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

Paper 2

Question 1
(i)
16 -\frac{1}{2} h \ge 0

\Rightarrow h \le 32

\therefore, \mathrm{maximum~h}=32m

(ii)
\frac{dh}{dt} = \frac{1}{10} \frac{\sqrt{32-h}}{\sqrt{2}}

\int \frac{1}{\sqrt{32-h}} ~dh = \int \frac{1}{10 \sqrt{2}}~dt

-2 \sqrt{32-h} + C = \frac{1}{10 \sqrt{2}}t

t = - 20 \sqrt{64-2h} + 10 \sqrt{2}C

When t = 0, h = 0, C = \frac{16}{\sqrt{2}}

\therefore, t = - 20 \sqrt{64-2h} + 160

When h = 16m, t = 46.9 years.


 

 

Question 2
(i)
\theta = \mathrm{cos^{-1}} |\frac{\begin{pmatrix}2\\3\\{-6}\end{pmatrix}\cdot \begin{pmatrix}1\\0\\{0}\end{pmatrix}}{\sqrt{49} \cdot 1}| = 73.4^{\circ}

(ii)
Let \vec{ON} be point on L that makes \sqrt{33} from P.

\vec{ON} = \begin{pmatrix}1\\{-2}\\{-4}\end{pmatrix} + \lambda \begin{pmatrix}2\\3\\{-6}\end{pmatrix} for some \lambda

\vec{PN} = \begin{pmatrix}{-1}\\{-7}\\{2}\end{pmatrix} + \lambda \begin{pmatrix}2\\3\\{-6}\end{pmatrix}

|\vec{PN}| = \sqrt{(-1+2\lambda)^2 + (-7+3\lambda)^2 + (2-6\lambda)^2}

33 = 49 \lambda^2 - 70 \lambda + 54

\lambda = 1 \mathrm{~or~} \frac{3}{7}

\vec{ON} = \begin{pmatrix}3\\{1}\\{-10}\end{pmatrix} \mathrm{~or~} \frac{1}{7}\begin{pmatrix}13\\{-5}\\{-46}\end{pmatrix}

L = |\vec{PN}|^2 = (-1+2\lambda)^2 + (-7+3\lambda)^2 + (2-6\lambda)^2 = 49 \lambda^2 - 70 \lambda +54

\frac{dL}{d\lambda} = 98 \lambda - 70

\frac{d^2L}{d\lambda ^2} = 98 > 0

So when \lambda = \frac{70}{98} = \frac{5}{7}, L is minimum.

\vec{ON} =  \frac{1}{7} \begin{pmatrix}{17}\\{1}\\{-58}\end{pmatrix}

(iii)
\begin{pmatrix}{-1}\\{-7}\\{2}\end{pmatrix} \times \begin{pmatrix}{2}\\{3}\\{-6}\end{pmatrix} = \begin{pmatrix}{36}\\{-2}\\{11}\end{pmatrix}

\begin{pmatrix}{1}\\{-2}\\{-4}\end{pmatrix} \bullet \begin{pmatrix}{36}\\{-2}\\{11}\end{pmatrix} = -4

\therefore, \pi : 36x - 2y +11z=-4

 


 
Question 3
(ai)
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.

(aii)
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 )


 

Question 4
(a)
Let P(n) be the statement \sum_{r=1}^n r(r+2)(r+5) = \frac{1}{12}n(n+1)(3n^2+31n+74) \forall n \in \mathbb{Z^+}

When n=1, LHS = 1 \times 3 \times 6 = 18
RHS = \frac{1}{12}\times 1 \times 2 \times (3+31+74) = 18

Since LHS = RHS, P(1) is true.

Assume P(k) is true for some k, k \in \mathbb{Z^+}

\sum_{r=1}^k r(r+2)(r+5) = \frac{1}{12}k(k+1)(3k^2+31k+74)

Want to show P(k+1) is true, \sum_{r=1}^{k+1} r(r+2)(r+5) = \frac{1}{12}(k+1)(k+2)(3k^3+43k^2+182k+216)

LHS
= \sum_{r=1}^{k+1} r(r+2)(r+5)

= \frac{1}{12}k(k+1)(3k^2+31k+74) + (k+1)(k+3)(k=6)

= \frac{1}{12}(k+1)[3k^3 + 321k^2 + 74k + 12(k^2 + 9k +18)]

= \frac{1}{12}(k+1)(3k^3+43k^2+182k+216)

= RHS

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

(bi)
A = 1, B = -1

(bii)
\sum_{r=1}^n \frac{1}{2r+1} - \frac{1}{2r+3}

= \frac{1}{3} - \frac{1}{5}

+ \frac{1}{5} - \frac{1}{7}

\ldots

+ \frac{1}{2n-1} - \frac{1}{2n+1}

+ \frac{1}{2n+1} - \frac{1}{2n+3}

= \frac{1}{3} - \frac{1}{2n+3}

(biii)

S_{\infty} = \frac{1}{3}

S_n - S_{\infty} = \frac{1}{2n+3} ~ \textless ~ 10^{-3}

Least n = 499


Question 5
(i)
Manager is unable to obtain an appropriate sampling frame.

