2024 H2 Mathematics Paper 2 Question 3

Functions

Answers

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

Sketch.

All horizontal lines $y = k,$ $k \in ℝ,$ cuts the graph of $y = g (x)$ at most once. Hence $g$ is a one-to-one function and $g^{- 1}$ exists.

$f g^{- 1} (1.5) = 7.$

(a)

From the graph of $f,$

R_{f} = (- \infty, 11] ∎

(b)

When $x = 0,$

\begin{aligned} y & = - (0^{2}) + 4 (0) + 7 \\ = 7 \end{aligned}

When $y = 0,$

\begin{aligned} x & = \frac{- (4) \pm \sqrt{4^{2} - 4 (- 1) (7)}}{2 (- 1)} \\ = 2 + \sqrt{11} or 2 - \sqrt{11} \end{aligned}

graph

(c)

All All horizontal lines $y = k,$ $k \in ℝ,$ cuts the graph of $y = g (x)$ at most once. Hence $g$ is a one-to-one function and $g^{- 1}$ exists $∎$

(d)

\begin{aligned} y & = \frac{2}{x} + 1 \\ y - 1 & = \frac{2}{x} \\ x & = \frac{2}{y - 1} \\ ∴ g^{- 1} (x) & = \frac{2}{x - 1} \end{aligned}

\begin{aligned} f g^{- 1} (1.5) \\ = f (\frac{2}{1.5 - 1}) \\ = f (4) \\ = - (4^{2}) + 4 (4) + 7 \\ = 7 ∎ \end{aligned}