Thoughts on A’levels H2 Mathematics 2016 Paper 2

I’ll keep this short since we are all busy. One thing about paper 1 we saw, there were many unknowns.

So topics which I think will come out…

Differentiation – I think a min/max problem will come out, possibly with r and h both not given and asked to express r in terms of h. But students should revise a on the properties of curves with differentiation; given a curve equation with an unknown, for instance $y=/frac{x^2+kx+1}{x-1}$, find the range of k such that there is stationary points.

Complex Number – Loci will definitely come out. I’m saying they will combine with trigonometry.

Integration – Modulus integration hasn’t really been tested. Else a question on Area/ Volume could be tested, and I’ll say they need students to do some
Conics too.

For statistics, my students should have gotten the h1 stats this year. And if it’s an indicator, then it should not be a struggle.

I expect PnC and probability to be combined. Conditional Probability in a poisson question should be tested too, so do revise it well. For hypothesis testing, students should be careful of their formula and read really carefully about the alternative hypothesis. Also, :9 know that the formulas for poison PDF and binompdf are both given in mf15. Lastly, know when to use CLT.

All the best!

Random Questions from 2016 Prelims #13

NYJC/2/11

On a typical weekday morning, customers arrive at the post office independently and at a rate of 3 per 10 minute period.

(i) State, in context, a condition needed for the number of customers who arrived at the post office during a randomly chosen period of 30 minutes to be well modelled by a Poisson distribution.

(ii) Find the probability that no more than 4 customers arrive between 11.00 a.m. and 11.30 a.m.

(iii) The period from 11.00 a.m. to 11.30 a.m. on a Tuesday morning is divided into 6 periods of 5 minutes each. Find the probability that no customers arrive in at most one of these periods.

The post office opens for 3.5 hours each in the morning and afternoon and it is noted that on a typical weekday afternoon, customers arrive at the post office independently and at a rate of 1 per 10 minute period. Arrivals of customers take place independently at random times.

(iv) Show that the probability that the number of customers who arrived in the afternoon is within one standard deviation from the mean is 0.675, correct to 3 decimal places.

(v) Find the probability that more than 38 customers arrived in a morning given that a total of 40 customers arrived in a day.

(vi) Using a suitable approximation, estimate the probability that more than 100 customers arrive at the post office in a day.

2016 A-level H1 Mathematics (8864) Paper 1 Suggested Solutions

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

As these workings and answers are rushed out asap, please pardon me for my mistakes and let me know if there is any typo. Many thanks.
I’ll try my best to attend to the questions as there is H2 Math Paper 2 coming up.

Question 1
(i)
$\frac{d}{dx} [2 \text{ln}(3x^2 +4)]$

$= \frac{12x}{3x^2+4}$

(ii)
$\frac{d}{dx} [\frac{1}{2(1-3x)^2}]$

$= \frac{6}{2(1-3x)^3}$

$= \frac{3}{(1-3x)^3}$

Question 2

$2e^{2x} \ge 9 - 3e^x$

$2u^2 + 3u - 9 \ge 0$

$(2u-3)(u+3) \ge 0$

$\Rightarrow u \le -3 \text{~or~} u \ge \frac{3}{2}$

$e^x \le -3$ (rejected since $e^x > 0$) or $e^x \ge \frac{3}{2}$

$\therefore x \ge \text{ln} \frac{3}{2}$

Question 3
(i)

(ii)
Using GC, required answer $= -1.606531 \approx -1.61$ (3SF)

(iii)
When $x = 0.5, y = 0.35653066$

$y - 0.35653066 = \frac{-1}{-1.606531}(x-0.5)$

$y = 0.622459 x +0.04530106$

$y = 0.622 x + 0.0453$ (3SF)

(iv)

$\int_0^k e^{-x} -x^2 dx$

$= -e^{-x} - \frac{x^3}{3} \bigl|_0^k$

$= -e^{-k} - \frac{k^3}{3} + 1$

Question 4
(i)
$y = 1 + 6x - 3x^2 -4x^3$

$\frac{dy}{dx} = 6 - 6x - 12x^2$

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

$6 - 6x - 12x^2 = 0$

$1 - x - 2x^2 = 0$

$2x^2 + x - 1 = 0$

$x = -1 or \frac{1}{2}$

When $x =-1, y = -4$

When $x = \frac{1}{2}, y = 2.75$

Coordinates $= (-1, -4) \text{~or~} (0.5, 2.75)$

(ii)
$\frac{dy}{dx} = 6 - 6x - 12x^2$

$\frac{d^2y}{dx^2} = - 6 - 24x$

When $x = -1, \frac{d^2y}{dx^2} = 18 > 0$. So $(-1, -4)$ is a minimum point.

