2015 H2 Maths Paper 1 Q1 — Worked Solution

Part (i)

4 marks

Since $C$ passes through $(1.6,\ - 2.4)$ ,

\frac{a}{{1.6}^{2}} + 1.6b + c = - 2.4

\Rightarrow 0.390625a + 1.6b + c = - 2.4.\quad(1)

Since $C$ passes through $( - 0.7,\ 3.6)$ ,

\frac{a}{( - 0.7)^{2}} - 0.7b + c = 3.6

\Rightarrow 2.040816a - 0.7b + c = 3.6.\quad(2)

Differentiating,

\frac{dy}{dx} = - \frac{2a}{x^{3}} + b.

Using the given gradient of $2$ at $x = 1$ ,

- 2a + b = 2.\quad(3)

By GC, solving (1), (2) and (3) simultaneously,

a = - 3.593,\ \ b = - 5.187,\ \ c = 7.303\ \ \text{(3 d.p.)}.

Part (ii)

2 marks

Approach: We use the unrounded values of $a$ , $b$ , $c$ stored on the GC (not the values rounded to 3 d.p.) to find where $C$ meets the $x$ -axis, then round the final answer only.

$C$ crosses the $x$ -axis where $y = 0$ , i.e.

\frac{a}{x^{2}} + bx + c = 0.

By GC, solving this numerically with the stored values of $a$ , $b$ , $c$ , the only real root is

x = - 0.589\ \ \text{(3 d.p.)}.

To check your answer: substituting

x = - 0.589

into

y = \frac{a}{x^{2}} + bx + c

with the unrounded constants gives

y \approx 0

. ✓

$\therefore$ $C$ crosses the $x$ -axis at $x = - 0.589$ .

Part (iii)

1 mark

As $x \rightarrow \pm \infty$ , $\frac{a}{x^{2}} \rightarrow 0$ , so $y \rightarrow bx + c$ .

$\therefore$ the other asymptote is $y = bx + c$ , i.e. $y = - 5.187x + 7.303$ .

2015 H2 Maths Paper 1, Question 1

Part (i)

Part (ii)

Part (iii)

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