If a real-valued function is given by: is an onto function, then find the co-domain for . If the…

$y = \sqrt{25 - x^2} \Rightarrow x = \sqrt{25 - y^2}$ $25 - y^2 \geq 0 \Rightarrow (5-y)(5+y) \geq 0 \Rightarrow -5 \leq y \leq 5$ But, $y \geq 0$ $\therefore$ Range is $[0, 5]$. Given, $f(x)$ is an onto function the range is equal to the co-domain. $\therefore$ co-domain of $f(x)$ is $[0, 5]$

Let $x_1, x_2 \in [-5, 5]$. If $f(x_1) = f(x_2)$ $\Rightarrow \sqrt{25 - x_1^2} = \sqrt{25 - x_2^2}$ $\Rightarrow x_1 = \pm x_2$ $\therefore f(x)$ is not one-one in the given domain.

$f(a) = 4 \Rightarrow \sqrt{25 - a^2} = 4 \Rightarrow 25 - a^2 = 16 \Rightarrow a^2 = 9$ $\therefore a = \{-3, 3\}$