When $x = 0.5, \frac{d^2y}{dx^2} = -18 \textless 0$. So $(0.5, 2.75)$ is a maximum point.

(iii)

x-intercept $= (-1.59, 0) \text{~and~} (1, 0) \text{~and~} (-0.157, 0)$

(iv)
Using GC, $\int_0.5^1 y dx = 0.9375$

Question 5
(i)
Area of ABEDFCA $= \frac{1}{2}(2x)(2x)\text{sin}60^{\circ} - \frac{1}{2}(y)(y)\text{sin}60^{\circ}$

$2\sqrt{3} = \sqrt{3}x^2 - \frac{\sqrt{3}}{4}y^2$

$2 = x^2 - \frac{y^2}{4}$

$4x^2 - y^2 =8$

(ii)

Perimeter $= 10$

$4x+2y + (2x-y) = 10$

$6x + y = 10$

$y = 10 - 6x$

$4x^2 - (10-6x)^2 = 8$

$4x^2 - 100 +120x -36x^2 = 8$

$32x^2 -120x+108=0$

$x=2.25 \text{~or~} 1.5$

When $x=2.25, y = -3.5$ (rejected since $y >0$)

When $x=1.5, y = 1$

Question 6
(i)
The store manager has to survey $\frac{1260}{2400} \times 80 = 42$ male students and $\frac{1140}{2400} \times 80 = 38$ female students in the college. He will do random sampling to obtain the required sample.

(ii)
Stratified sampling will give a more representative results of the students expenditure on music annually, compared to simple random sampling.

(iii)
Unbiased estimate of population mean, $\bar{x} = \frac{\sum x}{n} = \frac{312}{80} = 3.9$

Unbiased estimate of population variance, $s^2 = \frac{1}{79}[1328 - \frac{312^2}{80}] = 1.40759 \approx 1.41$

Question 7
(i)
[Venn diagram to be inserted]

(ii)
(a)
$\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B) = 0.8$
(b)
$\text{P}(A \cup B) - \text{P}(A \cap B) = 0.75$

(iii)
$\text{P}(A | B')$

$= \frac{\text{P}(A \cap B')}{\text{P}(B')}$

$= \frac{\text{P}(A) - \text{P}(A \cap B)}{1 - \text{P}(B)}$

$= \frac{0.6 - 0.05}{1-0.25}$

$= \frac{11}{15}$

Question 8
(i)
Required Probability $= \frac{1}{2} \times \frac{3}{10} \times \frac{2}{9} = \frac{1}{30}$

(ii)
Find the probability that we get same color. then consider the complement.

Required Probability $= 1 - \frac{1}{30} - \frac{1}{2} \times \frac{5}{10} \times \frac{4}{9} - \frac{1}{2} \times \frac{2}{10} \times \frac{1}{9} - \frac{1}{2} \times \frac{4}{6} \times \frac{3}{5} - \frac{1}{2} \times \frac{2}{6} \times \frac{1}{5} = \frac{11}{18}$

(iii)
$\text{P}(\text{Both~balls~are~red} | \text{same~color})$

$= \frac{\text{P}(\text{Both~balls~are~red~and~same~color})}{\text{P}(\text{same~color})}$

$= \frac{\text{P}(\text{Both~balls~are~red})}{\text{P}(\text{same~color})}$

$= \frac{1/30}{7/18}$

$= \frac{3}{35}$

Question 9
(i)
Let X denote the number of batteries in a pack of 8 that has a life time of less than two years.

$X \sim B(8, 0.6)$
(a)
Required Probability $=\text{P}(X=8) = 0.01679616 \approx 0.0168$

(b)
Required Probability $=\text{P}(X \ge 4) = 1 - \text{P}(X \le 3) = 0.8263296 \approx 0.826$

(ii)
Let Y denote the number of packs of batteries, out of 4 packs, that has at least half of the batteries having a lifetime of less than two years.

$Y \sim B(4, 0.8263296)$

Required Probability $=\text{P}(Y \le 2) = 0.1417924285 \approx 0.142$

(iii)
Let W denote the number of batteries out of 80 that has a life time of less than two years.

$W \sim B(80, 0.6)$

Since n is large, $np = 48 > 5, n(1-p)=32 >5$

$W \sim N(48, 19.2)$ approximately

Required Probability

$= \text{P}(w \ge 40)$

$= \text{P}(w > 39.5)$ by continuity correction

$= 0.9738011524$
$\approx 0.974$

Question 10
(i)
Let X be the top of speed of cheetahs.
Let $\mu$ be the population mean top speed of cheetahs.

