### 2016 A-level H1 Mathematics (8864) Paper 1 Suggested Solutions

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

As these workings and answers are rushed out asap, please pardon me for my mistakes and let me know if there is any typo. Many thanks.
I’ll try my best to attend to the questions as there is H2 Math Paper 2 coming up.

Question 1
(i)
$\frac{d}{dx} [2 \text{ln}(3x^2 +4)]$

$= \frac{12x}{3x^2+4}$

(ii)
$\frac{d}{dx} [\frac{1}{2(1-3x)^2}]$

$= \frac{6}{2(1-3x)^3}$

$= \frac{3}{(1-3x)^3}$

Question 2

$2e^{2x} \ge 9 - 3e^x$

$2u^2 + 3u - 9 \ge 0$

$(2u-3)(u+3) \ge 0$

$\Rightarrow u \le -3 \text{~or~} u \ge \frac{3}{2}$

$e^x \le -3$ (rejected since $e^x > 0$) or $e^x \ge \frac{3}{2}$

$\therefore x \ge \text{ln} \frac{3}{2}$

Question 3
(i)

(ii)
Using GC, required answer $= -1.606531 \approx -1.61$ (3SF)

(iii)
When $x = 0.5, y = 0.35653066$

$y - 0.35653066 = \frac{-1}{-1.606531}(x-0.5)$

$y = 0.622459 x +0.04530106$

$y = 0.622 x + 0.0453$ (3SF)

(iv)

$\int_0^k e^{-x} -x^2 dx$

$= -e^{-x} - \frac{x^3}{3} \bigl|_0^k$

$= -e^{-k} - \frac{k^3}{3} + 1$

Question 4
(i)
$y = 1 + 6x - 3x^2 -4x^3$

$\frac{dy}{dx} = 6 - 6x - 12x^2$

Let $\frac{dy}{dx} = 0$

$6 - 6x - 12x^2 = 0$

$1 - x - 2x^2 = 0$

$2x^2 + x - 1 = 0$

$x = -1 or \frac{1}{2}$

When $x =-1, y = -4$

When $x = \frac{1}{2}, y = 2.75$

Coordinates $= (-1, -4) \text{~or~} (0.5, 2.75)$

(ii)
$\frac{dy}{dx} = 6 - 6x - 12x^2$

$\frac{d^2y}{dx^2} = - 6 - 24x$

When $x = -1, \frac{d^2y}{dx^2} = 18 > 0$. So $(-1, -4)$ is a minimum point.

When $x = 0.5, \frac{d^2y}{dx^2} = -18 \textless 0$. So $(0.5, 2.75)$ is a maximum point.

(iii)

x-intercept $= (-1.59, 0) \text{~and~} (1, 0) \text{~and~} (-0.157, 0)$

(iv)
Using GC, $\int_0.5^1 y dx = 0.9375$

Question 5
(i)
Area of ABEDFCA $= \frac{1}{2}(2x)(2x)\text{sin}60^{\circ} - \frac{1}{2}(y)(y)\text{sin}60^{\circ}$

$2\sqrt{3} = \sqrt{3}x^2 - \frac{\sqrt{3}}{4}y^2$

$2 = x^2 - \frac{y^2}{4}$

$4x^2 - y^2 =8$

(ii)

Perimeter $= 10$

$4x+2y + (2x-y) = 10$

$6x + y = 10$

$y = 10 - 6x$

$4x^2 - (10-6x)^2 = 8$

$4x^2 - 100 +120x -36x^2 = 8$

$32x^2 -120x+108=0$

$x=2.25 \text{~or~} 1.5$

When $x=2.25, y = -3.5$ (rejected since $y >0$)

When $x=1.5, y = 1$

Question 6
(i)
The store manager has to survey $\frac{1260}{2400} \times 80 = 42$ male students and $\frac{1140}{2400} \times 80 = 38$ female students in the college. He will do random sampling to obtain the required sample.

(ii)
Stratified sampling will give a more representative results of the students expenditure on music annually, compared to simple random sampling.

