Question

If $f\left(x\right)=\{\begin{array}{c}\frac{1-\cos 4x}{x^2},x<0\\ a,x=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4},x>0\end{array}$, then the correct statement is-

• A. $f\left(x\right)$ is discontinuous at $x=0$ for any value of $a$
• B. $f\left(x\right)$ is continuous at $x=0$ when $a=8$
• C. $f\left(x\right)$ is continuous at $x=0$ when $a=0$
• D. None of these

Answer 1

The correct option is B $f\left(x\right)$ is continuous at $x=0$ when $a=8$

$\underset{x\to 0^{-}}{lim}\frac{1-\cos 4x}{x^2}=8$

$\underset{x\to 0^{+}}{lim}\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}=8$
$\therefore f\left(0\right)=8$
So $f\left(x\right)$ is continuous at $x=0$ when $a=8$