2010 Paper 2 Question 4

Sketch.

Least value of $k=0.$

$fg(x)=\frac{{(x-3)}^2}{(4-x)(x-2)}.$

$2<x<3$ or $3<x<4.$

$R_{fg}=(-\infty ,-1)\cup (0,\infty ).$

Least value of $k=0∎$

This is because, when $k=0,$ the domain of $f$ is $[0,1)\cup (1,\infty ).$
Under this domain, all horizontal lines $y=k,k\in ℝ,$ cuts the graph of $y=f(x)$ at most once so $f$ is a one-to-one function and the function $f^{-1}$ exists.

$$\begin{array}{rl}fg(x)& =f\left(\frac{1}{x-3}\right)\\ & =\frac{1}{{\left(\frac{1}{x-3}\right)}^2-1}\\ & =\frac{1}{\frac{1}{{(x-3)}^2}-1}\\ & =\frac{1}{\frac{-8-x^2+6x}{{(x-3)}^2}}\\ & =\frac{{(x-3)}^2}{-8-x^2+6x}\\ & =\frac{{(x-3)}^2}{(4-x)(x-2)}\end{array}$$

$$\frac{{(x-3)}^2}{(4-x)(x-2)}>0$$

$$2<x<3\text{ or }3<x<4$$

$$\begin{array}{rl}{D}_g& =(-\infty ,2)\cup (2,3)\cup (3,4)\cup (4,\infty )\\ & \downarrow g\\ R_g& =(-\infty ,-1)\cup (-1,0)\cup (0,1)\cup (1,\infty )\\ & \downarrow f\\ R_{fg}& =(-\infty ,-1)\cup (0,\infty )\end{array}$$