Functions — General Properties practice questions — A-Level Maths

$f$ is invertible iff its slope $a - 2 \ne 0$. So $f$ fails to be invertible exactly when $a = 2$ (it then becomes the constant $5$, neither injective nor surjective).

The graph of $f^{-1}$ is the reflection of the graph of $f$ across $y = x$. Their intersection lies on $y = x$, so we solve $f(x) = x$: $2x - 3 = x \Rightarrow x = 3$.

$y = 3x - 6 \Rightarrow x = \dfrac{y + 6}{3}$, so $f^{-1}(y) = \dfrac{y + 6}{3}$.

$f(-1) = 1 = f(1)$, so $f$ is not injective. The value $-1$ has no real preimage (since $x^2 \ge 0$), so $f$ is not surjective either.

$f(x) = x^2 + 1 \ge 1$, with equality only at $x = 0$. So $\operatorname{Im}(f) = [1, +\infty)$.

$g(3) = 3^2 = 9$.

Then $f(9) = 2 \cdot 9 + 1 = 19$.

$g(2) = 5$, then $f(5) = 2 \cdot 5 - 1 = 9$. Hence $(f \circ g)(2) = 9$.

For any $y \in \mathbb{R}$, the equation $2x + 3 = y$ has the unique solution $x = \dfrac{y - 3}{2}$. Existence proves surjectivity, uniqueness proves injectivity, so $f$ is bijective.

We need $x - 2 \ge 0$, i.e. $x \ge 2$.

The value at $x = 2$ is $\sqrt{0} = 0$, which is allowed, so the domain is $[2, \infty)$.

$x^2$, $|x|$ and the constant $3$ each give one value for every real $x$.

$\pm\sqrt{x}$ fails twice: it is not single-valued (two outputs) and is undefined for $x < 0$. So it is not a function $\mathbb{R} \to \mathbb{R}$.