### 2015 A-level H2 Mathematics (9740) Paper 2 Question 10 Suggested Solutions

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

(i)
Graph to be inserted
(ii)
(a) $r \approx -0.9807$ (4 DP)
(b) $r \approx -0.9748$ (4 DP)
(c) $r \approx -0.9986$ (4 DP)

(iii)
Using GC, $P = -0.14686 \sqrt{h} + 34.78895$
$\therefore, P = -0.147 \sqrt{h} + 34.8$ (3 SF)

(iv)
$\therefore, P = -0.147 \sqrt{3.28h_{metres}} + 34.8$

$\therefore, P = -0.266 \sqrt{h_{metres}} + 34.8$ (3 SF)

### 2015 A’level Suggested Solutions

Congratulations on the completion of A’levels for the 2015 batch!

As for those who still, are taking A’levels 2016, we hope you find this site helpful. Read the comments (ignore the trolls who have yet grown up) and learn from our mistakes.

It has been a pleasure for all us to interact with you guys. Do check back after your release of A’levels, as we will love to hear from you! In the meantime, you can always read through some of the posts regarding undergraduate & postgraduate courses in Mathematics (KS), Economics (KS), Physics (Casey) & Chemistry (Eric) as we share some of our past experiences, along with some of our works today.

You’ve come a long way, now party hard. We wish you all the best in your future endeavours.

### KS’ thoughts on H2 Math 2015

This are simply personal opinions.

Just to clarify, I am not spotting as I never believe in spotting for a Mathematics paper, the thought is pretty ambiguous.

Let us first look at the effect of having the papers being one week of each others first. So some students plan to hold back on Statistics a bit and focus more on Pure Mathematics since there is a week to work on Statistics. Please don’t bank on this thought too much and ignore Statistics too much. Lets be honest, we are not going to spend the next 6 days entirely on statistics as most of you have papers straight after Paper 2.

Next, will the paper be easier, you know the hearsay about paper being easy and hard every year. Lets not bank on that, since things are pretty erratic these days, people don’t do things the norm (just look at the haze and what some neighbours say). Furthermore, an easy paper is not exactly what you should pray for. Everybody can do an easy paper but not everybody don’t make a careless mistake. 🙂 I’ll advise students be really meticulous with their papers. And for the $n^{th}$ time, please present legibly. Its time to write “Let X denote…” for every single variables you introduce from system of linear equations to statistics. And lets not forget to get real with $\lambda$ in vectors.

Now, every year’s paper has some questions that springs a surprise and throws students off their feet. In 2013, we saw the distance with two planes. In 2011, we saw a trigonometry Mathematical Induction. In 2009, we saw periodic/ cyclical functions coupled with piecewise functions. These questions were not hard if students are aware of the correct technique to treat the questions. The marks are really generous given. Some of you who saw my suggested solutions did ask me why my working so short. Because its all about knowing and understanding, instead of rote-learning.

Surprise questions aside, basic questions will definitely appear. Students who still struggle with implicit differentiation, should really get some help already. Let’s be clear that you have learnt enough differentiation to be highly capable to differentiate every single thing that appears in A-levels, even the rate of gradient, i.e., $\frac{d}{dt}(\frac{dy}{dx})$. If you still can remember all the formulas for trigonometry differentiation, learn the trigonometry and just go back to basics, perform that quotient or product rule. Every trigonometry differentiation goes back to differentiating $\mathrm{sin}x, \mathrm{cos}x \mathrm{~and~tan}x$. Just saying. And please know your formulas in MF15, don’t need to memorise them but know what you have and how to apply/ use them CORRECTLY. Oh, and who is to say that we can’t have you differentiate piecewise functions. hmmm. One just have to be meticulous enough to check domains in functions.

Aside from being able to present solution well for vectors, students must know how to work with lines. The thing about H2 Math syllabus is that, we can always test concepts not in syllabus so long as there is sufficient pointers and guidance. For example, the shortest distance between two skewed lines (which is not required to be learnt in syllabus) can be solved by having students finding an angle & a projection. Students should also be careful with their drawing in exams, I’ve marked really good art pieces which unfortunately are misrepresentation of the questions. Take your drawing with a pinch of salt and don’t assume anything unless you are certain. Given that the line is on plane, is the perpendicular distance of a point to line same as the perpendicular distance from point to the plane? A question was designed during revision period for my students and many students assumed so. It is really convenient to assume this, but we should never jump to that conclusion. So please be careful and take that picture with a pinch of salt.