(iii)
Unbiased estimate of population mean, $\bar{x} = \frac{\sum x}{n} = \frac{312}{80} = 3.9$

Unbiased estimate of population variance, $s^2 = \frac{1}{79}[1328 - \frac{312^2}{80}] = 1.40759 \approx 1.41$

Question 7
(i)
[Venn diagram to be inserted]

(ii)
(a)
$\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B) = 0.8$
(b)
$\text{P}(A \cup B) - \text{P}(A \cap B) = 0.75$

(iii)
$\text{P}(A | B')$

$= \frac{\text{P}(A \cap B')}{\text{P}(B')}$

$= \frac{\text{P}(A) - \text{P}(A \cap B)}{1 - \text{P}(B)}$

$= \frac{0.6 - 0.05}{1-0.25}$

$= \frac{11}{15}$

Question 8
(i)
Required Probability $= \frac{1}{2} \times \frac{3}{10} \times \frac{2}{9} = \frac{1}{30}$

(ii)
Find the probability that we get same color. then consider the complement.

Required Probability $= 1 - \frac{1}{30} - \frac{1}{2} \times \frac{5}{10} \times \frac{4}{9} - \frac{1}{2} \times \frac{2}{10} \times \frac{1}{9} - \frac{1}{2} \times \frac{4}{6} \times \frac{3}{5} - \frac{1}{2} \times \frac{2}{6} \times \frac{1}{5} = \frac{11}{18}$

(iii)
$\text{P}(\text{Both~balls~are~red} | \text{same~color})$

$= \frac{\text{P}(\text{Both~balls~are~red~and~same~color})}{\text{P}(\text{same~color})}$

$= \frac{\text{P}(\text{Both~balls~are~red})}{\text{P}(\text{same~color})}$

$= \frac{1/30}{7/18}$

$= \frac{3}{35}$

Question 9
(i)
Let X denote the number of batteries in a pack of 8 that has a life time of less than two years.

$X \sim B(8, 0.6)$
(a)
Required Probability $=\text{P}(X=8) = 0.01679616 \approx 0.0168$

(b)
Required Probability $=\text{P}(X \ge 4) = 1 - \text{P}(X \le 3) = 0.8263296 \approx 0.826$

(ii)
Let Y denote the number of packs of batteries, out of 4 packs, that has at least half of the batteries having a lifetime of less than two years.

$Y \sim B(4, 0.8263296)$

Required Probability $=\text{P}(Y \le 2) = 0.1417924285 \approx 0.142$

(iii)
Let W denote the number of batteries out of 80 that has a life time of less than two years.

$W \sim B(80, 0.6)$

Since n is large, $np = 48 > 5, n(1-p)=32 >5$

$W \sim N(48, 19.2)$ approximately

Required Probability

$= \text{P}(w \ge 40)$

$= \text{P}(w > 39.5)$ by continuity correction

$= 0.9738011524$
$\approx 0.974$

Question 10
(i)
Let X be the top of speed of cheetahs.
Let $\mu$ be the population mean top speed of cheetahs.

$H_0: \mu = 95$

$H_1: \mu \neq 95$

Under $H_0, \bar{X} \sim N(95, \frac{4.1^2}{40})$

Test Statistic, $Z = \frac{\bar{X}-\mu}{\frac{4.1}{\sqrt{40}}} \sim N(0,1)$

Using GC, $p=0.0449258443 \textless 0.05 \Rightarrow H_0$ is rejected.

(ii)
$H_0: \mu = 95$

$H_1: \mu > 95$

For $H_0$ to be not rejected,

$\frac{\bar{x}-95}{\frac{4.1}{\sqrt{40}}} \textless 1.644853626$

$\bar{x} \textless 96.06 \approx 96.0$ (round down to satisfy the inequality)

$\therefore \{ \bar{x} \in \mathbb{R}^+: \bar{x} \textless 96.0 \}$

Question 11
(i)
[Sketch to be inserted]

(ii)
Using GC, $r = 0.9030227 \approx 0.903$ (3SF)

(iii)
Using GC, $y = 0.2936681223 x - 1.88739083$

$y = 0.294 x - 1.89$ (3SF)

(iv)
When $x = 16.9, y = 3.0756 \approx 3.08$(3SF)

Time taken $= 3.08$ minutes

Estimate is reliable since $x = 16.9$ is within the given data range and $|r|=0.903$ is close to 1.

(v)
Using GC, $r = 0.5682278 \approx 0.568$ (3SF)

(vi)
The answers in (ii) is more likely to represent since $|r|=0.903$ is close to 1. This shows a strong positive linear correlation between x and y.

Question 12
Let X, Y denotes the mass of the individual biscuits and its empty boxes respectively.

