Question

If \left\\{ \begin{matrix} \frac{(1-\cos \,4x)}{{{x}^{2}}}, & if & x<0 \\\ a, & if & x=0, \\\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4}, & if & x>0 \\\ \end{matrix} \right. then $f(x)$ is continuous at $x=0,$ for $a = ?$

• A. $4$
• B. $\sqrt{32}$
• C. $8$
• D. $16$

Answer 1

Answer: (C)

$8$

Answer 2

Given, f(x)=\left\\{ \begin{matrix} \frac{(1-\cos 4x)}{{{x}^{2}}}, & if & x<0 \\\ a, & if & x=0 \\\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})-4}}, & if & x > 0 \\\ \end{matrix} \right.
LHL= $f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)$
$\underset{h\to 0}{\mathop{\lim }}\,\frac{1-\cos \,4(0-h)}{{{(0-h)}^{2}}}$
=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left\\{ \frac{{{(-4h)}^{2}}}{2!}-\frac{{{(-4h)}^{4}}}{4!}+..... \right\\}}{{{(-h)}^{2}}}
=\underset{h\to 0}{\mathop{\lim }}\,\,\left\\{ \frac{\frac{16{{h}^{2}}}{2!}-\frac{{{(-4h)}^{4}}}{4!}+....}{{{(-h)}^{2}}} \right\\}
$=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{16}{2!}-\frac{{{4}^{2}}{{h}^{2}}}{4!}+....$
RHL $=f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,\,f(0+h)$
$=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{\sqrt{h}}{\sqrt{16+\sqrt{h}}-4}$
$\left( \frac{0}{0}\,form \right)$
Using L- Hospital rule,
$=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{1}{2\sqrt{h}}\times \frac{1}{2\sqrt{16+\sqrt{h}\frac{1}{2\sqrt{h}}}}$
$=\underset{h\to 0}{\mathop{\lim }}\,\,2\sqrt{16+\sqrt{h}}=2\sqrt{16+0}$
$=2\times 4=8$
Since, $f(x)$ is continuous at $x=0.$
$\therefore$ $LHL=RHL=f(0)$
$\Rightarrow$ $a=8$