Statistics is like a free marks for many of my students. Students love them cos its all about Graphing Calculator. But lets be honest, in Graphing Calculator we trust and we will copy that $0.159002476$ on the screen and round it off appropriately. But what if we have butter fingers and typed wrong. Will we key it twice and check if we made a human error. And don’t get me started on square rooting the variance please. My students find me too naggy already, yes, thou shalt hear my voice in exam. If you finish the paper without continuity correction, you need to go back and do some correction. And not to be forgotten, PRESENTATION PRESENTATION PRESENTATION! Round off when you need, know when to use 3SF and 5SF. 🙂 This is very important in statistics, we don’t have time to try to understand your variable $X$.

Lastly, please have enough rest. Drink water and get enough rest daily. A handful of students have fallen ill due to the haze and that’s time lost. That 3 hours more of studying which will cause you to fall sick for a day, not worth it. And don’t tell yourself, you can do this and will not fall sick. Even the A-levels 2013 Paper 2 Question 12 knows that the probability of a person falling is not constant. :p

All the best, and I will share my suggested solutions shortly after your paper.

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 12 Suggested Solutions

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

(i)
Let X denote the number of people joining the queue in 4 minutes.

$X \sim \mathrm{Po}(4.8)$

$\mathrm{P}(X \ge 8) = 1 - \mathrm{P}(X \le 7) \approx 0.113$

(ii)
Let Y denotes number of people joining the queue in t seconds.

$Y \sim \mathrm{Po}(0,02t)$

$\mathrm{P}(Y \le 1) = 0.7$

$\mathrm{P}(Y = 0) + \mathrm{P}(Y = 1) = 0.7$

$\frac{e^{-0.02t}(0.02t)^0}{0!} + \frac{e^{-0.02t}(0.02t)^1}{1!}=0.7$

$e^{-0.02t}(1+0.02t) = 0.7$

Using the Graphing Calculator, t = 55

(iii)
Let W denote the number of people joining the queue in 15 minutes.

$W \sim \mathrm{Po}(18)$

Since $\lambda = 18 > 10, ~W \sim \mathrm{N}(18, 18)$ approximately.

Let V denote the number of people leaving the queue in 15 minutes.

$V \sim \mathrm{Po}(27)$

Since $\lambda = 27 > 10, ~V \sim \mathrm{N}(27, 27)$ approximately

$\Rightarrow W - V \sim \mathrm{N}(-9,45)$ approximately

$\mathrm{P}(W -V \ge -11)$

$= \mathrm{P}(W-V > -11.5)$ by continuity correction

$\approx 0.645$

(iv)
The average number of people joining the queue per minute may not remain constant throughout a period of long hours as they have different traveling time. There might be no flight departing too. Thus, it would not appropriate.

The question is pretty standard Poisson. So long as students are careful with continuity correction, I don’t see much problems with this questions. They should answer (iv) contextually.

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 5 Suggested Solutions

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

$X \sim \mathrm{N}(\mu, \sigma^2)$

$\mathrm{P}(X \textless 40.0) = 0.05$

$\Rightarrow \frac{40 - \mu}{\sigma} = -1.6449 \rightarrow (1)$

$\mathrm{P}(X \textless 70.0) = 0.05$

$\Rightarrow \frac{70 - \mu}{\sigma} = 1.9600 \rightarrow (1)$

Solving, $\mu = 53.7$ and $\sigma = 8.32$

A very straightforward question here, as it didn’t even try to trick unsuspecting students who often forget to change the signs before applying inverse norm function. Please use a Graphing Calculator to solve if you aren’t good with simultaneous equations.

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 3 Suggested Solutions

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

(i)
Let $y = \mathrm{ln} (2x+1) + 3$

$x = \frac{e^{y-3}-1}{2}$

$\therefore, f^{-1}(x) = \frac{e^{x-3}-1}{2}, x \in \mathbb{R}$

$D_f = \mathbb{R}$

$R_{f^{-1}} = D_f = (-\frac{1}{2}, \infty)$

(ii)

(iii)
$f(x) = f^{-1}(x) = x$

$\Rightarrow f(x) = x$

$\mathrm{ln}(2x+1) + 3 = x$

$\mathrm{ln}(2x+1) = x - 3$

Using Graphing calculator, $x = -0.4847 \mathrm{~or~} 5.482$

Students should be careful to draw the curves according to the given domain. And since the question requests for equations of asymptotes, we should have label them too.

The last part, students can also use the inverse function to solve.

This question, in general did not pose much of a problem for students.

