2015 A-level H2 Mathematics (9740) Paper 2 Question 2 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\theta = \mathrm{cos^{-1}} |\frac{\begin{pmatrix}2\\3\\{-6}\end{pmatrix}\cdot \begin{pmatrix}1\\0\\{0}\end{pmatrix}}{\sqrt{49} \cdot 1}| = 73.4^{\circ}$

(ii)
Let $\vec{ON}$ be point on L that makes $\sqrt{33}$ from P.

$\vec{ON} = \begin{pmatrix}1\\{-2}\\{-4}\end{pmatrix} + \lambda \begin{pmatrix}2\\3\\{-6}\end{pmatrix}$ for some $\lambda$

$\vec{PN} = \begin{pmatrix}{-1}\\{-7}\\{2}\end{pmatrix} + \lambda \begin{pmatrix}2\\3\\{-6}\end{pmatrix}$

$|\vec{PN}| = \sqrt{(-1+2\lambda)^2 + (-7+3\lambda)^2 + (2-6\lambda)^2}$

$33 = 49 \lambda^2 - 70 \lambda + 54$

$\lambda = 1 \mathrm{~or~} \frac{3}{7}$

$\vec{ON} = \begin{pmatrix}3\\{1}\\{-10}\end{pmatrix} \mathrm{~or~} \frac{1}{7}\begin{pmatrix}13\\{-5}\\{-46}\end{pmatrix}$

$L = |\vec{PN}|^2 = (-1+2\lambda)^2 + (-7+3\lambda)^2 + (2-6\lambda)^2 = 49 \lambda^2 - 70 \lambda +54$

$\frac{dL}{d\lambda} = 98 \lambda - 70$

$\frac{d^2L}{d\lambda ^2} = 98 > 0$

So when $\lambda = \frac{70}{98} = \frac{5}{7}$, L is minimum.

$\vec{ON} = \frac{1}{7} \begin{pmatrix}{17}\\{1}\\{-58}\end{pmatrix}$

(iii)
$\begin{pmatrix}{-1}\\{-7}\\{2}\end{pmatrix} \times \begin{pmatrix}{2}\\{3}\\{-6}\end{pmatrix} = \begin{pmatrix}{36}\\{-2}\\{11}\end{pmatrix}$

$\begin{pmatrix}{1}\\{-2}\\{-4}\end{pmatrix} \bullet \begin{pmatrix}{36}\\{-2}\\{11}\end{pmatrix} = -4$

$\therefore, \pi : 36x - 2y +11z=-4$

2015 A-level H2 Mathematics (9740) Paper 2 Question 1 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$16 -\frac{1}{2} h \ge 0$

$\Rightarrow h \le 32$

$\therefore, \mathrm{maximum~h}=32m$

(ii)
$\frac{dh}{dt} = \frac{1}{10} \frac{\sqrt{32-h}}{\sqrt{2}}$

$\int \frac{1}{\sqrt{32-h}} ~dh = \int \frac{1}{10 \sqrt{2}}~dt$

$-2 \sqrt{32-h} + C = \frac{1}{10 \sqrt{2}}t$

$t = - 20 \sqrt{64-2h} + 10 \sqrt{2}C$

When $t = 0, h = 0, C = \frac{16}{\sqrt{2}}$

$\therefore, t = - 20 \sqrt{64-2h} + 160$

When $h = 16m, t = 46.9$ years.

Straightforward question.

2011 A-level H2 Mathematics (9740) Paper 2 Question 9 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
(a)
P(faulty)
= P(made by A and faulty) + P(made by B and faulty)
= 0.6(0.05) + 0.4(0.07)
= 0.058

(b)
= $\frac{\mathrm{P(made~by~A~and~faulty)}}{\mathrm{P(faulty)}}$
= $\frac{0.6(0.05)}{0.058}$
= $\frac{15}{29}$

(ii)
(a)
P(exactly one of them is faulty)
= $0.058 \times (1 - 0.058) \times 2!$
= 0.109272 (exact)

(b)
P(both were made by A | exactly one is faulty)
= $\frac{\mathrm{P(both~were~made~by~A~and~exactly~one~is~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}$
= $\frac{\mathrm{P(one~is~made~by~A~and~faulty,~the~other~is~made~by~A~and~not~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}$
= $\frac{0.6(0.05) \times 0.6(0.95) \times 2!}{0.109272}$
= $\frac{1425}{4553}$

Question can be easily solved by drawing a tree diagram. Do take note that we only wrong off NON exact answers to 3sf, so for (iia), we keep the full exact answer.

