2020 H2 Math Paper 2 Full Solutions

Let

{y=ax^2+bx+c}

{\frac{\mathrm{d}y}{\mathrm{d}x} = 2ax + b}

When

{x=1, y=-2}

\begin{equation} a + b + c = -2 \end{equation}

When

{x=1, \frac{\mathrm{d}y}{\mathrm{d}x}=0}

\begin{equation} 2 a + b = 0 \end{equation}

When

{x=2, \frac{\mathrm{d}y}{\mathrm{d}x}=5}

\begin{equation} 4 a + b = 5 \end{equation}

Solving using a G.C.,

{a=2.5, b=-5, c=0.5.}

Equation of curve:

{y = \frac{5}{2} x^2 - 5 x + \frac{1}{2}.}

(A)

{u_1 = 7, } \allowbreak { u_2 = 9, } \allowbreak { u_3 = 13, \ldots}

The sequence increases and diverges.
(B)

{u_1 = 5, } \allowbreak { u_2 = 5, } \allowbreak { u_3 = 5, \ldots}

The sequence is a constant sequence

{u_n = 5}

for all

{n \in \mathbb{ Z }^+.}

{u_5 = 101}

{2 u_4 - 5 = 101 \; \Rightarrow \; u_4 = 53}

{2 u_3 - 5 = 53 \; \Rightarrow \; u_3 = 29}

{2 u_2 - 5 = 29 \; \Rightarrow \; u_2 = 17}

{2 u_1 - 5 = 29 \; \Rightarrow \; u_1 = 11}

{p=11.}

{v_4 = v_2 + 2v_3 - 7}

{2 v_3 = v_2 + 2v_3 - 7}

{v_2 = 7}

{b=7}

\begin{aligned} v_5 &= v_3 + 2 v_4 - 7 \\ &= v_3 + 2 (2 v_3) - 7 \\ &= 5 v_3 - 7 \\ &= 5(v_1 + 2 v_2 - 7) - 7 \\ &= 5 (a + 2 b - 7) - 7 \\ &= 5(a + 2(7) - 7) - 7 \\ &= 5 a + 28 \end{aligned}

\begin{aligned} u_n &= S_n - S_{n-1} \\ &= (n^3 - 11 n^2 + 4 n) - \Big( (n-1)^3 - 11(n-1)^2 + 4(n-1) \Big) \\ &= (n^3 - 11 n^2 + 4 n) - (n^3 - 14 n^2 + 29 n - 16) \\ &= 3 n^2 - 25 n + 16. \end{aligned}

\begin{aligned} & S_m = S_3 \\ & m^3 - 11 m^2 + 4 m = (3)^3 - 11(3)^2 + 4(3) \\ & m^3 - 11 m^2 + 4 m = - 60 \\ & n^3 - 11 n^2 + 4 n + 60 = 0 \end{aligned}

Using G.C.,

{m=-2, 3}

or

{10}

Since

{m > 3}

,

{m = 10.}

{\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{6}{6 t} = \frac{1}{t} }

At

{(14, 11),}

{t = 2}

{\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}}

Gradient of normal

{= - 2}

{y - 11 = - 2 (x - 14)}

Cartesian equation of

{N}

:

{2 x + y = 39.}

The normal cuts the

{x}

-axis at

{\frac{39}{2}}

Area of triangle

{ = \frac{1}{2} \times (\frac{39}{2} - 14) \times 11 = \frac{121}{4}}

The curve cuts the

{x}

-axis at

{t = \frac{1}{6}, x = \frac{25}{12}}

\begin{aligned} \textrm{Area under curve } &= \int_{\frac{25}{12}}^{14} y \, \mathrm{d} x \\ &= \int_{\frac{1}{6}}^{2} (6 t - 1) 6 t \, \mathrm{d} t \\ &= \frac{3025}{36} \end{aligned}

Area required

{ = \frac{121}{4} + \frac{3025}{36} = \frac{2057}{18} \textrm{ units}^2.}

Area

{= \frac{2057}{18} \times 2 \times 3 = \frac{2057}{3} \textrm{ units}^2.}

Replacing

{y}

with

{\frac{y}{3},}

{\frac{y}{3} = 6 t - 1 \; \Rightarrow \; t = \frac{y+3}{18}}

Replacing

{x}

with

{\frac{x}{2},}

{\frac{x}{2} = 3 t^2 + 2}

{x = 2 \left( 3 \left( \frac{y+3}{18} \right)^2 + 2 \right)}

Cartesian equation of

{D}

:

{x = \frac{1}{54} \left( y + 3 \right)^{ 2 } + 4.}

{a+2h = 30 \; \Rightarrow \; h = \frac{30-a}{2}}

\begin{aligned} H^2 &= h^2 - \frac{a}{2}^2 \\ &= \frac{(30-a)^2 - a^2 }{4} \\ &= 225 - 15a \end{aligned}

{V = \frac{1}{3} a^2 H}

{V^2 = \frac{1}{9} a^4 H^2}

{V^2 = \frac{1}{9} a^4 (225-15a) = \frac{1}{9} (225a^4 - 15a^5)}

Differentiating w.r.t.