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 8 Suggested Solutions

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

(i)

(ii)
Using the Graphing Calculator, $r = -0.99232 \approx -0.992$

The product moment correlation coefficient measures the strength and direction of a linear relationship between two variables. Thus, the r-value only indicate correlation, but not causality. It is possible for the relation to be non-linear, and yet have a value close to 1. We need to look at the scatter diagram to determine whether the relationship is truly linear.

(iii)
Using $y = a + bx^2, r = -0.99998$

Using $y = c + dx, r = -0.99232$

Since $|-0.99998|$ is close to 1. We conclude that $y = a + bx^2$ is the better model.

(iv)
Using Graphing Calculator, $y = 22.2 - 0.856x^2$

When $x = 3.2, y = 22.230 - 0.85621(3.2)^2 \approx 13.5$

Some students actually drew the scatter diagram for the given data, which was really strange. The meaning of r-value has to be well understood to answer (ii) well. Some actually don’t know what does least squares regression refer to.

### 2011 A-level H2 Mathematics (9740) Paper 1 Question 2 Suggested Solutions

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

(i)
$f(-1.5) = (-1.5)^2 a + (-1.5) b + c = 4.5$

$2.25 a -1.5 b + c = 4.5 \rightarrow (1)$

$f(2.1) = (2.1)^2 a + (2.1) b + c = 3.2$

$4.41 a + 2.1 b + c = 3.2 \rightarrow (2)$

$f(3.4) = (3.4)^2 a + (3.4) b + c = 4.1$

$11.56 a + 3.4b + c = 4.1 \rightarrow (3)$

Using the Graphing Calculator, $a = 0.215, b = -0.490, c = 3.281$

(ii)

$f \prime (x) = 2ax + b > 0$ for f to be increasing

$\therefore, x > 1.14$

Firstly, students must be attentive to leaving the answers to 3 decimal places. For f(x) is an increasing function, we know that implies that $f'(x) > 0.$

### 2011 A-level H2 Mathematics (9740) Paper 1 Question 4 Suggested Solutions

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

(i)
$cos ^6 x$

$= (cos x)^6$

$\approx (1 -\frac{x^2}{2} + \frac{x^4}{24})^6$

$= 1 + {6 \choose 1}(1)(-\frac{x^2}{2} + \frac{x^4}{24}) + {6 \choose 2}(1)(-\frac{x^2}{2} + \frac{x^4}{24})^2 + \ldots$

$= 1 + 6(-\frac{x^2}{2} + \frac{x^4}{24}) + 15(-\frac{x^2}{2} + \frac{x^4}{24})^2 + \ldots$

$\approx 1 - 3x^2 + 4x^4$

(ii)
(a)

$\int_0^a g(x) dx$

$= \int_0^a 1 - 3x^2 + 4x^4 dx$

$= x - x^3 + \frac{4x^5}{5} \biggl|_0^a$

$= a - a^3 + \frac{4a^5}{5}$

When $a = \frac{\pi}{4}, ~ \int_0^a g(x) dx \approx 0.540$

(b)
Using the Graphing Calculator, $\int_0^{\frac{\pi}{4}} g(x) dx = 0.475$

The approximation is not very good since $\frac{\pi}{4}$ is not very close to zero.
We only considered up to $x^4$, thus the approximation can be improved by taking into consideration more terms.

Just don’t be careless for (i)! Some students still can be unsure of how to evaluate a definite integral using Graphing Calculator. Lastly, maclaurin’s series expansion are most accurate when value is close to one since they are centred at x = 0.

### 2012 A-level H2 Mathematics (9740) Paper 1 Question 2 Suggested Solutions

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

(i)
$\int \frac{x^{3}}{1+x^{4}}dx = \frac{1}{4} \mathrm{ln} (1 + x^{4}) + C$

(ii)
Given $u = x^{2}, dx = \frac{1}{2x} du$

$\int \frac{x}{1+x^{4}} dx$

$= \int \frac{x}{1 +u^{2}} \frac{1}{2x} du$

$= \frac{1}{2} \int \frac{1}{1+u^{2}} du$

$= \frac{1}{2} tan^{-1}u + C$

$= \frac{1}{2} tan^{-1} (x^{2}) + C$

(iii)
Using a Graphing Calculator, $\int_0^1 (\frac{x}{1+x^{4}})^{2} dx = 0.186$

Do remember to change the $dx$ when doing substitution. Students are reminded that the integration formula can be found in MF15. The last part can be solved using GC since there is no mention of exact value.