2015 A-level H2 Mathematics (9740) Paper 2 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

2015 A-level H2 Mathematics (9740) Paper 1 Question 11 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\frac{dy}{d\theta} = 6\mathrm{sin}\theta \mathrm{cos^2}\theta - 3 \mathrm{sin^3} \theta$

$\frac{dx}{d\theta} = 3 \mathrm{sin^2}\theta \mathrm{cos}\theta$

$\frac{dy}{dx} = \frac{6\mathrm{sin}\theta \mathrm{cos^2}\theta3 - \mathrm{sin^3} \theta}{3 \mathrm{sin^2}\theta \mathrm{cos}\theta}$

$\frac{dy}{dx} = \frac{3\mathrm{cos^2}\theta - \mathrm{sin^2} \theta}{ \mathrm{sin^2}\theta \mathrm{cos}\theta}$

$\frac{dy}{dx} = 2 \frac{\mathrm{cos}\theta}{\mathrm{sin}\theta} - \frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}$

$\frac{dy}{dx} = 2 \mathrm{cot} \theta - \mathrm{tan} \theta$

(ii)
Let $\frac{dy}{dx} = 0, \frac{2}{\mathrm{tan}\theta} - \mathrm{tan}\theta = 0$

$\mathrm{tan^2} \theta = 2$

$\mathrm{tan} \theta = \sqrt{2} or - \sqrt{2}$ (reject since $\theta \in [0, \frac{\pi}{2}]$

$\therefore, k = 2$

When $\mathrm{tan} \theta = \sqrt{2}$, using trigonometry identity, we find

$\mathrm{cos} \theta = \frac{\sqrt{3}}{3} ~and~ \mathrm{sin} \theta = \frac{\sqrt{6}}{3}$

From $\frac{dy}{dx}, \frac{d^2y}{dx^2} = -2\mathrm{cosec^2} \theta \frac{d\theta}{dx} - \mathrm{sec^2} \theta \frac{d\theta}{dx}$

For $\theta \in [0, \frac{\pi}{2}], \frac{d^2y}{dx^2} \textless 0$ since $\mathrm{sin} \theta > 0$ and $\mathrm{cos} \theta >0$ for principal values of $\theta$

We have that when $\mathrm{tan} \theta = \sqrt{2}$, its is a maximum point.

$x = \mathrm{sin^3} \theta = \frac{2\sqrt{6}}{9}$

$y = 3 \mathrm{sin^2}\theta \mathrm{cos} \theta = \frac{2 \sqrt{3}}{3}$

$\therefore, \text{Required}~\text{coordinates}~ = (\frac{2\sqrt{6}}{9}, \frac{2\sqrt{3}}{3})$

(iii)

$\int_0^1 y dx$

$= \int_0^{\frac{\pi}{2}} 3 \mathrm{sin^2}\theta \mathrm{cos} \theta 3 \mathrm{sin^2}\theta \mathrm{cos}\theta d\theta$

$= \int_0^{\frac{\pi}{2}} 9 \mathrm{sin^4}\theta \mathrm{cos^2} \theta d\theta$

Using GC, Area $= 0.88357 \approx 0.884 \mathrm{units^2}$

(iv)
At P, $y = ax$
$\Rightarrow 3 \mathrm{sin^2}\theta \mathrm{cos} \theta = a \mathrm{sin^3} \theta$

$\mathrm{tan} \theta = \frac{3}{a}$

At maximum point, $\mathrm{tan} \theta = \sqrt{2}$

$\Rightarrow \sqrt{2} = \frac{3}{a}$

$\therefore, a = \frac{3\sqrt{2}}{2}$

To find the exact coordinates, students need to be able to either identify the correct trigonometry identify, or draw the right angled triangle to find $\mathrm{sin} \theta$ and $\mathrm{cos} \theta$ carefully. As for the show part for the parametric integration, students need to start from the definition of area then proceed carefully, not forgetting to change the limits.