{a,}

{2V \frac{\mathrm{d}V}{\mathrm{d}a} = \frac{1}{9} (900a^3 - 75a^4)}

At maximum volume

{\frac{\mathrm{d}V}{\mathrm{d}a} = 0}

{900a^3 - 75a^4 = 0}

{a = 12}

Maximum volume

{ = \frac{1}{3} (12)^2 \sqrt{225 - 15(12)} = 144 \sqrt{5} \textrm{ cm}^3.}

Let

{A}

be the total surface area of the four triangular faces of the pyramid

{A = 4 \cdot \frac{1}{2} a h = 2a \left( \frac{30-a}{2} \right) = 30a - a^2}

{\frac{\mathrm{d}A}{\mathrm{d}a} = 30 - 2a}

At maximum area

{\frac{\mathrm{d}A}{\mathrm{d}a} = 0}

{a = 15.}

The net forms a square with sides

{15 \sqrt{2} \textrm{ cm}.}

{0, 4, 10, 25.}

Let

{T}

denote Tina's score.

{\mathrm{P}(T=4) = \frac{2r}{3r+1} \times \frac{2r-1}{3r} = \frac{2(2r-1)}{3(3r+1)}}

{\mathrm{P}(T=10) = \frac{r}{3r+1} \times \frac{2r}{3r} \times 2! = \frac{4r}{3(3r+1)}}

{\mathrm{P}(T=25) = \frac{r}{3r+1} \times \frac{r-1}{3r} = \frac{r-1}{3(3r+1)}}

Let

{p}

denote

{\mathrm{P}(T=0) = 1 - \mathrm{P}(T=4)} \allowbreak {- \mathrm{P}(T=10) - \mathrm{P}(T=25)}

The probability distribution of

{T}

is

${t}$	0	4	10	25
${P(T=t)}$	${p}$	${\frac{2(2r-1)}{3(3r+1)}}$	${\frac{4r}{3(3r+1)}}$	${\frac{r-1}{3(3r+1)}}$

\begin{aligned} \mathrm{E}(T) &= \sum t \, \mathrm{P}(T=t) \\ &= 4 \cdot \frac{2(2r-1)}{3(3r+1)} + 10 \cdot \frac{4r}{3(3r+1)} + 25 \cdot \frac{r-1}{3(3r+1)} \\ &= \frac{ 16r - 8 + 40r + 25r - 25 }{3(3r+1)} \\ &= \frac{ 81r - 33 }{3(3r+1)} \\ &= \frac{ 27r - 11 }{3r+1} \end{aligned}

\begin{aligned} \mathrm{E}(T^2) &= \sum t^2 \, \mathrm{P}(T=t) \\ &= 16 \cdot \frac{2(2r-1)}{3(3r+1)} + 100 \cdot \frac{4r}{3(3r+1)} + 625 \cdot \frac{r-1}{3(3r+1)} \\ &= \frac{ 64r - 32 + 400r + 625r - 625 }{3(3r+1)} \\ &= \frac{ 1089r - 657 }{3(3r+1)} \\ &= \frac{ 363r - 219 }{3r+1} \end{aligned}

\begin{aligned} \mathrm{Var}(T) &= \mathrm{E}(T^2) - \mathrm{E}(T) \\ &= \frac{ 363r - 219 }{3r+1} - \left( \frac{ 27r - 11 }{3r+1} \right)^2 \\ &= \frac{(363r-219)(3r+1) - (27r-11)^2 }{(3r+1)^2} \\ &= \frac{ 360 r^2 + 300 r - 340 }{(3r+1)^2} \end{aligned}

\frac{360 r^2 + 300 r - 340}{(3 r + 1)^2} = 38

{r = 3.}

{T \sim N( 5, 1.44 )}

{\textrm{P}(T>6) = 0.202.}

{W \sim N( 21, 9 )}

{T+W \sim N( 26, 10.44 )}

{\textrm{P}(T+W > 30) = 0.10786}

The probability that James is late for work when he walks is

{0.108}

(to 3 s.f.).

