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)

Graph for 3i
Graph for 3i

(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)

Graph for 4iii
Graph for 4iii

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


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