$H_0: \mu = 95$

$H_1: \mu \neq 95$

Under $H_0, \bar{X} \sim N(95, \frac{4.1^2}{40})$

Test Statistic, $Z = \frac{\bar{X}-\mu}{\frac{4.1}{\sqrt{40}}} \sim N(0,1)$

Using GC, $p=0.0449258443 \textless 0.05 \Rightarrow H_0$ is rejected.

(ii)
$H_0: \mu = 95$

$H_1: \mu > 95$

For $H_0$ to be not rejected,

$\frac{\bar{x}-95}{\frac{4.1}{\sqrt{40}}} \textless 1.644853626$

$\bar{x} \textless 96.06 \approx 96.0$ (round down to satisfy the inequality)

$\therefore \{ \bar{x} \in \mathbb{R}^+: \bar{x} \textless 96.0 \}$

Question 11
(i)
[Sketch to be inserted]

(ii)
Using GC, $r = 0.9030227 \approx 0.903$ (3SF)

(iii)
Using GC, $y = 0.2936681223 x - 1.88739083$

$y = 0.294 x - 1.89$ (3SF)

(iv)
When $x = 16.9, y = 3.0756 \approx 3.08$(3SF)

Time taken $= 3.08$ minutes

Estimate is reliable since $x = 16.9$ is within the given data range and $|r|=0.903$ is close to 1.

(v)
Using GC, $r = 0.5682278 \approx 0.568$ (3SF)

(vi)
The answers in (ii) is more likely to represent since $|r|=0.903$ is close to 1. This shows a strong positive linear correlation between x and y.

Question 12
Let X, Y denotes the mass of the individual biscuits and its empty boxes respectively.

$X \sim N(20, 1.1^2)$

$Y \sim N(5, 0.8^2)$

(i)
$\text{P}(X \textless 19) = 0.181651 \approx 0.182$

(ii)
$X_1 + ... + X_12 + Y \sim N(245, 15.16)$

$\text{P}(X_1 + ... + X_12 + Y > 248) = 0.2205021 \approx 0.221$

(iii)
$0.6X \sim N(12, 0.66^2)$

$0.2Y \sim N(1, 0.16^2)$

Let $A=0.6X$ and $B = 0.2Y$

$A_1+...+A_12+B \sim N(145, 5.2528)$

$\text{P}(142 \textless A_1+...+A_12+B \textless 149) = 0.864257 \approx 0.864$

2016 A-level H2 Mathematics (9740) 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: $0.0251 \text{~m/min}$
Question 2: $\frac{x^2}{2} \text{sin}nx + \frac{2x}{n^2} \text{cos}nx - \frac{2}{n^3} \text{sin}nx + C;~ a = 2 \text{~or~} 6;~ \pi (\frac{2}{5} - \text{ln} \frac{3}{\sqrt{5}})$
Question 3: $a + \frac{3}{4} - \text{sin}a - \text{cos}a + \frac{1}{4} \text{cos}2a;~ \frac{1}{4}(\pi + 1)^2$
Question 4: $z = 2.63 + 1.93i, ~ 3.37 + 0.0715i ;~8^{\frac{1}{6}}e^{i(-\frac{\pi}{12})}, ~8^{\frac{1}{6}}e^{i(-\frac{3 \pi}{4})}, ~8^{\frac{1}{6}}e^{i(\frac{7\pi}{12})};~ n=7$
Question 5: $\frac{11}{42};~ \frac{4}{11};~ \frac{4}{1029}$
Question 6: $60;~ 10;~ \{ \bar{x} \in \mathbb{R}, 0 \textless \bar{x} \le 34.8 \};~ \{ \alpha \in \mathbb{R}, 0 \textless \alpha \le 8.68 \}$
Question 7: $24;~ 576;~ \frac{1}{12};~ \frac{5}{12}$
Question 8: $r = -0.980;~ c = -17.5;~ d = 91.8;~ y = 85.9$
Question 9: $a = 7.41;~ p = 0.599 \text{~or~} 0.166;~ 0.792$
Question 10: $0.442;~ 0.151 ;~ 0.800;~ \lambda = 1.85$

MF15

June Revision Exercise

You can find the solutions of all ten sets of the June Revision Exercise we did in class.

Have fun!