{T+D \sim N( 24, 37.44 )}

{P(T+D > 30) = 0.16340}

\begin{aligned} \textrm{P}(\textrm{fine} \mid \textrm{late}) &= \frac{\textrm{P}(\textrm{fine} \cap \textrm{late})}{\textrm{P}(\textrm{late})} \\ &= \frac{0.7 \times 0.10786}{0.7 \times 0.10789 + 0.3 \times 0.16340} \\ &= 0.606. \end{aligned}

To be updated

{\textrm{P}(R=1) = \frac{{17 \choose 1} {11 \choose 11} }{{28 \choose 12}} = \frac{1}{1789515}}

{\textrm{P}(R=2) = \frac{{17 \choose 2} {11 \choose 10} }{{28 \choose 12}} = \frac{88}{1789515}}

Hence

{\textrm{P}(R=1) < \textrm{P}(R=2).}

\begin{aligned} \mathrm{P}(R=4) &= 15 \, \mathrm{P}(R=3) \\ \frac{{17+r \choose 4} {11 \choose 8}}{{28+r \choose 12}} &= \frac{15 {17+r \choose 3} {11 \choose 9}}{{28+r \choose 12}} \\ \frac{165(17+r)!}{4!(13+r)!} &= \frac{825(17+r)!}{3!(14+r)!} \\ 14+r &= 20 \\ r=6. \end{aligned}

Each pen from the box has an equal probability of being selected into the sample of 10.

Let

{X}

denote the r.v. of the number of pens that are faulty out of 10.

{X \sim \mathrm{B}(10, 0.06)}

{\textrm{P}(X \leq 2) = 0.981.}

Let

{Y}

denote the r.v. of the number of boxes that are rejected out of 75.

{Y \sim \mathrm{B}(75, 1-0.98116)}

{\mathrm{P}(Y > 0.75 \cdot 75 ) = 1 - \mathrm{P}(Y \leq 3)} \allowbreak {= 0.0535.}

Let

{W}

denote the r.v. of the number of pens that are faulty out of 5.

{W \sim \mathrm{B}(5, 0.06)}

\begin{aligned} \mathrm{P}(\textrm{accepted}) &= \mathrm{P}(W=0) + \mathrm{P}(W=1) \cdot \mathrm{P}(W \leq 1) \\ & \qquad + \mathrm{P}(W=2) \cdot \mathrm{P}(W=0) \\ &= 0.983. \end{aligned}

Both testing procedures have a similar probability of having a randomly chosen box accepted for sale, but the alternative testing procedure tests less pens on average so it may be cheaper to implement.

Let

{Y}

be the r.v. of the amount of carbon in each bar, by weight, in percentage.

{H_0: \mu = 1.5}

{H_1: \mu \neq 1.5}

Under

{H_0,}

Test statistic

Z = \frac{\overline{Y}-1.5}{\frac{0.09}{15}} \sim N(0,1)

For the critical region of the test at the 5% level of significance,

\frac{\overline{y}-1.5}{\frac{0.09}{15}} \leq -1.3520

or

\frac{\overline{y}-1.5}{\frac{0.09}{15}} \geq 1.3520

Critical region:

{\{ \overline{y} \in \mathbb{R}: \allowbreak {\overline{y} \leq 1.47 \textrm{ or } \overline{y} \geq 1.53 \}.}}

In the earlier test, the percentage of carbon in the steel bars,

{Y,}

is known to be normally distributed so

{\overline{Y}}

will still be normally distributed.
In the new test, we no longer know if the amount of carbon in each bar is normally distributed. Moreover, the population standard deviation is not know. Hence a large sample size of 40 ensures that the sample mean amount of carbon in each bar

{\overline{X}}

will be normally distributed approximately by the Central Limit Theorem.

Unbiased estimate of population mean

{= \overline{x} = 0.254}

Unbiased estimate of population variance

{= s ^ 2 = 0.000146.}

Unbiased estimate of population mean

{= \overline{x} = 0.254}

Unbiased estimate of population variance

{= s ^ 2 = 0.000146.}

{H_0: \mu = 0.25}

{H_1: \mu > 0.25}

Under

{H_0,}

Test statistic

Z = \frac{\overline{X}-0.25}{\sqrt{\frac{0.00014621}{40}}} \sim N(0,1)

{p \textrm{-value} = 0.0182 \leq 0.025} \allowbreak \, \Rightarrow \, \allowbreak {H_0 \textrm{ rejected}.}

Hence there is sufficient evidence at the 2.5% level of significance to conclude that the mean amount of carbon in the flat bars is more than 0.25%.