The functions $u$ and $v$ are defined as $u(x) = x - 3,{\text{ }}v(x) = 2x$ where $x \in \mathbb{R}$.

(i)     State the range of the function $u \circ f$.

(ii)     State the range of the function $u \circ v \circ f$.

(iii)     Find the largest possible domain of the function $f \circ v \circ u$.

(i)     Explain why $f$ does not have an inverse.

(ii)     The domain of $f$ is restricted to define a function $g$ so that it has an inverse ${g^{ - 1}}$.

State the largest possible domain of $g$.

(iii)     Sketch a graph of $y = {g^{ - 1}}(x)$, showing clearly the $y$-intercept and stating the coordinates of the endpoints.

Consider the function defined by $h(x) = \frac{{2x - 5}}{{x + d}}$, $x \ne  - d$ and $d \in \mathbb{R}$.

(i)     Find an expression for the inverse function ${h^{ - 1}}(x)$.

(ii)     Find the value of $d$ such that $h$ is a self-inverse function.

For this value of $d$, there is a function $k$ such that $h \circ k(x) = \frac{{2x}}{{x + 1}},{\text{ }}x \ne  - 1$.

(iii)     Find $k(x)$.

Note:     For Q12(a) (i) – (iii) and (b) (ii), award A1 for correct endpoints and, if correct, award A1 for a closed interval.

Further, award A1A0 for one correct endpoint and a closed interval.

(i)     $- 4 \le y \le  - 2$     A1A1

(ii)     $- 5 \le y \le  - 1$     A1A1

(iii)     $- 3 \le 2x - 6 \le 5$     (M1)

Note:     Award M1 for $f(2x - 6)$.

$3 \le 2x \le 11$

$\frac{3}{2} \le x \le \frac{{11}}{2}$     A1A1

[7 marks]

(i)     any valid argument eg $f$ is not one to one, $f$ is many to one, fails horizontal line test, not injective     R1

(ii)     largest domain for the function $g(x)$ to have an inverse is $[ - 1,{\text{ }}3]$     A1A1

(iii)     

$y$-intercept indicated (coordinates not required)     A1

correct shape     A1

coordinates of end points $(1,{\text{ }}3)$ and $( - 1,{\text{ }} - 1)$     A1

Note:     Do not award any of the above marks for a graph that is not one to one.

[6 marks]

(i)     $y = \frac{{2x - 5}}{{x + d}}$

$(x + d)y = 2x - 5$     M1

Note:     Award M1 for attempting to rearrange $x$ and $y$ in a linear expression.

$x(y - 2) =  - dy - 5$     (A1)

$x = \frac{{ - dy - 5}}{{y - 2}}$     (A1)

Note:     $x$ and $y$ can be interchanged at any stage

${h^{ - 1}}(x) = \frac{{ - dx - 5}}{{x - 2}}$     A1

Note:     Award A1 only if ${h^{ - 1}}(x)$ is seen.

(ii)     self Inverse $\Rightarrow h(x) = {h^{ - 1}}(x)$

$\frac{{2x - 5}}{{x + d}} \equiv \frac{{ - dx - 5}}{{x - 2}}$     (M1)

$d =  - 2$     A1

(iii)     METHOD 1

$\frac{{2k(x) - 5}}{{k(x) - 2}} = \frac{{2x}}{{x + 1}}$     (M1)

$k(x) = \frac{{x + 5}}{2}$     A1

METHOD 2

${h^{ - 1}}\left( {\frac{{2x}}{{x + 1}}} \right) = \frac{{2\left( {\frac{{2x}}{{x + 1}}} \right) - 5}}{{\frac{{2x}}{{x + 1}} - 2}}$     (M1)

$k(x) = \frac{{x + 5}}{2}$     A1

[8 marks]

Total [21 marks]