2015 A-level H2 Mathematics (9740) Paper 1 Question 10 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$A_1 = \int_0^{\frac{\pi}{4}} \mathrm{sin}x~ dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{cos}x~ dx$

$= 2 - \sqrt{2}$

$A_2 = \int_0^{\frac{\pi}{4}} \mathrm{cos}x - \mathrm{sin}x~ dx$

$= \sqrt{2} - 1$

$\frac{A_1}{A_2} = \frac{2-\sqrt{2}}{\sqrt{2}-1} = \frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1} = \sqrt{2}$

(ii)
$y = \mathrm{sin} x \Rightarrow x = \mathrm{sin^{-1}}y$

Vol $= \pi \int_0^{\sqrt{2}/2} x^2~ dy$

$= \pi \int_0^{\sqrt{2}/2} (\mathrm{sin^{-1}}y)^2~ dy$

(iii)
When $y = \frac{\sqrt{2}}{2}, u = \frac{\pi}{4}$

When $y = 0, x = 0$

$y = \mathrm{sin}u$

$\mathrm{sin^{-1}}y = u \Rightarrow \frac{dy}{dx}=\mathrm{cos}u$

Vol $= \pi \int_0^{\sqrt{2}/2} (\mathrm{sin^{-1}}y)^2~ dy$

$= \pi \int_0^{\pi / 4} u^2 \mathrm{cos}u~du$

$\therefore, a = 0, b = \frac{\pi}{4}$

$\pi \int_0^{\pi / 4} u^2 \mathrm{cos}u~du$

$= \pi \{u^2 \mathrm{sin}u\bigl|_0^{\pi /4} - \int_0^{\pi /4} 2u \mathrm{sin} u ~ du\}$

$= \pi \{\frac{\sqrt{2}\pi ^2}{32} + 2u \mathrm{cos}u \bigl|_0^{\pi / 4} - \int_0^{\pi /4} 2\mathrm{cos}u~du\}$

$= \pi \{\frac{\sqrt{2}\pi ^2}{32} + \frac{\sqrt{2} \pi}{4} - 2 \mathrm{sin}u \bigl|_0^{\pi /4} \}$

$= \pi \{\frac{\sqrt{2}\pi ^2}{32} + \frac{\sqrt{2} \pi}{4} - \sqrt{2}\}$

2015 A-level H2 Mathematics (9740) Paper 1 Question 9 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(a)
$w^2 = (a+bi)^2 = a^2 - b^2 + 2abi$

$\frac{w^2}{w^*} = \frac{a^2 - b^2 + 2abi}{a-bi} \times \frac{a+bi}{a+bi}$

$= \frac{1}{a^2 + b^2}(a^3 - 3ab^2 + 3a^2bi - b^3 i)$

For $\frac{w^2}{w^*}$ to be purely imaginary, $Re{\frac{w^2}{w^*}} = 0$

$\Rightarrow a^3 - 3ab^2 = 0$

$a = 0 \mathrm{(NA)}~ \mathrm{or}~ a^2 - 3b^2 = 0$

$b= \pm \frac{a}{\sqrt{3}}$

$\therefore w = a \pm \frac{a}{\sqrt{3}}i$

(bi)
$z^5 = -32i = 2^5 e^{-i\frac{\pi}{2}}$

$z^5 = 2^5 e^{i(-\frac{\pi}{2}+2k\pi)}$ for $k = 0, \pm 1, \pm 2$

$z = 2e^{i(\frac{-\pi + 4k \pi}{10})}$

$\Rightarrow |z| = 2$

$\mathrm{arg}(z) = - \frac{\pi}{10}, - \frac{\pi}{2}, - \frac{9 \pi}{10}, \frac{3 \pi}{10}, \frac{7 \pi}{10}$

(bii)
$z_1 = 2e^{i\frac{7\pi}{10}}$ and $z_2 = 2e^{i\frac{-9\pi}{10}}$

$\angle OZ_2Z_2 = \frac{1}{2}(\pi - \frac{2\pi}{5}) = \frac{3\pi}{10}$

$\mathrm{arg}(Z_1 - Z_2) = \frac{3\pi}{10} + \frac{\pi}{10} = \frac{2\pi}{5}$

By Cosine Rule,

$|Z_1 - Z_2|^2 = 2^2 + 2^2 - 2(2)(2)\mathrm{cos} (\frac{2\pi}{5})$

$|Z_1 - Z_2|^2 = 8 - 8\mathrm{cos} (\frac{2\pi}{5})$

$|Z_1 - Z_2|^2 = 8[1 - \mathrm{cos} (\frac{2\pi}{5})]$

$|Z_1 - Z_2|^2 = 8[2\mathrm{sin^2} (\frac{\pi}{5})]$

$|Z_1 - Z_2|^2 = 16\mathrm{sin^2} (\frac{\pi}{5})$

$|Z_1 - Z_2| = 4\mathrm{sin} (\frac{\pi}{5})$

The first part of the question is rather interesting, since we don’t use the traditional argument method to solve it here. But that approach works fine too.
The next part, students just need to be really carful with the root finding, its a standard tutorial question. What follows is a bit tedious, but with an aid of a simple diagram and understanding the properties of roots on an argand diagram, students should not struggle that much.