June Revision Exercise 1
June Revision Exercise 2
June Revision Exercise 3
June Revision Exercise 4
June Revision Exercise 5
June Revision Exercise 6
June Revision Exercise 7
June Revision Exercise 8
June Revision Exercise 9
June Revision Exercise 10

Statistics related articles

Here is a compilation of all the Statistics articles KS has done. Students should read them when they are free to improve their mathematics skills. They will come in handy! 🙂

JC2 Statistics 2016

Most JCs by now should be embarking on their Statistics Journey. I know a school that is finishing up really fast, and I think they will easily finish the syllabus by May. So I notice many students are thrown off the track with Statistics. It is slightly out of their comfort zone, especially Permutation and Combinations. The topic itself is really sneaky. Here’s are some advices for Statistics.

1. Read every question carefully. Words like, “at least”, “not more than”, “if” and “given”.
2. Many students seem to struggle with the idea of conditional probability and that should not the case. Students should learn what it means and how it works, instead of it being simply $\mathrm{P}(A | B) = \frac{\mathrm{P}(A \Cap B)}{\mathrm{P}(B)}$. Understand how this came about will really save your 6-9 marks probability question.
3. Learn to break the question down to something you’re familiar with. As what we review in class, write probabilities in a format that you understand easily. Probability is often counter-intuitive, stick to the formulas.
4. Yes, the whole statistics is one big topic by itself. What you learnt in Permutation & Combination is used in probability (e.g. A’levels 2013 P2 Q11), and probability is used in all the distributions you will be learning, lastly normal distributions is used in hypothesis testing too.
5. Statistics is rather qualitative, so students should show understanding in what they learn. Topics such as sampling requires you to think rationally and consider what limitations we face. A clear understanding of your contents and definitions will go a long way.

2010 A-level H2 Mathematics (9740) Paper 2 Question 7 Suggested Solutions

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

(i)
$\mathrm{P}(A|B') = \frac{\mathrm{P} (A \cap B')}{\mathrm{P}(B')}$
$\Rightarrow \mathrm{P} (A \cap B') = 0.8 \times (1-0.6) = 0.32$

(ii)
$\mathrm{P}(A \cup B) = \mathrm{P}(A \cap B') + \mathrm{P}(B) = 0.32 + 0.6 = 0.92$

(iii)
P(B’|A) $= \frac{\mathrm{P} (B' \cap A)}{\mathrm{P}(A)} = \frac{0.32}{0.7}= 0.457$

(iv)
$\mathrm{P} (A' \cap C) = \mathrm{P} (A')\mathrm{P} (C)$ since A and C are independent
$\Rightarrow \mathrm{P} (A' \cap C) = 0.15$

(v)
$0 \le \mathrm{P} (A' \cap B \cap C) \le 0.15$

2015 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 and Y denote the mass of a apple and pear in grams respectively.
$\mathbb{E}(X_1 + \ldots X_5) = 5(300) = 1500$

$\mathrm{Var}(X_1 + \ldots X_5) = 5(20^2) = 2000$

$X_1 + \ldots X_5 \sim \mathrm{N}(1500, 2000)$

$\mathrm{P}(X_1 + \ldots X_5 > 1600) \approx 0.0127$ (3 SF)

(ii)
$\mathbb{E}[X_1 + \ldots X_5 - (Y_1 + \ldots Y_8)] = 5(300) - 8(200) = -100$

$\mathrm{Var}[X_1 + \ldots X_5 - (Y_1 + \ldots Y_8)] = 5(20^2) + 8(15^2) = 3800$

$X_1 + \ldots X_5 - (Y_1 + \ldots Y_8) \sim \mathrm{N}(-100, 3800)$

$\mathrm{P}(X_1 + \ldots X_5 - (Y_1 + \ldots Y_8) > 0) \approx 0.0524$ (3 SF)

(iii)
Let A and B denote the mass of a peeled apple and peeled pear in grams respectively.
$\mathbb{E}(A)= \mathbb{E}(0.85X) = 0.85(300)=255$

$\mathbb{E}(B)= \mathbb{E}(0.90Y) = 0.90(200)=180$

$\mathrm{Var}(A) = \mathrm{Var}(0.85X) = 0.85^2(20^2) = 289$

$\mathrm{Var}(B) = \mathrm{Var}(0.90Y) = 0.90^2(15^2) = 182.25$

$\mathbb{E}[A_1 + \ldots A_5 + B_1 + \ldots B_8)] = 5(255) + 8(180) = 2715$

$\mathrm{Var}[A_1 + \ldots A_5 + B_1 + \ldots B_8)] = 5(289) + 8(182.25) = 2903$

$A_1 + \ldots A_5 + B_1 + \ldots B_8 \sim \mathrm{N}(2715, 2903)$

$\mathrm{P}(A_1 + \ldots A_5 + B_1 + \ldots B_8 ~ \textless ~ 2750) \approx 0.742$ (3 SF)