If $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1-\cos 2 x}}{x \sqrt{2}}, & x \neq 0 \\ k, & x=0\end{array}\right.$,

then the value of k will make function f continuous at x = 0 is :

1

$f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1-\cos 2 x}}{x \sqrt{2}}, & x \neq 0 \\ k, & x=0\end{array}\right.$

for f to be continuous 

$f(0) = \lim\limits_{x → 0} f(x) \Rightarrow \lim\limits_{x → 0} f(x) = K$

so $\lim\limits_{x → 0} \frac{\sqrt{1-\cos 2x}}{x \sqrt{2}} = \frac{0}{0}$  form

so $\lim\limits_{x → 0} \frac{\sqrt{2\sin 2x}}{x \sqrt{2}}$

as $\cos 2x = 1 - 2 \sin^2 x$

so $2 \sin^2 x = 1 - \cos 2x$

$\lim\limits_{x → 0} \frac{\sin x}{x} = 1$

$=\lim\limits_{x → 0} \frac{\sqrt{2 \sin x}}{\sqrt{2} \sqrt{x}}$

= 1 = k