Question

If $f(x)=\{\begin{array}{ll}\frac{1-cos4x}{x^2},& x<0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4},& x>0\\ a,& x=0\end{array}$
is continuous at $x=0$, then $a=$

• A. $4$
• B. $6$
• C. $8$
• D. none of these

Answer 1

Answer: (C)

Since $f(x)$ is continuous at $x=0$
$\therefore L$.$H$.$L$. at $x=0=f(0)$ = $R$.$H$.$L$. at $x=0$
$\underset{x\to 0}{\lim }$ $\frac{1-cos4x}{x^2}=a$
$=\underset{x\to 0}{\lim }$ $\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$
$\underset{x\to 0}{\lim }$ $\frac{4sin4x}{2x}=a$
$=\underset{x\to 0}{\lim }$ $\sqrt{16+\sqrt{x}}+4$
$\underset{x\to 0}{\lim }$ $\frac{4sin4x}{4x}$ $\times \frac{4x}{2x}=a$
$=\underset{x\to 0}{\lim }$ $\sqrt{16+\sqrt{x}}+4$
$\therefore 8=a=8$