2021 H2 Math Paper 2 Answers

{a = - \frac{27}{4}, b = 5.}

Other roots:

{x = 1 - \frac{1}{2} \mathrm{i}, x = -4.}

(a)

{h=\frac{4}{5}.}

(b)

\sum_{n=1}^{5} \big( f(1+nh) \big) h.

(c)

{5.61 < A < 6.57.}

(a)

(i)

{gh(2) = \frac{1}{4}.}

(ii)

{x = \frac{2}{5}.}

(b)

(i)

{k = -\frac{b}{2}}

since

{f}

is undefined at

{x=-\frac{b}{2}.}

(ii)

{b=-1, \; a \in \mathbb{R}, a \neq -\frac{1}{2}.}

(iii)

{f^{-1}(-4) = \displaystyle \frac{ 4 - a }{ 9 }.}

(a)

{325 \textrm{ seconds}.}

(b)

{1.17 \textrm{ m/s}.}

(c)

Away from starting point

(a)

\frac{\tan 5x}{5} - x + C.

(b)

- \frac{1}{2} \left( \frac{\sin 5b}{5} - \sin b \right) .

(c)

\ln \left( \frac{\ln b}{\ln a} \right).

(d)

- \frac{1}{4(1+\mathrm{e}^{2x})^2} + C.

{\frac{27}{10}.}

(a)

{83160.}

(b)

{144.}

(c)

{\frac{1}{66}.}

(a)

He should carry out a 1-tail test since he is interested to test if the life span is greater than 20 000 miles.

Let

{\mu}

be the population mean life span of front tyres on the particular model of car,

{H_0}

be the null hypothesis and

{H_1}

be the alternative hypothesis.

{H_0: \mu = 20000}

{H_1: \mu > 20000}

(b)

Unbiased estimate of population mean,

{\overline{x} = 20188}

,
Unbiased estimate of population variance

{s^2 = 754955}

(c)

{{p \textrm{-value} = 0.0630 > 0.05} \allowbreak \, \Rightarrow \, \allowbreak {H_0 \textrm{ not rejected}.}}

Hence there is insufficient evidence at the 5% level of significance to conclude whether the mean life span of front tyres is more than 20 000 miles.

(d)

Since the distribution of the life span of front tyres is unknown, we require a large sample size (of at least 30) to ensure that the sample mean life span of front tyres is normally distributed approximately by the Central Limit Theorem.

To be updated.

(a)

{\mu = 2.03.}

(b)

154.

(c)

0.427.

(d)

0.969.

(e)

0.182.

(a)

(i)

They should be selected randomly such that each light fitting in the population has an equal probability to be selected into the sample. Each light fitting should be selected independently.

(ii)

{0.0200.}

(iii)

{34.1.}

(b)

(i)

The probability that a heating element is faulty is the same for each heating element.
Whether a heating element is faulty is independent of all other heating elements.

(ii)

{0.779.}

(iii)

{0.0505.}

(iv)

{0.593.}