### 2014 A-level H2 Mathematics (9740) Paper 1 Question 6 Suggested Solutions

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

(a)
(i)
Let P(n) be the preposition, $p_{n}=\frac{1}{3}(7-4^{n})$ for $n\in {\mathbb{Z}}^{+}$
When $n=1, \mathrm{LHS}=p_{1}=1$ and $\mathrm{RHS}=\frac{1}{3}(7-4)=1$
Since $\mathrm{LHS}=\mathrm{RHS}$, P(1) is true.
Assume P(k) is true for some $k \in {\mathbb{Z}}^{+}$.
To prove that P(K+1) is true, i.e., $p_{k+1}=\frac{1}{3}(7-4^{k+1})$
$\mathrm{LHS}$
$=p_{k+1}$

$=4p_{k}-7$

$=4(\frac{1}{3}(7-4^{k}))-7$

$= \frac{1}{3}(4\times{7}-4\times{4^{k}}-21)$

$= \frac{1}{3}(7-4^{k+1})$
$= \mathrm{RHS}$
Therefore, P(k) is true $\Rightarrow$ P(k+1) is also true.
Since P(1) is true and P(k) is true $\Rightarrow$ P(k+1) is true, by principle of mathematical induction, P(n) is true for all $n\in {\mathbb{Z}}^{+}$.

(ii)
$\sum_{r=1}^{n} p_{r}$

$= \sum_{r=1}^{n} {\frac{1}{3}}(7-4^{r})$

$= \frac{1}{3}[\sum_{r=1}^{n}7 - \sum_{r=1}^{n} 4^{r}]$

$= \frac{1}{3}\times{7n} - \frac{1}{3}[\frac{4(4^{n}-1)}{4-1}]$

$= \frac{7n}{3} - \frac{4}{9}(4^{n}-1)$

(b)
(i)
As $n\rightarrow \infty, \frac{1}{(n+1)!} \rightarrow 0$
Therefore, the series converges and the $\mathrm{Sum~to~Infinity}=1$.

(ii)
$u_{n}=S_{n}-S_{n-1}$

$u_{n}=1-\frac{1}{(n+1)!}-(1-\frac{1}{n!})$

$u_{n}=\frac{-1}{(n+1)!}+\frac{1}{n!}$

$u_{n}=\frac{-1+n+1}{(n+1)!}$

$u_{n}=\frac{n}{(n+1)!}$

Firstly, students should know that (a) and (b) are unrelated, it is how A-level questions are denoted. For the MI, it is direct and a simple one on recurrence relation. We are then told to find the sum which tests us on our abilities to identify Arithmetic Progression and Geometric Progression in summation forms.
Convergence of a series can be proved in a few ways, the given is the most direct, which is to show that the sum to infinity is a finite value. Question wants students to write down, so please write it explicitly. It is surprising that some students confuse the formulas and write $u_{n}=S_{n+1}-S_{n}$, this is totally wrong.

### 2014 A-level H2 Mathematics (9740) Paper 1 Question 5 Suggested Solutions

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

(i)
$z^{2}=(1+2i)^{2}$

$z^{2}=1+4i-4=-3+4i$

$\frac{1}{z^{3}}=\frac{1}{(1+2i)(-3+4i)}$

$\frac{1}{z^{3}}=\frac{1}{-11-2i}\times \frac{-11+2i}{-11+2i}$

$\frac{1}{z^{3}}=\frac{-11+2i}{121+4}$

$\frac{1}{z^{3}}=\frac{-11}{125}+\frac{2i}{125}$

(ii)
Using (i)
$pz^{2}+\frac{q}{z^{2}}=p(-3+4i)+q(\frac{-11+2i}{125})$

Since it is real, $4p+\frac{2q}{125}=0$

We have $q=-250p$,

$pz^{2}+\frac{q}{z^{2}}=-3p-(\frac{11}{125})(-250p)=19p$

(i) tests students on their understanding of manipulation and also how to rationalise a complex number. I strongly encourage students to still use the calculator to check their answers.
(ii) is quite comfortable so long as students realise the meaning of being real.

