Answer:
Option C
Explanation:
We have,
$f(x)=\begin{cases}a\left(\frac{1-\cos 2x}{x^{2}}\right), & for x < 0\\b ,& x = 0\\\frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}&for x>0\end{cases}$
f(x) is continuous at x=0
$\therefore$ $\lim_{x \rightarrow{0^{-}}}f(x)= f(0)$
$\lim_{x \rightarrow{0^{-}}}\frac{a(1- \cos 2x)}{x^{2}}=b$
$\lim_{x \rightarrow{0^{}}}\frac{a(2 \sin^{2} x)}{x^{2}}=b$
$\Rightarrow$ 2a=b
Also, $\lim_{x \rightarrow{0^{+}}}f(x)=b$
$\therefore$ $\lim_{x \rightarrow{0^{}}}\frac{\sqrt{x}}{\sqrt{4+\sqrt{x-2}}}=b$
$\Rightarrow\lim_{x \rightarrow{0^{}}}\frac{\sqrt{x}(\sqrt{4+\sqrt{x}}+2)}{\sqrt{x}}=b$
$\Rightarrow$ 4=b , a=2
$\therefore$ a+b=2+4=6
