CUET PG 2024 — Mathematics PYQ

The function is continuous at x=0
${\lim }_{x\to 0}f(x)$ $={\lim }_{r\to 0^{+}}f(x)=$f(a)

${\lim }_{x\to 0^{-}}f(x)={\lim }_{x\to 0^{-}}\frac{1-\cos 4x}{x^2}$

${\lim }_{x\to 0^{-}}\frac{2si{n}^{2}2x.4}{(2x{)}^2}$ = 8

$$\begin{array}{rl}& amp;\underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }f(x)\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}-4}}\\ & amp;\underset{x\to 0^{+}}{\lim }f(x)\sqrt{16+\sqrt{x}}+4\\ & amp;=8\\ & amp;\underline{\mathrm{So}}\mathrm{f}(\mathrm{a})=8\\ & amp;\underline{\mathrm{So}}\mathrm{a}=8\end{array}$$