### Thoughts on A’levels H2 Mathematics 2016 Paper 2

I’ll keep this short since we are all busy. One thing about paper 1 we saw, there were many unknowns.

So topics which I think will come out…

Differentiation – I think a min/max problem will come out, possibly with r and h both not given and asked to express r in terms of h. But students should revise a on the properties of curves with differentiation; given a curve equation with an unknown, for instance $y=/frac{x^2+kx+1}{x-1}$, find the range of k such that there is stationary points.

Complex Number – Loci will definitely come out. I’m saying they will combine with trigonometry.

Integration – Modulus integration hasn’t really been tested. Else a question on Area/ Volume could be tested, and I’ll say they need students to do some
Conics too.

For statistics, my students should have gotten the h1 stats this year. And if it’s an indicator, then it should not be a struggle.

I expect PnC and probability to be combined. Conditional Probability in a poisson question should be tested too, so do revise it well. For hypothesis testing, students should be careful of their formula and read really carefully about the alternative hypothesis. Also, :9 know that the formulas for poison PDF and binompdf are both given in mf15. Lastly, know when to use CLT.

All the best!

### Thoughts about 2016 A’levels H2 Mathematics Paper

I’ve covered some things in classes, with sufficient revisions and final lap papers set. So I thought we have a little breakdown. And of course, we should review what was weeded away in the 9740 H2 Mathematics Syllabus. After all, this is the very LAST time the can test them.

1. Recurrence Relations. I’ve harped on this last year too. Conjectures! Conjectures! You can read more about it here. An a question involving conjecture should start with a recurrence relation, then a conjecture, ending with a recurrence MI. Students should know how to do both $\Sigma$ and recurrence MI. Yes, they are different.

2. Loci. You guys are the lucky last batch to do Loci. So please buy a protractor and compass. Draw them, as one of my student put it, surgically. If need be, use a graph paper (why not?). Harder loci for example can require students to draw for example, $\text{arg}(z-1+2i) = \text{tan}^{-1}(\frac{4}{3})$. You should not have trouble measuring this angle, because you could not even be able to do it. Students should be able to draw such angles with ease. One little note about Loci, will definitely their geometrical descriptions. Many students can draw this, but stumble to describe them.

3. Vectors. Truth be told, I’m still waiting for a question involving vectors in 3D, to land in A’levels. An example can be the HCI Prelims Paper 1 Question 6, which can be found here.

4. Poisson Distribution. I don’t know what this topic has been axed. So students should ready for one big Poisson Distribution questions, I say give it 12-14 marks. And it should be tested with conditional probability. I’ll practice either Demand & Supply or Inflow & Outflow questions. An example can be the one found in NYJC Prelims Paper 2 Question 11, which can be found here.

5. Correlation & Regression. Ever wondered what the $r^2$ means in the GC? Well, $r^2 = bd$ is being removed from the syllabus as well. It hasn’t surfaced before, so maybe it shall finally make its one and only LAST presence felt this year. Students should familiarise themselves with the use of $y- \bar{y} = b ( x - \bar{x})$ equation, which can be found in the MF15. I know many of you probably have not seen it before.

6. Hypothesis Testing. Students should review definitions of level of significance and p-value. Also understand what you may conclude from a Z-Test, using the results of a T-test. A little small part that students can think about, is why use a small sample size? After all, we know that have a sufficiently large $n$ allows us to perform CLT and then use a Z-Test.

7. Trigonometry. After it appeared in 2011 for a trigonometry MI, the product to sum formulas is still a problem for most students. I highly doubt its coming out again with MI, but its can easily come out again with complex numbers. An example can be this.

More examples and discussion will be made in class.

### 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

### June Revision Exercise

You can find the solutions of all ten sets of the June Revision Exercise we did in class.

Have fun!

June Revision Exercise 1
June Revision Exercise 2
June Revision Exercise 3
June Revision Exercise 4
June Revision Exercise 5
June Revision Exercise 6
June Revision Exercise 7
June Revision Exercise 8
June Revision Exercise 9
June Revision Exercise 10

### 2010 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.

### 2010 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)

Direction vector of $l \text{~is~} \begin{pmatrix}{-3}\\6\\9\end{pmatrix} = -3 \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}$

Normal vector of $p \text{~is~} \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}$

Since $l$ is parallel to the normal vector of $p, l$ is perpendicular to $p$.

(ii)
$l: r = \begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}$

$[\begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}] \bullet \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix} = 0$

$10 + \lambda + 2 4 \lambda + 9 + 9 \lambda = 0$

$\lambda = - \frac{3}{2}$

Point of intersection $= (\frac{17}{2}, 2, \frac{3}{2})$.

(iii)
$\begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix} = \begin{pmatrix}{-2}\\23\\33\end{pmatrix}$

$10 + \lambda = - 2$ — (1)

$-1 - 2 \lambda = 23$ — (2)

$-3 - 3 \lambda = 33$ — (3)

Since $\lambda = -12$ satisfies all 3 equations, A lies on l.

Since $= (\frac{17}{2}, 2, \frac{3}{2})$ is the midpoint of A & B, by ratio theorem,

$\begin{pmatrix}{\frac{17}{2}}\\2\\{\frac{3}{2}}\end{pmatrix} = \frac{\vec{OA} + \vec{OB}}{2}$

$\vec{OB} = 2 \begin{pmatrix}{\frac{17}{2}}\\2\\{\frac{3}{2}}\end{pmatrix} - \begin{pmatrix}{-2}\\23\\33\end{pmatrix} = \begin{pmatrix}19\\{-19}\\{-30}\end{pmatrix}$

$B(19, -19, -30)$

(iv)
Area

$= \frac{1}{2} |\vec{OA} \times \vec{OB}|$

$= \frac{1}{2} |\begin{pmatrix}{-2}\\23\\33\end{pmatrix} \times \begin{pmatrix}19\\{-19}\\{-30}\end{pmatrix}|$

$= \frac{1}{2} |\begin{pmatrix}{-63}\\{567}\\{-399}\end{pmatrix}|$

$= \frac{1}{2} \sqrt{63^2 +567^2 + 399^2}$

$= 348$ to the nearest whole number.

Students must give answers in coordinates, rather than as a position vector.

### 2010 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.

Total Volume $= 3x(x)y = 300$

$x^2y = 100$

$y = \frac{100}{x^2}$

Total Surface Area, $A = 3x(x) + 8x(y) + 3x(x) + 8x(ky)$

$= 6x^2 + (k+1) \frac{800}{x}$

$\frac{dA}{dx} = 12x - (k+1)\frac{800}{x^2} = 0$

$x^3 = \frac{200}{3} (k+1)$

$x = ^3 \sqrt{\frac{200}{3}(k+1)}$

$\frac{d^2A}{dx^2} = 12 + (k+1) \frac{1600}{x^3} = 12 + (k+1)\frac{1600}{\frac{200}{3}(k+1)} = 36 > 0$

Thus, $x=^3 \sqrt{\frac{200}{3}(k+1)}$ gives a minimum total surface area.

(ii)
$\frac{y}{x} = \frac{100}{x^3} = \frac{3}{2(k+1)}$

(iii)
$0 \textless k \le 1$

$1 \textless k+1 \le 2$

$\frac{1}{2} \le \frac{1}{k+1} \textless 1$

$\frac{3}{4} \le \frac{3}{2(k+1)} \textless \frac{3}{2}$

$\frac{3}{4} \le \frac{y}{x} \textless \frac{3}{2}$

(iv)
If the box has square ends, then $y=x$

$\frac{3}{2(k+1)} = 1$

$k = \frac{1}{2}$