### 2017 A-level H2 Mathematics (9758) Suggested Solutions

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

Paper 1

Paper 2

MF26

### 2017 A-level H2 Mathematics (9758) 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:
Question 2:
Question 3:
Question 4:
Question 5:
Question 6:
Question 7:
Question 8:
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Question 11:

MF26

### Trigonometry Formulae & Applications (Part 2)

I meant to share more on factor Formulae today. However, a few students are not so sure how to get the R-formulae correctly during their preliminary exams recently. So I thought that I’ll share how they can derive the R-Formulae from the MF26.

The following is the R-Formulae which students should have memorised. It is under assumed knowledge, just saying…

$a \text{cos} \theta \pm b \text{sin} \theta = R \text{cos} (\theta \mp \alpha)$

$a \text{sin} \theta \pm b \text{cos} \theta = R \text{sin} (\theta \pm \alpha)$

where $R = \sqrt{a^2 + b^2}$ and $\text{tan} \alpha = \frac{b}{a}$ for $a > 0, b > 0$ and $\alpha$ is acute.

So here, I’ll write the addition formulae that’s found in MF26.

$\text{sin}(A \pm B) \equiv \text{sin}A \text{cos} B \pm \text{cos} A \text{sin} B$

$\text{cos}(A \pm B) \equiv \text{cos}A \text{cos} B \mp \text{sin} A \text{sin} B$

I’ll use an example I discussed previously.

$f(x) = 3 \text{cos}t - 2 \text{sin}t$

Write $f(x)$ as a single trigonometric function exactly.

Lets consider the formulae from MF26.

$\text{cos}(A \pm B) \equiv \text{cos}A \text{cos} B \mp \text{sin} A \text{sin} B$

$R\text{cos}(A \pm B) \equiv R \text{cos}A \text{cos} B \mp R \text{sin} A \text{sin} B$

We can let

$3 = R \text{cos} B ---(1)$

$2 = R \text{sin} B ---(2)$

$\Rightarrow \sqrt{ 3^2 + 2^2 } = \sqrt{ R^2 \text{cos}^2 B + R^2 \text{sin}^2 B}$

$\Rightarrow \sqrt{13} = R$

$\Rightarrow \frac{R \text{sin} B}{R \text{cos} B} = \frac{2}{3}$

$\Rightarrow \text{tan} B = \frac{2}{3}$

Putting things together, we have that

$f(x) = \sqrt{13} \text{cos} ( t + \text{tan}^{\text{-1}} (\frac{2}{3}))$

### Trigonometry Formulae & Applications (Part 1)

Upon request by some students, I’ll discuss a few trigonometry formulae here and also some of their uses in A’levels. I’ve previously discussed the use of factor formulae here under integration.

I’ll start with the R-Formulae. It should require no introduction as it is from secondary Add Math. This formulae is not given in MF26, although students can derive it out using existing formulae in MF26.

$a \text{cos} \theta \pm b \text{sin} \theta = R \text{cos} (\theta \mp \alpha)$

$a \text{sin} \theta \pm b \text{cos} \theta = R \text{sin} (\theta \pm \alpha)$

where $R = \sqrt{a^2 + b^2}$ and $\text{tan} \alpha = \frac{b}{a}$ for $a > 0, b > 0$ and $\alpha$ is acute.

Here is a quick example,

$f(x) = 3 \text{cos}t - 2 \text{sin}t$

Write $f(x)$ as a single trigonometric function exactly.

Here, we observe, we have to use the R-Formulae where

$R = \sqrt{3^2 + 2^2} = \sqrt{13}$

$\alpha = \text{tan}^{\text{-1}} (\frac{2}{3})$

We have that

$f(x) = \sqrt{13} \text{cos} ( t + \text{tan}^{\text{-1}} (\frac{2}{3}))$.

I’ll end with a question from HCI Midyear 2017 that uses R-Formulae in one part of the question.

A curve D has parametric equations

$x = e^{t} \text{sin}t, y = e^{t} \text{cos}t, \text{~for~} 0 \le t \le \frac{\pi}{2}$

(i) Prove that $\frac{dy}{dx} = \text{tan} (\frac{\pi}{4} - t)$.

I’ll discuss about Factor Formulae soon.  And then the difference and application between this two formulae.

