Explanation

(a)
Let $y = f(x) = \sqrt{(x - 1)(3 - x)}$
$\Rightarrow \ \ x^{2} - 4x + 3 + y^{2} = 0$
This is a quadratic in $x$, we get
$x = \frac{+ 4 \pm \sqrt{16 - 4(3 + y^{2})}}{2(1)} = \frac{4 \pm 2\sqrt{1 - y^{2}}}{2(1)}$
Since, $x$ is real, then $1 - y^{2} \geq 0 \Rightarrow - 1 \leq y \leq 1$
But $f(x)$ attains only non-negative values.
Hence, $y = f(x) = \lbrack 0,\ 1\rbrack$