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 H1 Mathematics (8864) Paper 1 Suggested Solutions […]

• Elston

Hello! I’m not sure whether it’s something wrong with my phone or some other problem like that but I can’t see any of the answers!

• hi, sorry. but h1 has to wait. H2 was more fun. So I did H2 first..

• Elston

• Helloooo

Hello roughly how much do you need to get in order to get an A in h1 math?

• asdfghjkl

what time will the answers be out? ESTIMATE PLS

• Hello

• lydia

thanks a lot for trying to upload the ans ASAP!

• jam

hello! for 6iii), the question mentioned in hundreds of dollars, is there a need to take all the answers x100?

• no need. since its already defined to be X

• Elston

This is an odd question but if you leave the answer as mean=$3.9, and variance=$1.41, will I not get the answer mark because of the units? Or is it alright?

• the mean is fine. but variance will need to be $\^2$ which is rather awkward.

• Elston

I see. Thanks! So I guess that’s one answer mark gone, sighs. Thanks agn!

Hello! Will the rest of the answers be uploaded as well? Thanks very much for getting all these answers out for us 🙂

• yea. haha cos I’m having lessons now. But will be out.

• millie

hi! based on the difficulty of this paper, what mark range do you think A and B will be? 🙂

• asdfghjkl

hey how can 8ii be a negative probability

• its a typo. you can just key the answer in yourself.

• Jack

The probability question is wrg? How can u have a negative probability

• just key it into the GC yourself. I’m having lessons now for H2 math.

• Juju

what’s the rough marks to get an A for h1 maths in A levels?

• Hi, I think there is an error in 8(iii), how can it be that your answer in 8(ii) is 11/18 (different balls) and yet you use 11/18 for same balls? I think it should be 1/30 divided by 7/18, giving you an answer of 3/35. Thanks for uploading the answers, greatly appreciate it 🙂

• hi, thanks for that. cos i’m just blindly typing them on the phone as I’m having H2 Math class now. Too careless.

• eeeeee

Hello! For Q4iii, isn’t there 3 x intersections?

• yea, my bad. didnt read the question. sorry, i’ll typing the h1 on the phone while having h2 math classes now. cos i only planned on doing h1 at 10pm lol
Thanks for pointing it out!

• Q4 asking for coordinates that cut x-axis and it should be 3 points (1,0), (-0.157,0), (-1.593,0). Cutting at y-axis is not required

• oh right. didnt read the question. sorry, i’ll typing the h1 on the phone while having h2 math classes now. cos i only planned on doing h1 at 10pm lol.
Thanks for pointing it out!

• test

getting 95/100 will definitely be A even if the bell curve is high right?

• haha idk. it seems like everybody is getting >90 for this paper… tough moderation.:/

• random passerby

Just a minor mistake for 11(v) when you rounded it off to 3 s.f the decimal point went missing. Omg I just realised my careless mistakes and how I didn’t x 1/2 for the probability qn I am feeling q depressed now LOL ohwell thanks for uploading the answers for all of us! 🙂 gladly appreciate it 🙂

• Oops! Thanks so much for pointing that out! Its over! Focus on the papers to come 🙂

• why 8(i) need to multiply by 1/2

• random passerby

You have to pick the box first that’s why x 1/2 I guess! I didn’t do that too bc I was afraid it’ll lead to “double counting” 🙁

• Yup, passerby is right. you need to consider which box you’re picking cos they said a box is picked at random. 🙂

• Juju

For 10ii why do we test the claim at less than 95 instead of more than 95?

• J

Does anybody know how h1 bellcurves are usually like? Is 70 an A, or really harsh like >90?

• its harsh. especially given this paper, tough moderation

• It is a question that confused us with words. The last part stated ” he finds that the mean top speed….is NOT greater than 95″, which have to reject the Ho that is greater than 95.

• Juju

oh I see thank you! 🙂

• Daphne

Hello, I think for Q10ii, you should test for u>95 since Probability of X being more than 95 is not achieved (insufficient evidence to prove H1 is true) and hence H0 is not rejected, and is more than 0.05 (critical region)

P(X > x) > 0.05, hence find X values to satisfy this using sample mean distribution.

No? 🙂

• Daphne

I took “he finds that mean top speed is not greater than 95…” to refer to the rejection of H1 and to work from this rejection

• Daphne

*sry last comment*
You rejected H0: u = 95 to derive the range of sample mean values.
But question only stated not more than 95, so there is a possibility of u=95 and u95 and thus not reject H0: u=95 to derive our sample values from there 🙂

• Daphne

*Possibility of u=95 and u<95 (less than 95) oh god sorry for the spam just very worried

• Hi, yup you’re right. I didnt read carefully enough. thanks! and sorry for the confusion

• Lmao

For 10 ii, can my alternate hypothesis more than 95? After which I show that Ho is valid, hence mean is not greater than 95.

And questiom 11vi, isnt the p value in part v more accurate? Because the increase in sample size will provide a more accurate representation of the entire population?

• you’re correct. sorry I forgot to switch the inequality in 10ii

• Daphe, thanks for your sharing. Then my mark gone-lah ;-(

• yea, this question is really tricky. Must really read precisely. :/

• Hellohello

For 10ii shouldn’t Zcal>Zcritical since we do not reject Ho? Then shouldn’t the answer be 93.sth if I’m not wrong?

• Hi, this is a right tailed test. So to not reject $H_0, ~ Z_{calc} \textless Z_{crit}$