$X \sim N(20, 1.1^2)$

$Y \sim N(5, 0.8^2)$

(i)
$\text{P}(X \textless 19) = 0.181651 \approx 0.182$

(ii)
$X_1 + ... + X_12 + Y \sim N(245, 15.16)$

$\text{P}(X_1 + ... + X_12 + Y > 248) = 0.2205021 \approx 0.221$

(iii)
$0.6X \sim N(12, 0.66^2)$

$0.2Y \sim N(1, 0.16^2)$

Let $A=0.6X$ and $B = 0.2Y$

$A_1+...+A_12+B \sim N(145, 5.2528)$

$\text{P}(142 \textless A_1+...+A_12+B \textless 149) = 0.864257 \approx 0.864$

### 2016 A-level H2 Mathematics (9740) Paper 2 Suggested Solutions

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

Numerical Answers (click the questions for workings/explanation)

Question 1: $0.0251 \text{~m/min}$
Question 2: $\frac{x^2}{2} \text{sin}nx + \frac{2x}{n^2} \text{cos}nx - \frac{2}{n^3} \text{sin}nx + C;~ a = 2 \text{~or~} 6;~ \pi (\frac{2}{5} - \text{ln} \frac{3}{\sqrt{5}})$
Question 3: $a + \frac{3}{4} - \text{sin}a - \text{cos}a + \frac{1}{4} \text{cos}2a;~ \frac{1}{4}(\pi + 1)^2$
Question 4: $z = 2.63 + 1.93i, ~ 3.37 + 0.0715i ;~8^{\frac{1}{6}}e^{i(-\frac{\pi}{12})}, ~8^{\frac{1}{6}}e^{i(-\frac{3 \pi}{4})}, ~8^{\frac{1}{6}}e^{i(\frac{7\pi}{12})};~ n=7$
Question 5: $\frac{11}{42};~ \frac{4}{11};~ \frac{4}{1029}$
Question 6: $60;~ 10;~ \{ \bar{x} \in \mathbb{R}, 0 \textless \bar{x} \le 34.8 \};~ \{ \alpha \in \mathbb{R}, 0 \textless \alpha \le 8.68 \}$
Question 7: $24;~ 576;~ \frac{1}{12};~ \frac{5}{12}$
Question 8: $r = -0.980;~ c = -17.5;~ d = 91.8;~ y = 85.9$
Question 9: $a = 7.41;~ p = 0.599 \text{~or~} 0.166;~ 0.792$
Question 10: $0.442;~ 0.151 ;~ 0.800;~ \lambda = 1.85$

MF15

### 2016 A-level H2 Mathematics (9740) Paper 1 Suggested Solutions

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

As these workings and answers are rushed out asap, please pardon me for my mistakes and let me know if there is any typo. Many thanks. As these workings and answers are rushed out asap, please pardon me for my mistakes and let me know if there is any typo. Many thanks.

I’ll try my best to attend to the questions as there is H2 Math Paper 2 coming up and I have a lot of prepare. I’ll try my best to attend to the questions as there is H2 Math Paper 2 coming up and I have a lot of prepare.

Side note: I think the paper is tedious, but definitely manageable. Hard ones, could have been q3, q7a, 10, 11ib. So if you did your tutorials and past year papers well, with proper time management and no careless, 70 is manageable. To get an A, you need to fight for that 30 marks which really test you on your comprehension skills and precisions. And these questions should distinct the students who deserve an A.

Numerical Answers (click the questions for workings/explanation)

Question 1: $x \textless -2, \text{~or~} \frac{1}{3} \textless x \textless 4$
Question 2: $0, -\text{ln}2;~ y=2, y = -x\text{ln}2 + \frac{\pi \text{ln}2}{2} + 1;~ (\frac{\pi}{2} - \frac{1}{\text{ln}2}, 2)$
Question 3: $m = b,~ l = a, ~ kl^4 + m = c$
Question 4: $0.74;~ \frac{b(0.74^n)}{0.26}$
Question 5: $\begin{pmatrix}{2b}\\{4b-4a}\\{-2a}\end{pmatrix};~ \pm \frac{1}{6\sqrt{2}};~ 3$
Question 6: $4,~ 14,~ 44;~ \frac{1}{4}n(n+1)(n^2+n+2)+2$
Question 7: $2+3i;~ a=3,~ k=-30$
Question 8: $f(x)=1+2ax+2a^2x^2+\frac{8}{3}a^3x^3 ;~ \text{tan}2x = 2x + \frac{8}{3}x^3$
Question 9: $y = 5-5e^{-2t}; x = 5t + \frac{5}{2}e^{-2t} - \frac{5}{2};~ x = 5t^2 + 20\text{sin}\frac{t}{2}-10t;~ 1.47s,~ 1.05s$
Question 10: $f^{-1}(x) = (x-1)^2, x \in \mathbb{R}, x \ge 1;~ x = 2.62;~6,~ 8,~ 9;~\text{No}$
Question 11: $t = -\frac{5}{9},~ \lambda = -\frac{8}{9}, ~ \mu = \frac{19}{18};~ -2x+y +2z=35 \text{~or~} -2x+y+2z = -37;~a = 4.5$

