### 2013 A-level H1 Mathematics (8864) Question 3 Suggested Solutions

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

(i)
Perimeter $= 3x + 4x + 2y + 5x = 20$

$\Rightarrow y = 10 -6x$

Area, $S = 4xy + \frac{1}{2}(3x)(4x) = 40 x - 18x^{2}$

(ii)
$\frac{dS}{dx} = 40 - 36x = 0$

$\Rightarrow x = \frac{10}{9}$

Since $latex \frac{d^{2}S}{dx^{2}} = -36 < 0$, S is maximum when $latex x = \frac{10}{9}$ Required Area $latex = 40(\frac{10}{9}) - 18(\frac{10}{9})^{2} = \frac{200}{9}$

Students should bare in mind to check the second order to substantiate that its a maximum area.

### 2013 A-level H1 Mathematics (8864) Question 2 Suggested Solutions

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

(i)
$\frac{d}{dx} \mathrm{ln} (1+2x^{2}) = \frac{4x}{1+2x^{2}}$

(ii)
$\int_{-1}^{0} \frac{1}{{1-3x}^{4}} dx$

$= -\frac{1}{3} [-\frac{1}{3(1-3x)^{3}}]\biggl|_{-1}^0$

$= \frac{1}{9} [\frac{63}{64}]$

$= \frac{7}{64}$

Be careful when you resolve the definite integral. Students should use the graphing calculator to check their answers!

### 2013 A-level H1 Mathematics (8864) Question 1 Suggested Solutions

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Since equation has no real roots, $latex Discriminant < 0$ $latex (k-2)^{2} - 4(2k+1) < 0$ $latex k^{2} - 12k < 0$ $latex \therefore, 0 < k < 12$

A very simple question that test student on their understanding of discriminant.

### 2013 A-level H1 Mathematics (8864) Question 11 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 mass of food content in a type A packet, in grams
Let Y denote the mass of food content in a type B packet, in grams.

$X ~\sim~ \mathrm{N}(1000, \sigma^{2})$
$Y ~\sim~ \mathrm{N}(1010,428)$

$latex \mathrm{P} (X < 990) = 0.2$ $latex \mathrm{P} (Z < \frac{990 - 1000}{\sigma}) = 0.2$ $latex -\frac{10}{\sigma} = -0.84162$ $latex \sigma = 11.8$ (ii) $latex \mathrm{P} (Y < 1000) = 0.314$ (iii) $latex Y - X ~\sim~ \mathrm{N}(10, 569.18)$ $latex \mathrm{P} (Y > X) = \mathrm{P} (Y – X > 0) = 0.662$

Students are expected to show the workings for expectation and variance clearly here.

### 2013 A-level H1 Mathematics (8864) Question 7 Suggested Solutions

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

Unbiased estimate of mean $= \frac{305}{250} + 75 = 76.22$

Unbiased estimate of variance $= \frac{1}{249} (29555 - \frac{306^{2}}{250}) = 117$

$H_0: \mu = 75$

$H_1: \mu > 75$

Under $H_0$, $\bar{X} ~\sim~ \mathrm{N} (75 \frac{117.2004}{250})$.

From graphing calculator, p-value $= 0.0374 > 0.025$, we do not reject $H_0$.

Thus, there is insufficient evidence at 2.5% level of significance that this type of device can retain information for more than 75 hours.

This question is fairly manageable.

### 2013 A-level H2 Mathematics (9740) Paper 2 Question 12 Suggested Solutions

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(i)
The mean number of absences per day is constant.

(ii)

Let A be the number of absentees from Admin Department on n days.

$A~\sim~\mathrm{Po}(1.2n)$

$latex \mathrm{P}(A=0) = e^{-1.2n} < 0.01$ $latex n > 3.8$

$\therefore, \mathrm{smallest~number~of~days} = 4$

(iii)

Let T be the number of absentees from two Departments on 5 days.

$T~\sim~\mathrm{Po}(19.5)$

$\mathrm{P}(T>20) = 1- \mathrm{P}(T \le 20) = 0.397$

(iv)

Let S be the number of absentees from two Departments on 60 days.

$S~\sim~\mathrm{Po}(234)$

Since $\lambda = 234 > 10, S ~\sim~N(234, 234)$ approximately.

