### 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.

### 2015 A Level H1 Chemistry (8872) Paper 1 Suggested Solutions

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

Many thanks to the student who sent us the question paper. ðŸ™‚

You can post any questions regarding the answers but please do not ask me how many marks to secure an A for the examination. Nobody will know so there is no point in discussing this.

### Post- 2015 H2 Mathematics Paper 1 Thoughts

Easier than 2014. Tricky. Very tricky. Tedious.

When I did it, it felt a bit like a AJC paper slightly, so I believe AJC students might find the paper quite comfortable. Paper is not super conceptual.

Highlights: Question 3 which test students on the definition of integration, not a difficult question, but some may struggle to express their answers. Piecewise function was expected, as I told my students. Truck load of integration, technique no good means you struggle already.

### So what should we expect in paper 2?

#### Differential Equation

Start you off with a differential equation then have you find the equation, perhaps draw a graph and comment about the long run? Remember to state if k is positive or negative constant.
How it can be hard, getting you to find the equation and link it with conics. Try finding the equation of curve whose normal to the curve passes through $(4, 8)$.

Probably give it a 7-10 marks.

#### Vectors

Clearly about planes, projection and perpendicular distance. Students should familiarise themselves with the conditions for various situations for plane (intersecting at a common line or point). And if they want a set of values, be alert. Same for coordinates.
How it can be hard, when they start reflecting the lines and giving you an angle, and ask you to hence.

Probably give it a 10 marks at least.

#### Mathematical Induction

I doubt a trigonometry one will repeat itself. Paper 1 did test student enough trigonometry in the last question. What remains is either a conjecture, or recurrence MI. So please go revise how to key a sequence into the GC, how to show increasing/ decreasing sequence. For conjecture, students may find this post helpful. ðŸ™‚

How it can be hard, you can’t do your favourite MI if you can’t even do a conjecture.

Probably give it a 10 marks

#### Functions

I really hope they start putting some effort and put the fun in functions. Functions has been quite boring for the past years.

How it can be hard, put in some conics or ask you to find the range of composite, or restrict some domains such that the composite exists.

#### Complex Numbers

We know it will definitely still come out, and its gonna be Loci already. Be careful and if possible, learn how to check if your bisector or half-line should cut the origin or the centre of a circle, etc. Er don’t forget compass and protractor please.
How it can be hard, finding cartesian equations. Or simply giving you the $\mathrm{arg}(z)=\mathrm{tan}^{-1}{\frac{3}{4}}$ and then ask you for exact coordinates.

Probably just 8 marks.

### Conclusion

Okay I know they don’t add up to 40 marks, but I’m not god, else I can tell you what the actual population mean mark will be. Give and take.

The paper should be tricky, so be careful with the unit, like $x$ (thousands) of people, ya.

As for statistics, definitions like the p-value, unbiased estimate, level of significance, necessary conditions for approximations and assumptions should be well memorised. Sampling, last time already la.

Students should know how to find things like $P(T+F=5| F=2)$ given the distribution of T and F. Find the range of level of significance is an easy task but eludes some students still, partly because they overlook that we only reject $h_0$ when $p-value \le \alpha$, yes it is less than equal, not just less than. And we do not reject when $p-values > \alpha$, no more equal; I wrote this intuitive line cos just in case some blindly write equal again. Please take note, definitions are important in Mathematics.

For regression, I hope students know when to use $(\bar{x}, \bar{y})$ to calculate sample values, and the intuition behind it. Also understand that r-value indicates correlation, and not causality, know that its independent of scaling and translation. You can read a bit here.

For a normal distribution, they should know that its centred about $\mu$ ALWAYS. So learn to abuse the fact. Try a question here.

For PnC, just pray ðŸ™‚

Be precise and leave all answers like the regression line in 3SF, but use the 5SF to estimate values, etc.

And please be careful when you round off something in an inequality. :/ A good example will be the APGP question in 2015 H2 Math.

Now I need to get back to my lessons. But I do hope these help.

### A little on bell curve

Since many students were asking, I thought I’ll roughly explain the bell curve and its distribution properties.