### 2014 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)

(ii) They are parallel to the y-axis.

This question is not tough, students just need to be cautious of adjusting the concavity at $x=-a$ and $x=b$. Since the value of d is not given, the concavity is quite debatable too, but one can be safe just to follow the given shape. For part (ii), some students gave really long answers and should learn to condense it to a simple “parallel to x or y axis”.

### 2014 A-level H2 Mathematics (9740) Paper 1 Question 3 Suggested Solutions

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

(i)
a and b are either zero vectors or parallel to each other.

(ii)
From (i)
$\hat{n}= \frac{1}{\sqrt{1+2^{2}+2^{2}}}\begin{pmatrix}1\\2\\-2\end{pmatrix}$

(iii)
$cos{\theta}=\frac{|\begin{pmatrix}1\\2\\-2\end{pmatrix}{\bullet}\begin{pmatrix}0\\0\\1\end{pmatrix}|}{\sqrt{9}}=\frac{2}{3}$

(i) tests students on understanding. Most students were able to deduce they are parallel. However, it did not cross their minds that the vectors could be zero vectors. They should take note that the question is worth 2 marks and it is far to generous for a simple answer.
Some students forgot and went ahead to evaluate the answer, disregarding that the question asks for COSINE of the acute angle so there is no need. Modulus is required since they are asking for an acute angle.

### 2014 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.

$x^{2}y+xy^{2}+54=0$
differentiating with respect to x,
$2xy+x^{2}(\frac{dy}{dx})+y^{2}+2xy(\frac{dy}{dx})=0$
When $\frac{dy}{dx}=-1$
$-x^{2}+2xy+y^{2}-2xy=0$
$y^{2}=x^{2}$
$y=\pm{x}$

When $y=x$,
$y^{3}+y^{3}+54=0$
$y^{3}=-27$
$y=-3$
$x=-3$

When $y=-x$
$-y^{3}+y^{3}+54=0$
There is no solutions.

Therefore, required coordinates are $(-3,-3)$

This question tests students on their understanding of basic differentiations. First pitfall was when students forget to introduce $\pm{ }$. And to SHOW there is only one such point baffled many as they didn’t know how to go about. But the most direct way is to solve for them and find there is only one. Lastly, read the question carefully and you will see they require the answers to be COORDINATES.

### 2014 A-level H2 Mathematics (9740) Paper 1 Question 1 Suggested Solutions

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

(i)
$f^{2}(x)$
$=f[f(x)]$
$=f(\frac{1}{1-x})$
$=\frac{1}{1-\frac{1}{1-x}}$
$=\frac{1}{\frac{1-x-1}{1-x}}$
$=\frac{x-1}{x}. x\neq{0}, x\neq{1}$

let $y=f(x)$
$y(1-x)=1$
$x=\frac{y-1}{y}$
$f^{-1}(x)=\frac{x-1}{x}, x\neq{1}, x\neq{0}$

Since $D_{f^{-1}}=R_{f}=\mathbb{R} \backslash \{0,1\}$

We have that $f^{2}(x)=f^{-1}(x)$

(ii)
$f^{2}(x)=f^{-1}(x)$
$f[f^{2}(x)]=f[f^{-1}(x)]$
$f^{3}(x)=x, x\neq{1}, x\neq{0}$

Students must understand what it means for two functions to be equal. It isn’t just about the rule (loosely put, formula) the same, but they require the same domain, and as a result the same range too. Stating the domains in (i) is important to show true understanding of what equal functions mean. It is saddening that some students also thought that $f^{2}(x)=f(x)f(x)$ which is not the case. $f^{2}(x)$ is a composite function, “f in f” while $f(x)f(x)$ is a product of two functions.