Math Repository
We have digitized the answers and solutions from 2007 to 2022 at
Math Repo.
In particular, you can start navigating the 2018 answers and full worked solutions from
2018P1Q1.
They are designed to be digital and mobile friendly: we hope you will have a good experience there.
Math Pro
Done with your TYS but Prelims still too difficult? Try out the questions
at Math Pro where we tweak the past TYS questions.
We are still under construction, but 6 pure math topics have been completed so far, with more to come!
Paper 1
- (i) $1-\frac{1-\ln x}{x^2}$.
(ii) $1-\frac{2}{\mathrm{e}}$. -
(i) $\frac{1}{2}$ and $3$.
(ii) $\frac{125\pi}{6} \textrm{ units}^3$. -
(i) $f(x) = -\frac{6}{x^3}$.
(ii) $y=3-x^2$. -
(i) $x=-1-\sqrt{3}, -1, 0$ or $-1+\sqrt{3}$.
(ii) $-1-\sqrt{3} < x <-1$ or $0 < x <-1+\sqrt{3}$. -
$b=-1$.
$f^{-1}(x) = \frac{x+a}{x-1}$. - (ii) $\pm 2 \sqrt{31}$.
- (ii) $N \left (-\frac{1}{17},0\right )$.
-
(i) $A=5, u_3=40$.
(ii) $a=\frac{15}{2}, b=-5, c=-5$.
(iii) $15(2^n-1) – \frac{5n}{2} (n + 3)$. -
(ii) $x=2\alpha$.
(iii) $8$ units. -
(i) Conditions on $V$: $V$ is a constant.
(iii) $I = \frac{3A}{2\mathrm{e}}$. -
(i)(a) $102.43.
(i)(b) $1215.71.
(i)(c) First day of June 2018.
(ii)(a) $$(100+12b)$.
(ii)(b) $b=\frac{4}{3}$.
(iii) $b=1.23$.
Paper 2
- (i) $f(x)=\frac{2\sqrt{2}}{9}(x+18)^{\frac{3}{2}} + 45$.
(ii) $(54,237)$. -
(a)(i) $x=2+3i, 2-3i$ or $\frac{1}{2}$.
(b)(i) $\frac{-3+3\sqrt{3}i}{2}$ and $\frac{-3-3\sqrt{3}i}{2}$.
(b)(ii) $3\mathrm{e}^{0i}, 3\mathrm{e}^{\frac{2\pi}{3}i},3\mathrm{e}^{\frac{-2\pi}{3}i}.$
(b)(iii) Sum = $0$, product = $27$. -
(i) $D(-5,-4,3)$.
(ii) $4x+45y+20z=200$.
(iii) $58.6^{\circ}.$
(iv) $6.88$ units. -
(i) $-2x^2-\frac{4}{3}x^4-\frac{64}{45}x^6$.
Expansion not valid when $x=\frac{\pi}{4}$.
(ii) $-2x-\frac{4}{9}x^3-\frac{64}{225}x^5+c$.
$-1.0644$.
(iii) $-1.0670$. -
(i) The sample size should be at least 30 so that $n$ is large and Central Limit Theorem will apply. Hence $\overline{X}$ will be normally distributed approximately.
The fans should be randomly chosen.
(ii) Let $X$ be time to failure and $\mu$ the population mean of $X$.
Null hypothesis $H_0: \mu = 65000$
Alternative hypothesis $H_1: \mu < 65000$.
(iii) $\sigma^2 > 9420000$. -
(i) Hint: use a binomial.
(ii) $\frac{5}{9} < p < \frac{2}{3}$.
(iii) $0.430$. -
(i) $P(A’ \cap B’) = 1-a-b+ab$.
Since $P(A’) \cdot P(B’) = P(A’ \cap B’)$, $A’$ and $B’$ are independent.
(ii) $P(A’ \cap B’) = 1-a-c$.
(iii) $\frac{2}{15} \leq P(A \cap B) \leq \frac{1}{3}$. -
(i) $P(S=6) = \frac{2}{(n+4)(n+5)},P(S=7) = \frac{12}{(n+4)(n+5)}$,
$P(S=8) = \frac{4n+6}{(n+4)(n+5)},P(S=9) = \frac{6n}{(n+4)(n+5)}$,
$P(S=10) = \frac{n^2-n}{(n+4)(n+5)}$.
(ii) $P(S=10)=0$.
This is because it is impossible to obtain $S=10$ when there is only one ball numbered 5 (at least two balls numbered 5 are needed to obtain $S=10$.
(iii) $Var(S)=\frac{22n^2+78n+36}{(n+4)(n+5)^2}$. -
(i) It is unlikely to be modelled by $P=aR+b$ because the scatter diagram shows that the rate of increase of $P$ with respect to $R$ is increasing as $R$ increases.
(ii) For $P$ and $R$, $r=0.969$.
For $P$ and $R^2$, $r=0.993$.
Since the $|r|$ value for $P$ and $R^2$ is closer to 1, $P=aR^2+b$ is the better model.
$P=2.85\times 10^{-8}R^2-0.283$.
(iii)$R=6450$. The estimate is reliable since $P=0.9$ is within the given data range and $|r|\approx 1$.
(iv) $P=0.0273$. The estimate is not reliable since $R=3300$ is outside the given data range.
(v) $P = 1.02 \times 10^{-4} R^2 – 0.283$. -
(ii) $0.605$.
(iii) $0.773$.
(iv) $0.126$.
(v) $136$.
(vi) $0.185$.