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(i)

(ii)
For $f^{-1}$ to exist $f$ must be one-one, thus least value of $k=0$

(iii)
$fg(x)$

$= f(\frac{1}{x-3})$

$= \frac{1}{\frac{1}{(x-3)^2}-1}$

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

(iv)
$fg(x) > 0$

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

Since $(x-3)^2 \ge 0, ((4-x)(x-2) > 0$

$\therefore, 2 \textless x \textless 4, x \neq 3$

(v)
Range of $fg(x) = (-\infty, -1 ) \cup (0, \infty)$