Mathematics

(b) $n(A)+n(B)$

For disjoint sets $n(A\cap B)=0$, so $n(A\cup B)=n(A)+n(B)-n(A\cap B)=n(A)+n(B)$.

(a) $(y-2)(y-3)\geq0$

The product $(y-2)(y-3)$ is non-negative when $y\le2$ or $y=3$, which is the region the given condition is intended to select (the expression is negative only for $2<y<3$), so the inequality with $\geq0$ is the expected choice.

(d) $(-\infty,-1]$

$\log(1-x)$ needs $1-x>0$, so $x<1$. $\sqrt{x^2-1}$ needs $x^2\ge1$, so $x\le-1$ or $x\ge1$. The intersection is $x\le-1$, i.e. $(-\infty,-1]$.

(b) Odd function

$f(-x)=(-x)^3+\tan(-x)=-x^3-\tan x=-f(x)$, so $f$ is odd.

(a) 0

$i+i^2+i^3+i^4=i-1-i+1=0$.

(b) $4x^2-1=0$

Sum of roots $=0$ and product $=-\tfrac14$, so $x^2-\tfrac14=0$, i.e. $4x^2-1=0$.

(b) 10

This is a G.P. with $a=5$ and $r=\tfrac12$, so $S_\infty=\dfrac{a}{1-r}=\dfrac{5}{1/2}=10$.

(a) $\frac\pi2$

Let the angles be $a-d,\ a,\ a+d$. Their sum is $3a=180^\circ$, so $a=60^\circ$. The least angle is $60^\circ-d=30^\circ$, so $d=30^\circ$ and the greatest angle is $90^\circ=\dfrac{\pi}{2}$ radian.

(a) inside triangle

The circumcentre of an acute-angled triangle lies inside it (on the hypotenuse for a right triangle, outside for an obtuse one).

(a) $-\cos\theta$

$\sin(270^\circ-\theta)=\sin270^\circ\cos\theta-\cos270^\circ\sin\theta=-\cos\theta$.

(a) 0

The slopes are $\frac{3-2}{2-1}=1$ and $\frac{4-3}{3-2}=1$, so the points are collinear and the area is 0.

(c) $x+y+3=0$

Slope $=\tan135^\circ=-1$ and the y-intercept is $-3$, so $y=-x-3$, i.e. $x+y+3=0$.

(a) $\left(\frac cm,\frac{2a}{m}\right)$

Write the line as $mx-y+c=0$. The pole of $lx+my+n=0$ with respect to $y^2=4ax$ is $\left(\dfrac nl,-\dfrac{2am}{l}\right)$. Here $l=m$, the y-coefficient is $-1$ and $n=c$, giving $\left(\dfrac cm,\dfrac{2a}{m}\right)$.

(d) $-\cot\theta$

$\dfrac{d(\sin\theta)}{d(\cos\theta)}=\dfrac{\cos\theta}{-\sin\theta}=-\cot\theta$.

(d) Symmetric matrix

$(A+A^T)^T=A^T+A=A+A^T$, so the sum is symmetric.

(c) 5040 All 7 letters of SOLVING are distinct, so the number of arrangements is $7! = 5040$.

(c) 3.75 $0.8x = 3 \Rightarrow x = \frac{3}{0.8} = 3.75$

(b) 4 $\frac{\sqrt{96}}{\sqrt{k}} = 2\sqrt6 \Rightarrow \frac{96}{k}=24 \Rightarrow k=4$

(b) 1/6 A die has 6 equally likely outcomes and only one shows 2, so $P = \frac16$.

(c) (p+r)/2 is an even integer Let $p=2n+1$, $q=2n+3$, $r=2n+5$. Then $p+q=4n+4$ is even, $p+q+r=6n+9$ is odd, and $(p+q)/2=2n+2$ is even, so (a), (b) and (d) are true. But $(p+r)/2=2n+3=q$ is odd, so statement (c) is not true.

(d) None of these

Let the number be $10a+b$. Then $(10a+b)-(10b+a)=9(a-b)=36$, so $a-b=4$. The digit sum is not fixed: $84$ gives $12$, $73$ gives $10$, $62$ gives $8$. So no single value among 24, 18 or 12 is forced.

(b) 5

A number is divisible by 3 if its digit sum is, and by 9 if the digit sum is divisible by 9.

• 2133 (9), 2646 (18), 4131 (9), 6336 (18), 8118 (18) are divisible by 9, so they are excluded.
• 3471 (15), 5286 (21), 5340 (12), 7347 (21), 9276 (24) are divisible by 3 but not by 9.

So there are 5 such numbers.

(b) 71

Three years ago each of the three was 3 years younger, so the total was $80-3\times3=71$.

(a) 5 kmph

Let Abhay's speed be $a$ and Sameer's be $s$. $\frac{30}{a}-\frac{30}{s}=2,\qquad \frac{15}{a}-\frac{30}{s}=-1$ Subtracting gives $\frac{15}{a}=3$, so $a=5$ kmph (and $s=7.5$ kmph).

(b) 11

As printed, 72 is the only number in range divisible by 72, which matches no option, so the intended question is "divisible by 7". The multiples of 7 between 11 and 90 are $14,21,\dots,84$, which number $\frac{84-14}{7}+1=11$.