Functions — General Properties practice questions — A-Level Maths

$f$ is invertible iff its slope $a-2\mathrm{\ne }0$. 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 $f(x)=x$: $2x-3=x\Rightarrow x=3$.

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

$f(-1)=1=f(1)$, 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 $\mathrm{Im}(f)=[1,+\mathrm{\infty })$.

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

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 $x={\displaystyle \frac{y-3}{2}}$. 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 $[2,\mathrm{\infty })$.

$x^2$, $\mathrm{\mid }x\mathrm{\mid }$ 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 $\mathbb{R}\to \mathbb{R}$.