2018 H2 Math Solutions

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Paper 1

  1. (i) $1-\frac{1-\ln x}{x^2}$.
    (ii) $1-\frac{2}{\mathrm{e}}$.
  2. (i) $\frac{1}{2}$ and $3$.
    (ii) $\frac{125\pi}{6} \textrm{ units}^3$.
  3. (i) $f(x) = -\frac{6}{x^3}$.
    (ii) $y=3-x^2$.
  4. (i) $x=-1-\sqrt{3}, -1, 0$ or $-1+\sqrt{3}$.
    (ii) $-1-\sqrt{3} < x <-1$ or $0 < x <-1+\sqrt{3}$.
  5. $b=-1$.
    $f^{-1}(x) = \frac{x+a}{x-1}$.
  6. (ii) $\pm 2 \sqrt{31}$.
  7. (ii) $N \left (-\frac{1}{17},0\right )$.
  8. (i) $A=5, u_3=40$.
    (ii) $a=\frac{15}{2}, b=-5, c=-5$.
    (iii) $15(2^n-1) – \frac{5n}{2} (n + 3)$.
  9. (ii) $x=2\alpha$.
    (iii) $8$ units.
  10. (i) Conditions on $V$: $V$ is a constant.
    (iii) $I = \frac{3A}{2\mathrm{e}}$.
  11. (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

  1. (i) $f(x)=\frac{2\sqrt{2}}{9}(x+18)^{\frac{3}{2}} + 45$.
    (ii) $(54,237)$.
  2. (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 = $27$.
  3. (i) $D(-5,-4,3)$.
    (ii) $4x+45y+20z=200$.
    (iii) $58.6^{\circ}.$
    (iv) $6.88$ units.
  4. (i) $-2x^2-\frac{4}{3}x^4-\frac{64}{45}x^6$.
    Expansion not valid when $x=\frac{\pi}{4}$.
    (ii) $-2x-\frac{4}{9}x^3-\frac{64}{225}x^5+c$.
    (iii) $-1.0670$.
  5. (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) $\sigma^2 > 9420000$.
  6. (i) Hint: use a binomial.
    (ii) $\frac{5}{9} < p < \frac{2}{3}$.
    (iii) $0.430$.
  7. (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) $\frac{2}{15} \leq P(A \cap B) \leq \frac{1}{3}$.
  8. (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) $Var(S)=\frac{22n^2+78n+36}{(n+4)(n+5)^2}$.
  9. (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) $P = 1.02 \times 10^{-4} R^2 – 0.283$.
  10. (ii) $0.605$.
    (iii) $0.773$.
    (iv) $0.126$.
    (v) $136$.
    (vi) $0.185$.