### H2 Mathematics (9740) 2016 Prelim Papers

So many students have been asking for more practice. I’ll put up all the Prelim Papers for 2016 here. Do note that the syllabus is 9740 so students should practice discretion and skip questions that are out of syllabus. 🙂

Here are the Prelim Paper 2016. Have fun!

Here is the MF26.

### Thinking [email protected] #5

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by KS. More of him can be found here.

This is a very challenging trigonometry question from AJC Mid Years P1 Q10.

Using the formula for $\text{sin}(A+B)$, prove that $\text{sin}(3A) = 3 \text{sin} A - 4 \text{sin}^3 A$.

(i) Hence show that $\sum_{r=1}^n \dfrac{4}{3^r} \text{sin}^3 (3^r \theta) = \text{sin} (3 \theta) - \dfrac{1}{3^n} \text{sin}(3^{n+1} \theta)$.

(ii) Deduce the sum of the series $\sum_{r=1}^n \text{sin}^2 (3^r \theta) \text{cos} (3^r \theta)$

### June Crash Course

The team at The Culture SG has been really busy and we have a lot of things prepared to help you guys work for that A. First up! Crash course for June…

And we know it is a bit late to be announcing this on the site now, but we have really been caught up with preparing our students lately that we don’t have the time to properly update here. So here are the details for the Math Crash Course and the Chemistry Crash Course.

P.S. For SCIENCE students who wish to chiong in October, please take note that the H2 Chem/ Phy/ Bio Paper 4 (practical) is in October. So better start soon! Here are the details!

Click to view

For 3 hr lessons, they are priced at $105. For 2 hr lessons, they are priced at$70.

Lessons will be held at:
Newton Apple Learning Hub
Blk 131, Jurong Gateway Road #03-263/265/267 Singapore 600131
Tel: +65 6567 3606

For math enquiries, you may contact Mr. Teng at +65 9815 6827.

For chem enquiries, you may contact Ms. Chan at +65 93494384.

For GP enquiries, you may contact Ms. Chen at +65 91899133.

### Differentiation Question #1

Given that $y = \frac{8}{x^3} - \frac{6}{x^2} + \frac{5}{2x}$, find the approximate percentage change in $y$ when $x$ increases from 2 by 2%.

### Probability Question #4

A gambler bets on one of the integers from 1 to 6. Three fair dice are then rolled. If the gambler’s number appears $k$ times ($k = 1, 2, 3$), he wins $$k$. If his number fails to appear, he loses$1. Calculate the gambler’s expected winnings

### Random Sec 4 Differentiations

B6

$y = 3e^x + \frac{4}{e^x}$

$\frac{dy}{dx} = 3e^x - \frac{4}{e^x}$

$\frac{d^2y}{dx^2} = 3e^x + \frac{4}{e^x}$

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

$3e^x - \frac{4}{e^x} = 0$

$3e^{2x} = 4$

$2x = \mathrm{ln} \frac{4}{3}$

$x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$

Sub $x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$ to $\frac{d^2y}{dx^2}$

$\frac{d^2y}{dx^2} > 0$ Thus, it is a min point.

C7

$y = \mathrm{ln} \frac{5-4x}{3+2x}$

$y = \mathrm{ln} (5-4x) - \mathrm{ln} (3+2x)$

$\frac{dy}{dx} = \frac{-4}{5-4x} - \frac{2}{3+2x}$

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

$\frac{-4}{5-4x} - \frac{2}{3+2x} = 0$

$\frac{-4}{5-4x} = \frac{2}{3+2x}$

$-4(3+2x) = 2(5-4x)$

$-12 - 8x = 10 - 8x$

$-12 = 10$ (NA).

There are no stationary points for this curve.

C8

$x = \frac{1}{3}e^{y(2x+5)}$

$\mathrm{ln}(3x) = y(2x+5)$

$\frac{\mathrm{ln}(3x)}{2x+5} = y$

$y = \frac{\mathrm{ln}(3x)}{2x+5}$

$\frac{dy}{dx} = \frac{\frac{1}{x}(2x+5) - \mathrm{ln}(3x) \times 2}{(2x+5)^2}$

Let $x = e^2$

$\frac{dy}{dx} = \frac{\frac{1}{e^2}(2e^2+5) - \mathrm{ln}(3e^2) \times 2}{(2e^2+5)^2}$

Evaluate with a calculator…