A function $\displaystyle \mathrm{f}:[-4,4] \rightarrow[0,4]$ is given by $\displaystyle \mathrm{f}(\mathrm{x})=\sqrt{16-\mathrm{x}^{2}}$. Show that f is an onto function but not a one-one function. Further, find all possible values of ' $\displaystyle a$ ' for which $\displaystyle \mathrm{f}(a)=\sqrt{7}$. This section comprises $\displaystyle 3$ source based/case-based/passage based/integrated units of assessment questions of $\displaystyle 4$ marks each.

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