(ii)
Manager can consider taking equal number of respondents from adults and aged first. Then, he can stand outside the supermarket, interview whoever walks in, based on his personal preferences until he obtains the number of respondents he require.

(iii)
As he samples based on his own preferences, there might be selection bias.


Question 6
(i)
Let X denote the number of red sweets in a small packet of 10 sweets.

X \sim \mathrm{Bin} (10, 0.25)

\mathrm{P}(X \ge 4) = 1 - \mathrm{P}(X \le 3) \approx 0.224 (3 SF)

(ii)
Let Y denote the number of red sweets in a large packet of 100 sweets.

Y \sim \mathrm{Bin} (100, 0.25)

Since n is large, np = 25 > 5, n(1-p) = 75 > 5

Y \sim \mathrm{N}(25, 18.75) approximately

\mathrm{P}( Y \ge 30) = \mathrm{P}(X > 29.5) by continuity correction
\approx 0.149 (3 SF)

(iii)
Let W denote the number of packets out of 15 packets that contain at least 30 red sweets.

Y \sim \mathrm{Bin}(15, 0.1493487984)

\mathrm{P}( W \le 3) \approx 0.825 (3 SF)


Question 7
(i)
Firstly, the average number of errors per page is constant.
Secondly, the number of errors in a page is independent of the number of errors in another page.

(ii)
Let X denote the number of errors per page.
X \sim \mathrm{Po}(1.3)

X_1 \ldots X_6 \sim \mathrm{Po}(7.8)

\mathrm{P}(X_1 \ldots X_6 > 10)

= 1 - \mathrm{P}(X_1 \ldots X_6 \le 10)

\approx 0.165 (3 SF)

(iii)
X_1 \ldots X_n \sim \mathrm{Po}(1.3n)

\mathrm{P}(X_1 \ldots X_n ~ \textless ~ 2) ~ \textless ~ 0.05

e^{-1.3n} + e^{-1.3n}(1.3n) ~ \textless ~ 0.05

e^{-1.3n} (1  + 1.3n) ~ \textless ~ 0.05

Using GC, n > 3.6491

Least n = 4


Question 8
Let X denote the mass of pineapple tarts and \mu denote the population mean mass of pineapple tarts.

H_0: \mu = 0.9

H_1: \mu ~ \textless ~ 0.9

Unbiased estimate of population mean = 0.8825

s^2 = 0.0747854073^2 \approx 0.00559 (3 SF)

Under H_0, \bar{X} \sim \mathrm{N} (\mu, \frac{s^2}{n})

Test Statistic, T = \frac{\bar{X}- \mu}{\frac{s}{\sqrt{n}}} \sim t{(7)} at 10% level of significance.

Using GC, p-value = 0.264619 > 0.1

\Rightarrow, do not reject H_0

There is insufficient evidence at 10% significance level to reject the stall owner’s claim.


Question 9
(i)
\mathrm{P}(B|A) = \mathrm{P}(B) since A and B are independent.

\therefore, \mathrm{P}(B|A) = \mathrm{P}(B) = 0.4

(ii)
\mathrm{P}(A' \cap B' \cap C')

= 1 - \mathrm{P}(C \cup B \cup A)

= 1 - (0.45 + 0.02 + 0.2 + 0.145)

= 0.185

(iii)

If \mathrm{P}(A' \cap B' \cap C) = 0,

\mathrm{P}(A' \cap B' \cap C')

=1 - [0.935 - (0.3 - 0.035 )]

= 0.33 (maximum)

If \mathrm{P}(A' \cap B \cap C) = 0,

\mathrm{P}(A' \cap B' \cap C')

= 1 - \mathrm{P}(A \cup B \cup C) - 0.22

= 0.165 (minimum)

0.165 \le \mathrm{P}(A' \cap B' \cap C') \le 0.33


Question 10
(i)
Graph later I plot
(ii)
(a) r \approx -0.9807 (4 DP)
(b) r \approx -0.9748 (4 DP)
(c) r \approx -0.9986 (4 DP)

(iii)
Using GC, P = -0.14686 \sqrt{h} + 34.78895
\therefore, P = -0.147 \sqrt{h} + 34.8 (3 SF)

(iv)
\therefore, P = -0.147 \sqrt{3.28h_{metres}} + 34.8

\therefore, P = -0.266 \sqrt{h_{metres}} + 34.8 (3 SF)


Question 11
(i)
Number of ways = \frac{8!}{2!2!} = 10080 ways

(ii)
Number of ways = 10080 -1 = 10079 ways

Side Note: Mr Wee provided a different and interesting perspective here, and suggested that it could be SGECBBAA. I think we both agree that Cambridge isn’t that evil and sneaky. I felt that my nursery rhyme should suffice. Food for thought. πŸ™‚

(iii)
Number of ways = 6! = 720 ways

(iv)
Case 1: 2 A’s together and B’s separated
5! \times ^6 C_2 = 1800
Case 2: 2 B’s together and A’s separated
5! \times ^6 C_2 = 1800
Case 3: 2 A’s together and 2 B’s together
720

Number of ways =10080 – 1800 – 1800 – 720 = 5760 ways


Question 12

(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)


 

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