$\mathrm{P}(200 \le S \le 250) = \mathrm{P}(199.5 \le S \le 250.5) = 0.848$

There are several answers to (i), please refer to school notes. Some students still forget to do continuity correction.

### 2013 A-level H2 Mathematics (9740) Paper 2 Question 11 Suggested Solutions

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

(i)

Required Probability $= \frac{26 \times 25 \times 24 \times 9 \times 8}{26 \times 26 \times 26 \times 9 \times 9}$

(ii)

Required Probability $= \frac{1-\mathrm{P}(\mathrm{same digits})}{2}$
$= \frac{1-\frac{9}{9^{2}}}{2}$
$= \frac{4}{9}$

(iii)
Case 1: 2 same letter and 2 different digits

$26 \times 25 \times \frac{3!}{2!} \times {^9\!P_2}=140400$

Case 2: different letter and 2 same digits

$9 \times {^26\!P_3}=140400$

Case 3:3 same letter and 2 same digits

$9 \times 26 =234$

Required Probability $= \frac{140400+140400+234}{26 \times 26 \times 26 \times 9 \times 9} = 0.197$

(iv)
Case 1: 2 different consonants, 1 vowel & 1 even digit

${^{21}\!C_2} \times {^5\!C_1} \times 3! \times {^4\!C_1} \times {^5\!C_1} \times 2! = 252000$

Case 2: 2 same consonant, 1 vowel & 1 even digit

${^{21}\!C_1} \times {^5\!C_1} \times \frac{3!}{2!} \times {^4\!C_1} \times {^5\!C_1} \times 2! = 12600$

Required Probability $= \frac{252000+12600}{26 \times 26 \times 26 \times 9 \times 9} = 0.186$

This question is not that easy. A lot of thinking is required; but that is the case with all permutation questions. I strongly suggest students to not spend too much time on permutation questions and come back to solve it later.

### 2013 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)

(ii)

(iii)
Since the date points exhibit a curvilinear relationship with a negative gradient and concaving downwards, Model A is the most appropriate.

From the Graphing Calculator, r $= -0.939$

(iv)
From the Graphing Calculator, required regression line: $y = 189.748 - 0.0046198 x^{2}$
$\therefore,~ \mathrm{Distance~travelled} = 134 \mathrm{km}$

The first part of this question tests students on their understanding of graphs. Students that have difficulties with this, should attempt to fit in simple values into graphing calculator.

### 2013 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.

9
(i)
From the graphing calculator,

unbiased estimate of $\mu = 12.8$

unbiased estimate of ${\sigma}^{2} = 2.31$

(ii)
We assume that 2.31 is a good estimate of the unknown population variance.
We assume the distance travelled per litre of fuel by a car is independent of another car.

$H_0: \mu = 13.8$
$H_1: \mu < 13.8$ Under $H_0$, perform a left-tailed T-test.

Test statistic, $T ~ = ~ \frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} ~ \sim ~ t(7)$

Using the graphing calculator, p-value $= 0.052397 > 0.05$, thus we do not reject $H_0$.

There is insufficient evidence at 5% significance level to say that the distance travelled per litre of fuel is too high.

This is a fairly manageable question that should be attempted. The only issue is probably with the assumption part. Students cannot use their regular “assume population mean followed a normal distribution” assumption since the question stated it. So they need to be more creative. As to what constitutes a good estimate, we consider if they are consistent and unbiased.

### 2013 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)
$\mathrm{P}(B|A') = \frac{\mathrm{P}(B \cap A')}{\mathrm{P}(A')}$

$0.8 (1- 0.7) = \mathrm{P}(B \cap A')$

$\mathrm{P}(B \cap A') = 0.24$

(ii)
$\mathrm{P}(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$

$P(A \cup B) = P(B \cap A') + P(A) = 0.94$

$\therefore, \mathrm{P}(A' \cap B') = 0.06$

(iii)
$\mathrm{P}(A|B') = \frac{\mathrm{P}(A \cap B')}{\mathrm{P}(B')}$

$0.88[\mathrm{P}(A \cap B') + \mathrm{P}(A' \cap B')] = \mathrm{P}(A \cap B')$

$0.12\mathrm{P}(A \cap B') = 0.0528$

$\mathrm{P}(A \cap B') = 0.44$

$\mathrm{P}(A ) = \mathrm{P}(A \cap B) + \mathrm{P}(A \cap B')$

$\mathrm{P}(A \cap B) = 0.26$