A) \[4\]
B) \[\sqrt{32}\]
C) \[8\]
D) \[16\]
Correct Answer: C
Solution :
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 \[x=0.\] \[\therefore \] \[LHL=RHL=f(0)\] \[\Rightarrow \] \[a=8\]
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