Question

If $f(x) = \begin{cases} \frac{1-cos\,4x}{x^2}, & \text{} x < 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text{} x > 0\\ a, & x = 0 \end{cases} $
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 $x = 0$
$\displaystyle \lim_{x \to 0}$ $\frac{1-cos\,4x}{x^2}=a$
$= \displaystyle \lim_{x \to 0}$ $\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$
$\displaystyle \lim_{x \to 0}$ $\frac{4\,sin\,4x}{2x}=a$
$= \displaystyle \lim_{x \to 0}$ $\sqrt{16+\sqrt{x}}+4$
$\displaystyle \lim_{x \to 0}$ $\frac{4\,sin\,4x}{4x}$ $\times \frac{4x}{2x}=a$
$ = \displaystyle \lim_{x \to 0}$ $\sqrt{16+\sqrt{x}}+4$
$\therefore 8=a=8$