2015 A-level H2 Mathematics (9740) Paper 1 Question 8 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let $A_{50}$ and $B_{50}$ denote the total time A & B take respectively, in seconds.

$A_{50} = \frac{50}{2}[2T+49(2)] = 50T + 2450$

$1.5H = 5400s$

$1.75H = 6300s$

$5400 \le 50T +2450 \le 6300$

$59 \le T \le 77$

$\therefore, \{ T\in \mathbb{R} | 59 \le T \le 77\}$

(ii)
$B_{50} = \frac{t(1.02^{50}-1)}{1.02-1} = 50t(1.02^{50}-1)$

$5400 \le 50t(1.02^{50}-1) \le 6300$

$63.84 \le t \le 74.483$

$63.9 \le t \le 74.4$ (3 SF)

$\therefore, \{ t\in \mathbb{R} | 63.9 \le t \le 74.4\}$

(iii)
For A: Time taken $= 59 + 49(2) = 157$
For B: Time taken $= 63.84533(1.02)^{49} = 168.475$

Difference $= 168.475 - 157 \approx 11 \mathrm{seconds}$ (nearest seconds)

• Be careful of the units, its in seconds.
• Be careful to write in sets.
• Be careful when you’re rounding off (ii) as you need to round off according to the range of values. If you rounded down to 63.8, this means you accept 63.81, which is out of range and will cause him to arrive before 1.5H. I reckon this part will get many students who are less careful. :/

2015 A-level H2 Mathematics (9740) Paper 1 Question 7 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$2 \vec{OC} = 3 \vec{CA} = 3 (\vec{OA} - \vec{OC})$

$5 \vec{OC} = 3 a$

$\vec{OC} = \frac{3}{5} a$

$11 \vec{OD} = 5 \vec{OB}$

$\vec{OD} = \frac{5}{11} b$

(ii)
$\vec{BC} = \frac{3}{5}a -b$

$l_{BC}: r = b + \lambda (\frac{3}{5}a - b)$

$l_{BC}: r = \frac{3}{5} \lambda a + (1-\lambda)b, \lambda \in \mathbb{R}$

$\vec{AD} = \frac{3}{5}a -b$

$l_{AD}: r = a + \mu (\frac{5}{11}b - a)$

$l_{AD}: r = \frac{5}{11} \mu b + (1-\mu)a, \mu \in \mathbb{R}$

(iii)
At E, $\frac{5}{11} \mu b + (1-\mu)a = \frac{3}{5} \lambda a + (1-\lambda)b$

$\frac{5}{11} \mu = 1 - \lambda \rightarrow(1)$

$\frac{3}{5} \lambda = 1 - \mu \rightarrow(1)$

Solving with GC, $\lambda = \frac{3}{4}$ and $\mu = \frac{11}{20}$

$\vec{OC} = \frac{9}{20}a + \frac{1}{4} b$

$\vec{AE} = \frac{1}{4}b - \frac{11}{20}a$

$\vec{ED} = \frac{9}{44}b - \frac{9}{20}a$

$= \frac{9}{11}(\frac{1}{4}b - \frac{11}{20}a)$

$\vec{ED} = \frac{9}{11} \vec{AE}$

$\therefore, AE: ED = 11: 9$

Students that faced difficulties with ratio (like me) might struggle a bit here. But if you guys do listen in class, you should know the above method I use, should help a lot.
(ii) should be manageable. For (iii), students need to be careful if they use a GC to solve the question. Its like statistics, solving $\mu$ and $\lambda$.

2015 A-level H2 Mathematics (9740) Paper 1 Question 6 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
From MF15,
$\mathrm{ln}(1+2x) = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} + \ldots$

$\approx 2x - 2x^2 + \frac{8}{3}x^3$

(ii)
$ax(1+bx)^c = ax(1 + cbx + \frac{c(c)-1}{2!}(bx)^2 + \ldots)$

Comparing coefficients,

$a = 2$

$abc = -2 \Rightarrow bc = -1$

$2[\frac{c(c-1)}{2!}b^2] = \frac{8}{3}$

$\Rightarrow 2[\frac{c^2-c}{2c^2}] = \frac{8}{3}$

$c = -\frac{3}{5}$

$b = \frac{5}{3}$

Coefficient of $x^4 = a[\frac{c(c-1)(c-2)}{3!}(\frac{5}{3})^3] = -\frac{104}{27}$

The next parts, just need to be really careful. It is quite a lot of variables to juggle around here. For the last part, dont forget the $a$ and $b^2$