2021 H2 Maths Paper 1 Q6 — Worked Solution

Part (a)

4 marks

y = \frac{1}{\sqrt{4ax - x^{2}}}

For $y$ to be defined, we need $4ax - x^{2} > 0$ , i.e. $x(4a - x) > 0$ , i.e. $0 < x < 4a$ .

Completing the square:

4ax - x^{2}

= - \left( x^{2} - 4ax \right)

= - \left\lbrack (x - 2a)^{2} - 4a^{2} \right\rbrack

= 4a^{2} - (x - 2a)^{2}

As $x \rightarrow 0^{\text{+}}$ or $x \rightarrow (4a)^{-}$ , $4ax - x^{2} \rightarrow 0^{\text{+}}$ , so $y \rightarrow + \infty$ : there are vertical asymptotes at $x = 0$ and $x = 4a$ .

Since $4ax - x^{2} = 4a^{2} - (x - 2a)^{2}$ is maximised (equal to $4a^{2}$ ) at $x = 2a$ , the denominator $\sqrt{4ax - x^{2}}$ is maximised there too, so $y$ is smallest at $x = 2a$ :

y = \frac{1}{\sqrt{4a^{2}}} = \frac{1}{2a}

To check your answer: taking

a = 1

, at

x = 2

(the midpoint of

0 < x < 4

),

y = \frac{1}{\sqrt{8 - 4}} = \frac{1}{2}

, matching

\frac{1}{2a} = \frac{1}{2}

; at

x = 0.1

or

x = 3.9

,

y

is much larger, consistent with the asymptotes at

x = 0

and

x = 4a

. ✓

$\therefore$ vertical asymptotes $x = 0$ and $x = 4a$ .

Part (b)

1 mark

From part (a), $y$ is smallest at $x = 2a$ .

$\therefore$ $y_{\min} = \frac{1}{2a}$ .

Part (c)

3 marks

Approach

From part (a), $C$ ’s equation is already written in completed-square form, $y = \frac{1}{\sqrt{4a^{2} - (x - 2a)^{2}}}$ , so we can read the transformation off by comparing this directly with the target equation, without differentiating or solving anything.

Let $g(x) = \frac{1}{\sqrt{4a^{2} - x^{2}}}$ be the target curve. From part (a), $C$ ’s equation is $y = g(x - 2a)$ .

Since $g(x - 2a)$ is the graph of $g$ shifted $2a$ units in the positive $x$ -direction, $C$ is $g$ translated $2a$ units right. Reversing this, $g$ is obtained from $C$ by a translation $2a$ units in the negative $x$ -direction.

To check your answer:

C

is defined on

0 < x < 4a

with minimum point

\left( 2a,\frac{1}{2a} \right)

; translating

2a

units in the negative

x

-direction gives domain

- 2a < x < 2a

with minimum point

\left( 0,\frac{1}{2a} \right)

. Directly,

y = \frac{1}{\sqrt{4a^{2} - x^{2}}}

is indeed defined on

- 2a < x < 2a

, with minimum value

\frac{1}{\sqrt{4a^{2}}} = \frac{1}{2a}

at

x = 0

. ✓

$\therefore$ a translation of $2a$ units in the negative $x$ -direction.

2021 H2 Maths Paper 1, Question 6

Part (a)

Part (b)

Part (c)

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