2024 H2 Maths Paper 2 Q3 — Worked Solution

Part (a)

2 marks

Completing the square,

f(x) = - x^{2} + 4x + 7

= - \left( x^{2} - 4x \right) + 7

= - (x - 2)^{2} + 4 + 7

= - (x - 2)^{2} + 11

Since $(x - 2)^{2} \geq 0$ for all real $x$ , $- (x - 2)^{2} \leq 0$ , so $f(x) \leq 11$ .

$\therefore$ $R_{f} = ( - \infty,\ 11\rbrack$ .

Part (b)

2 marks

$y$ -intercept: $f(0) = 7$ , giving $(0,7)$ .

$x$ -intercepts: solving $f(x) = 0$ ,

- x^{2} + 4x + 7 = 0

\Rightarrow x^{2} - 4x - 7 = 0

\Rightarrow x = \frac{4 \pm \sqrt{16 + 28}}{2}

= 2 \pm \sqrt{11}

giving $\left( 2 - \sqrt{11},0 \right)$ and $\left( 2 + \sqrt{11},0 \right)$ , with maximum turning point $(2,11)$ from part (a).

Part (c)

1 mark

Approach

A function has an inverse exactly when it is one-one on its domain; for a function given by a formula, it is enough to show it is strictly monotonic there.

g(x) = \frac{2}{x} + 1,\ \ 1 \leq x \leq 10

\Rightarrow g'(x) = - \frac{2}{x^{2}}

Since $x^{2} > 0$ on $\lbrack 1,10\rbrack$ , $g'(x) < 0$ throughout, so $g$ is strictly decreasing on $\lbrack 1,10\rbrack$ .

$\therefore$ $g$ is one-one on $\lbrack 1,10\rbrack$ , so $g^{- 1}$ exists.

Part (d)

3 marks

Approach

${fg}^{- 1}(x)$ means $f\left( g^{- 1}(x) \right)$ : we first find $g^{- 1}$ and its domain, then substitute $x = 1.5$ into $g^{- 1}$ , then apply f to the result.

Let $y = g(x) = \frac{2}{x} + 1$ and solve for $x$ in terms of $y$ :

y = \frac{2}{x} + 1

\Rightarrow y - 1 = \frac{2}{x}

\Rightarrow x = \frac{2}{y - 1}

Replacing $y$ by $x$ ,

g^{- 1}(x) = \frac{2}{x - 1}

Since $g$ is strictly decreasing on $\lbrack 1,10\rbrack$ (part (c)), its range runs from $g(10) = 1.2$ up to $g(1) = 3$ , so $D_{g^{- 1}} = R_{g} = \lbrack 1.2,\ 3\rbrack$ . This contains $x = 1.5$ , so $g^{- 1}(1.5)$ is defined.

g^{- 1}(1.5) = \frac{2}{1.5 - 1} = 4

Applying f,

fg^{- 1}(1.5) = f(4)

= - (4)^{2} + 4(4) + 7

= 7

To check your answer:

g^{- 1}(1.5) = 4

lies in

\lbrack 1,10\rbrack

, the domain of

g

(and hence the range of

g^{- 1}

), and

g(4) = \frac{2}{4} + 1 = 1.5

confirms

g^{- 1}(1.5) = 4

✓

$\therefore$ ${fg}^{- 1}(1.5) = 7$ .

2024 H2 Maths Paper 2, Question 3

Part (a)

Part (b)

Part (c)

Part (d)

This is how we mark, and how we teach.