$\text { Let } f(x)=\frac{1}{\sqrt{4-x^{2}}} \text {. The domain and range of the real valued function } f \text { is - }$

$f(x)=\frac{1}{\sqrt{4-x^{2}}} \in \mathbb{R}$ $\therefore-2<x<2$ $\therefore y^{2}=\frac{1}{4-x^{2}}$ $\Rightarrow 4-x^{2}=\frac{1}{y^{2}}$ $\Rightarrow 4-\frac{1}{y^{2}}=x^{2}$ $\Rightarrow 4-x^{2}=\frac{1}{y^{2}}$ $|x|$ এর $\text { value minimum }$ হলে $\text { y mimimum }$ $\therefore-2<x<2,|x|_{mI n}=0$ $\therefore 4=\frac{1}{y_{m \ln }^{2}}$ $\therefore y_{m l n}=\frac{1}{2}$ $\therefore y_{m l n}=\frac{1}{2}$ $\therefore y \geq \frac{1}{2}$