A function $\displaystyle \mathrm{f}:[-4,4] \rightarrow[0,4]$ is given by $\displaystyle \mathrm{f}(x)=\sqrt{16-x^{2}}$. Show that f is an onto function but not a one-one function. Further, find all possible values of ' $\displaystyle a$ ' for which $\displaystyle \mathrm{f}(a)=\sqrt{7}$.

Onto: Let $\displaystyle y=\sqrt{16-x^{2}} \Rightarrow y \geq 0$ Squaring we get, $\displaystyle x^{2}=16-y^{2} \Rightarrow x= \pm \sqrt{16-y^{2}}$ For each $\displaystyle y \in[-4,4]$, ' $\displaystyle x$ ' is a real number, $\displaystyle \therefore 0 \leq y \leq 4 \Rightarrow R_{\mathrm{f}}=[0,4]=C o-$ domain ∴ ' $\displaystyle \mathbf{f}$ ' is an onto function. One-One: $\displaystyle \mathbf{f}(\mathbf{- 1})=\mathbf{f}(\mathbf{1})=\sqrt{15}$ but $\displaystyle -\mathbf{1} \neq \mathbf{1}, \therefore$ ' $\displaystyle \mathbf{f}$ ' is not a one-one function. $\displaystyle \mathbf{f}(\mathbf{a})=\sqrt{7} \Rightarrow \sqrt{16-\mathbf{a}^{2}}=\sqrt{7} \Rightarrow \mathbf{a}= \pm 3$