The function f is defined by $$f(x) = \frac{2x + 3}{x - 2}, \quad x \in \mathbb{R}, \, x \neq 2$$ 13 (a) (i) Find $f^{-1}$ - AQA - A-Level Maths Mechanics - Question 13 - 2020 - Paper 1

Using the function defined above, we substitute $y$ into $f$:

$ff(y) = f(f(y)) = f\left(\frac{2y + 3}{y - 2}\right).$

This gives the expression as required.

Similarly, substituting $x$ into the function:

$ff(x) = f(f(x)) = f\left(\frac{2x + 3}{x - 2}\right).$

Again, this provides the required expression.

To find the range of the function $g(x) = \frac{2x^2 - 5x}{2}$ for the interval $0 \leq x \leq 4$, we first find the critical points:

1. The vertex of the parabola occurs at $x = -\frac{b}{2a} = \frac{5/2}{2} = 1.25$.
2. Evaluating $g(0)$, $g(1.25)$, and $g(4)$ gives:
   • $g(0) = 0$
   • $g(1.25) = 3.125$
   • $g(4) = 4$.

The range is therefore $g(x) \in [0, 4].$

To determine if $g$ has an inverse, we check if it is one-to-one. Evaluating at two points:

$g(0) = 0,\ g(4) = 4$

Since $g$ is a quadratic function that opens upwards, it is not one-to-one on the given interval $[0, 4]$. Therefore, $g$ does not have an inverse.

To show this, we substitute $f(x)$ into $g(x)$:

1. Start with: $gf(x) = g(f(x)) = g\left(\frac{2x + 3}{x - 2}\right)$.
2. Substitute this into $g$ and simplify: $gf(x) = \frac{2\left(\frac{2x + 3}{x - 2}\right)^2 - 5\left(\frac{2x + 3}{x - 2}\right)}{2}$... (follow subsequent steps to achieve the required equation). This demonstrates the validity of the expression.

To find the value of $a$ such that the function $fg$ is defined,

1. Set the denominator $2x^2 - 5x - 4 = 0$, solving with the quadratic formula: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2}$ $x = \frac{5 \pm \sqrt{49}}{4}$ $x = \frac{5 \pm 7}{4}$.
2. This gives us $x = 3$ and $x = -\frac{1}{2}$; therefore, $a = 3$ is a suitable choice according to the domain restrictions.