The function $f$ is defined by $$f(x) = \frac{3x - 7}{x - 2} \quad x \in \mathbb{R}, \ x \neq 2$$ (a) Find $f^{-1}(7)$ (b) Show that $f(f(x)) = \frac{ax + b}{x - 3}$ where $a$ and $b$ are integers to be found. - Edexcel - A-Level Maths Pure - Question 6 - 2020 - Paper 1

To show that $f(f(x))$ has the form $\frac{ax + b}{x - 3}$, we first substitute $f(x)$ into itself:

$f(f(x)) = f\left(\frac{3x - 7}{x - 2}\right)$

Now, we substitute rac{3x - 7}{x - 2} into the function $f$:

$= \frac{3\left(\frac{3x - 7}{x - 2}\right) - 7}{\left(\frac{3x - 7}{x - 2}\right) - 2}$

Simplifying the numerator:

$= \frac{\frac{9x - 21 - 7(x - 2)}{x - 2}}{\frac{3x - 7 - 2(x - 2)}{x - 2}}$

The numerator simplifies to:

$9x - 21 - 7x + 14 = 2x - 7$

And the denominator simplifies to:

$3x - 7 - 2x + 4 = x - 3$

Thus, combining the terms yields:

$f(f(x)) = \frac{2x - 7}{x - 3}$

From this, we see that $a = 2$ and $b = -7$, which are indeed integers.