## 2015 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.

(a)
(i)
$f(x) = \frac{1}{1-x^2}, x \in \mathbb{R}, x>1$

$f'(c) = \frac{2x}{1-x^2} > 0$ for all $x \in \mathbb{R}, x>1$

Since $f(x)$ is increasing function, it has no turning points, f is 1-1, its inverse will exist.

(a)
(ii)
Let $f(x) = y$

$x = \pm \sqrt{1- \frac{1}{y}}$ Reject $-\sqrt{1- \frac{1}{y}}$ since $x>1$

$\Rightarrow f^{-1}(x) = \sqrt{1- \frac{1}{x}}$

$D_{f^{-1}} = \{x \in \mathbb{R} | x ~ \textless ~ 0 \}$

(b)
Let $g(x) = y = \frac{2+x}{1-x^2}$

We need $b^2 - 4ac \ge 0$ for all real values of g.

$(1)^2 - 4(y)(2-y) \ge 0$

$1 - 8y + 4y^2 \ge 0$

$y = \frac{8 \pm \sqrt{48}}{8} = 1 \pm \frac{\sqrt{3}}{2}$

$y \ge 1 + \frac{\sqrt{3}}{2} \mathrm{~or~} y \le 1 - \frac{\sqrt{3}}{2}$

$R_g = (-\infty, 1- \frac{\sqrt{3}}{2} ] \cup [1 + \frac{\sqrt{3}}{2}, \infty )$

## 2011 A-level H2 Mathematics (9740) Paper 2 Question 9 Suggested Solutions

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

(i)
(a)
P(faulty)
= P(made by A and faulty) + P(made by B and faulty)
= 0.6(0.05) + 0.4(0.07)
= 0.058

(b)
= $\frac{\mathrm{P(made~by~A~and~faulty)}}{\mathrm{P(faulty)}}$
= $\frac{0.6(0.05)}{0.058}$
= $\frac{15}{29}$

(ii)
(a)
P(exactly one of them is faulty)
= $0.058 \times (1 - 0.058) \times 2!$
= 0.109272 (exact)

(b)
P(both were made by A | exactly one is faulty)
= $\frac{\mathrm{P(both~were~made~by~A~and~exactly~one~is~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}$
= $\frac{\mathrm{P(one~is~made~by~A~and~faulty,~the~other~is~made~by~A~and~not~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}$
= $\frac{0.6(0.05) \times 0.6(0.95) \times 2!}{0.109272}$
= $\frac{1425}{4553}$

Question can be easily solved by drawing a tree diagram. Do take note that we only wrong off NON exact answers to 3sf, so for (iia), we keep the full exact answer.

## 2010 A-level H2 Mathematics (9740) Paper 2 Question 2 Suggested Solutions

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

(i)
Let $P(n)$ be the statement: $\sum_{r=1}^n r(r+2) = \frac{1}{6}n(n+1)(2n+7), n \in \mathbb{Z}^+$

When $n = 1, ~\text{LHS} = 1(3)=3, \text{~and~RHS} = \frac{1}{6}(2)(9) = 3 = \text{LHS}$

$\therefore, P(n)$ is true.

Assume that $P(k)$ is true for some $k \in \mathbb{Z}^+$, i.e. $\sum_{r=1}^k r(r+2) = \frac{1}{6}k(k+1)(2k+7)$

Want to prove that $P(k+1)$ is true, i.e. $\sum_{r=1}^{k+1} r(r+2) = \frac{1}{6}(k+1)(k+2)(2k+9)$

LHS

$= \sum_{r=1}^{k+1} r(r+2)$

$= \sum_{r=1}^{k} r(r+2) + (k+1)(k+3)$

$= \frac{1}{6}k(k+1)(2k+7) + (k+1)(k+3)$

$= \frac{1}{6}(k+1)[k(2k+7)+6(k+3)]$

$= \frac{1}{6}(k+1)(2k^2+13k+18)$

$= \frac{1}{6}(k+1)(k+2)(2k+9)$

$= \text{RHS}$

Since $P(1)$ is true and $P(k) \text{~is~true} \Rightarrow P(k+1)$ is true, hence by Mathematical Induction, $P(n)$ is true for all $n \in \mathbb{Z}^+$

(ii)
(a)

$\frac{1}{r(r+2)} = \frac{1}{2r} - \frac{1}{2(r+2)}$

$\sum_{r=1}^n \frac{1}{r(r+2)}$

$= \sum_{r=1}^n \frac{1}{2r} - \frac{1}{2(r+2)}$

$= \frac{1}{2} - \frac{1}{2(3)}$

$+ \frac{1}{2(2)} - \frac{1}{2(4)}$

$+ \frac{1}{2(3)} - \frac{1}{2(5)}$

$+ \frac{1}{2(n-1)} - \frac{1}{2(n+1)}$

$+ \frac{1}{2n} - \frac{1}{2(n+2)}$

$= \frac{1}{2} + \frac{1}{2(2)} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)}$

$= \frac{3}{4} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)}$

(b)
As $n \rightarrow \infty, \frac{1}{2(n+1)} \rightarrow 0, \frac{1}{2(n+2)} \rightarrow 0$

Hence, $\sum_{r=1}^{\infty} \frac{1}{r(r+2)}$ is a convergent series, and the value of the sum to infinity is $\frac{3}{4}$

## 2010 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)
$x^2 -6x + 34 = 0$

$x = \frac{6 \pm \sqrt{36-136}}{2}$

$x = 3 \pm 5i$

(ii)
Since the coefficients are all real, another root of the equation is $-2-i$

$[x-(-2+i)][x-(-2-i)]$

$= x^2 + 4x +5$

By comparing coefficients,

$x^4 + 4x^3 + x^2 + ax + b = (x^2+4x+5)(x^2-4)$

$\therefore, a=-16, b = -20$

The other three roots are $-2 - i, 2, -2$