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

The Correct Option is C

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 $