Before that, I thought I should mention that I’m not really sure myself if Private Candidates and School Candidates share a different bell curve. I doubt so, since all take the same subject code, there should be no differentiation as there will be bias. And I do not really see a particular need for them if all the students take the same paper. There is no differentiation between taking the paper again and first time too, it doesn’t mean that you take it twice, you have a higher chance/ lower chance. They should be competing in all fairness still.

A bit of intuition of how this bell curve thing works. We can’t have too many A’s. If ALL the JC students were to score zero mark and no student deviate (cheat), then all of you will get A. Similarly, if God answered all your prayers and ALL JC student were to score full marks, then all will get A. The curve attempts to distribute proportionally the results of everybody based on the marks, and place you in the correct percentile (probability).

Normal Distribution Generator

Logistics wise, you can click the link above which gives us a normal distribution. Here, you can input the Mean and Variance so you can create your own bell curve. I can’t really advise the mean and variance score since I have no data. Its really why we only see letter grade.

Next we look at the bell curve

First of all, this is a standardised normal, the $Z \sim \mathrm{N}(0,1)$ curve that you all know. To simplify things, just consider the cumulative percent at the bottom. 80% means you 80th percentile, and that in general warrants an A I’ll say. So consider we have 20 or 21 JC (I don’t know), each having approximately 700 students taking, we can find the population size. From here we can then take a percentage and figure out how many A’s are available.

Next, what does it mean when the bell curve shifts?

What we see above is a Normal curve centred about 0 (mark), and we know that Normal curve is centred about $\mu$ (mean/ modal mark). So if you can figure out the mean mark, by taking a sufficient sample size of students and how they did (we lack the information here since we only receive letter grades), we can that find the unbiased estimate of the population mean. And then the unbiased estimate of the population variance, which is crucial too.

Now since we all squeeze into the same bell curve and due to some friends scoring full marks or near full marks, yes there are definitely students that have full marks in a few particular JCs, I know a handful of my students above 90 actually… This results in the mean mark shifting to the right. And in general, the curve should not shift left, at least not in our society. We are Asians.

Here is why, it is really impossible to say what’s the range for A unless you can confidently tell me the population mean and variance. But I do consider the case of 25% of students should get an A.

I hope this clarifies a bit. And yes, this is what you guys actually learnt in A-levels, if you understood your content of course as what I’ve explained in class before. To spice things up, we can also perform hypothesis testing (simple case can be just with $\alpha$, while we should also look at $\beta$ )to see if the $\mu$ we found is the best unbiased estimate. And something out of syllabus will be to test if $\sigma^2$ is also best unbiased estimate. The word Best adds one more criterion for our estimate actually.

Now you can go at play with the above link and fit in what you believe the mean and standard deviation is, and then you can calculate the percentile (probability) you are in by putting the score you think you have.

Take note the standard deviation should be not go too big, as we can’t have people scoring more than 100. 100 marks should be $3.1 \sigma$ away from $\mu$.

Again, please don’t ask me whats the $\mu$ and $\sigma$. I don’t want to advise such things. You can open your sampling notes, perform stratified sampling then collect unbiased estimates of population mean and variance yourself, and put what you learnt to real (finally) use.

If you’re lazy, then just check what’s your school percentage of A then consider the population of your school, that roughly how well you will do. ðŸ™‚ I usually say this in class, that if your school delivers 50%, then you need to find a friend to not do well at least.

My take for the bell curve is that the bell curve will always help the best students. As for it helping the weaker students, it is highly dependent on the cohort on the whole and the difficulty of the paper. Consider me scoring 90/100 marks and everybody else scores above 95 marks, I’ll be the weakest and will end up failing. I hope these does not discourage or crush the confidence. Remember that there is a paper 2 still, with equal weightage.

So I took this image of google, it does not represent the A-levels. Please take note. But this is what happens when too many students do well. If you’re keen, you can read more on skewed normal distribution independently. I might do more on Stat 101 here when I’m free.

### 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 H1 Mathematics (8864) Paper Suggested Solutions

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

Students are highly encouraged to discuss freely for alternative solutions too.

I typed this, so pardon me if got errors. There should be a handful of errors as I just type and solve as I go. Just refresh as you go along as I’m typing with wifi in a cafe. AND I SKIP A LOT OF STEPS, including denoting my random variables. I’ll get to them probably tomorrow morning.

### 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)