MF15

### Random Questions from 2016 Prelims #8

ACJC P2 Q3

The function g is defined by

$g: x \mapsto \begin{cases} 2x, & \text{for }0 \le x \le \frac{1}{2} \\ 2-2x, & \text{for } \frac{1}{2} \le x \le 1 \end{cases}$

(ii) Explain why the composite function $gg$ exist.

(iii) Sketch the graph of $gg(x)$.

### Random Questions from 2016 Prelims #7

VJC P1 Q9

(i) Sketch the graph with equation $x^2 +(y-r)^2 = r^2$, where $r >0$ and $y \le r$

A hemispherical bowl of fixed radius $r$ cm is filled with water. Water drains out from a hole at the bottom of the bowl at a constant rate. When the depth of water if $h$cm (where $h \le r$).

(ii) Use your graph in (i) to show that the volume of water in the bowl is given by $V = \frac{\pi h^2}{3} (3r-h)$.

(iii) Find the rate of decrease of the depth of water in the bowl, given that a full bowl of water would become empty in 24 s,

(iv) without any differentiation, determine the slowest rate at which the depth of water is decreasing.

### Random Questions from 2016 Prelims #6

HCI P1 Q5

Sketch on a single Argand diagram, the loci defined by $-\frac{\pi}{4} \textless \text{arg}(z+1+2i) \le \frac{\pi}{4}$ and $|(2+i)w+5| \le \sqrt{5}$

(i) Find the minimum value of $\text{arg}(w)$

(ii) Find the minimum value of $|z-w|$

(iii) Given that $\text{arg}(z-w) \textless \theta, - \pi \textless \theta \le \pi$, state the minimum value of $\theta$

### Random Questions from 2016 Prelims #5

NYJC P1 Q4

Referred to the origin , the points A and B have position vectors a and b respectively. A point C is such that OACB forms a parallelogram. Given that M is the mid-point of AC, find the position vector of point N if M lies on ON produced such that OM:ON is in ratio 3:2. Hence show that A, B, and N are collinear.

Point P is on AB such that MP is perpendicular to AB. Given that angle AOB is $60^{\circ}, |a|=2 \text{~and~} |b|=3$, find the position vector of P in terms of a and b.

### 100 days more…

So the Midyear results are all out and Prelims are known to be in 4-5 weeks’ time. Many students are frantically searching for help and attempting to salvage their results. We are sorry to say that we aren’t able to take any more private students due to time constraints, and only the group classes are available. Our classes are all held in Newton Apple Jurong East.

### Question of the Day #18

Here is a very very interesting question involving probability that a student saw in her tutorial and asked me. Here it is 🙂

A student is concerned about her car and does not like dents. When she drives to school, she has a choice of parking it on the street in one space, parking it on the street and taking up two spaces, or parking in the lot.
If she parks on the street in one space, her car gets dented with probability 0.1.
If she parks on the street and takes two spaces, the probability of a dent is 0.02 and the probability of a $15 ticket is 0.3. Parking in a lot costs$5, but the car will not get dented.
If her car gets dented, she can have it repaired, in which case it is out of commission for 1 day and costs her $50 in fees and cab fares. She can also drive her car dented, but she feels that the resulting loss of value and pride is equivalent to a cost of$9 per school day.
She wishes to determine the optimal policy for where to park and whether to repair the car when dented in order to minimize her (long-run) expected average cost per school day. What should the student to maximise her utility (minimise her cost)?

This is an interesting question, I guess its good to know some JCs are trying to introduce decision making process in teaching probability.

I’ll post a solution here soon. But to start off, we observe that we have two states here and student has 4 decisions. Have fun! 🙂