9758/2021/P2/Q03

(a) The function $h$ is defined by $h: x \mapsto \frac{1}{2} x^2+3$, for $x \in \mathbf{R}$.
The function $g$ is defined by $g: x \mapsto \frac{x+1}{5 x-1}$, for $x \in \mathbf{R}, x \neq 0.2$.
(i) Find $\operatorname{gh}(2)$.

[2]

(ii) Find the value of $x$ for which $g(x)=1.4$.

(b) The function $\mathrm{f}$ is defined by $\mathrm{f}: x \mapsto \frac{x+a}{2 x+b}$, for $x \in \mathbb{R}, x \neq k$.
(i) Give an expression for $k$ and explain why this value of $x$ has to be excluded from the domain of $f$,

[2]

The function $\mathrm{f}$ is such that $\mathrm{f}(x)=\mathrm{f}^{-1}(x)$ for all $x$ in the domain of $\mathrm{f}$.
(ii) Determine the possible values of $a$ and of $b$.

[3]

(iii) Find an expression for $f^{-1}(-4)$.

[1]

[Maximum marks: 9 marks]
(a) $(i)$
$g h(2)=g(5)=\frac{1}{4}$

(a)(ii)
$g(x)=1.4$
$x=0.4$ using the $\mathrm{GC}$.

(b) (i)
Consider $2 x+b=0 \Rightarrow x=-\frac{b}{2}$, then $\mathrm{f}$ is not defined when $x=-\frac{b}{2}$ since $x=-\frac{b}{2}$ is a vertical asymptote. Thus, $k=-\frac{b}{2}$.

(b)(ii) For functions to be self-inverse, we need both functions to have the same rule, domain and range. Observe that $f(x)=\frac{x+a}{2 x+b}=\frac{1}{2}+\frac{2 a-b}{2(2 x+b)}$ has a horizontal asymptote at $y=\frac{1}{2}$. Then $\frac{1}{2}=-\frac{b}{2} \Rightarrow b=-1$.
Observe that $a \in \mathbb{R}$, however the function will not be a $1-1$ function if $f$ is a horizontal line, thus $a \neq-\frac{1}{2}$.
$\therefore \quad b=-1$ and $a \in \mathbb{R}, a \neq-0.5$

(b)(iii) Since we have that it is self-inverse, then $f^{-1}(x)=\frac{x+a}{2 x-1}$.
$$
f^{-1}(-4)=\frac{a-4}{-9}=\frac{4-a}{9}
$$