value). Then $C=180^\circ-45^\circ-60^\circ=75^\circ$.

(c) $\frac12+i\frac{\sqrt3}{2}$ $\frac12+i\frac{\sqrt3}{2}=\cos\frac\pi3+i\sin\frac\pi3$, so by De Moivre the 7th power is $\cos\frac{7\pi}{3}+i\sin\frac{7\pi}{3}=\cos\frac\pi3+i\sin\frac\pi3=\frac12+i\frac{\sqrt3}{2}$.

(d) 0 Let the first term be $a$ and common difference $d$. From $p\,[a+(p-1)d]=q\,[a+(q-1)d]$: $a(p-q)+d(p^2-q^2-p+q)=0\Rightarrow (p-q)\,[a+(p+q-1)d]=0.$ Since $p\ne q$, $a+(p+q-1)d=0$, which is the $(p+q)^{th}$ term. Hence it is 0.

(c) 1260 MATHEMATICS has 11 letters; the vowels are A, E, A, I (4). Without vowels we arrange the 7 consonants M, T, H, M, T, C, S, where M and T each occur twice: $\frac{7!}{2!\,2!}=\frac{5040}{4}=1260.$

(d) $\frac xy$ $x^2-y^2=\left(t+\frac1t\right)^2-\left(t-\frac1t\right)^2=4$. Differentiating, $2x-2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=\dfrac xy$. (Check: $\dfrac{dy/dt}{dx/dt}=\dfrac{1+1/t^2}{1-1/t^2}=\dfrac{t^2+1}{t^2-1}=\dfrac xy$.)

(a) 1 $\int\frac{dx}{1+\sin x}=\int\frac{1-\sin x}{\cos^2x}dx=\tan x-\sec x.$ $\left[\tan x-\sec x\right]_0^{\pi/2}=\lim_{x\to\pi/2}\frac{\sin x-1}{\cos x}-(0-1)=0+1=1.$

(d) $(\hat i-\hat j-\hat k)/\sqrt3$ $(\hat i+\hat j)\times(\hat i+\hat k)=\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&0\\1&0&1\end{vmatrix}=\hat i-\hat j-\hat k.$ Its magnitude is $\sqrt3$, so the unit vector is $(\hat i-\hat j-\hat k)/\sqrt3$.

(b) $\frac{-b}{1+a}$ Put $y=mx$ in $ax^2-bxy-y^2=0$: $m^2+bm-a=0$. The slopes are $\tan\alpha,\tan\beta$ with sum $-b$ and product $-a$. $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{-b}{1+a}.$

(c) $e^{-2}$ $\left(\frac{x+1}{x+2}\right)^{2x+1}=\left(1-\frac1{x+2}\right)^{2x+1}\to e^{\lim\,-\frac{2x+1}{x+2}}=e^{-2}.$

(b) $\frac13$ The curves $y=x^2$ and $y=|x|$ meet at $x=-1,0,1$ and the figure is symmetric about the y-axis. $A=2\int_0^1(x-x^2)\,dx=2\left(\frac12-\frac13\right)=\frac13.$

(a) $x+y=3$ At $t=1$ the point is $(1,2)$. $\dfrac{dy}{dx}=\dfrac{2}{2t}=1$, so the normal has slope $-1$. Normal: $y-2=-(x-1)\Rightarrow x+y=3$.

(a) 924 The general term is $\binom{12}{r}x^{2r}$. For $x^{12}$, $2r=12$, $r=6$, so the coefficient is $\binom{12}{6}=924$.