A) 1
B) \[-1\]
C) 2
D) \[-2\]
E) 3
Correct Answer: E
Solution :
Since, \[f(x)=\left\{ \begin{matrix} \frac{{{({{e}^{x}}-1)}^{2}}}{\sin \left( \frac{x}{a} \right)\log \left( 1+\frac{x}{4} \right)} & ,x\ne 0 \\ 12 & ,x=0 \\ \end{matrix} \right.\] is continuous at\[x=0\]. \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{({{e}^{x}}-1)}^{2}}}{\sin \left( \frac{x}{a} \right)\log \left( 1+\frac{x}{4} \right)}=f(0)\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{({{e}^{x}}-1)}^{2}}}{\frac{x}{a}.\left( \frac{\sin \left( \frac{x}{a} \right)}{\frac{x}{a}} \right).\frac{x}{4}\left( \frac{\log \left( 1+\frac{x}{4} \right)}{\frac{x}{4}} \right)}=12\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\left( 1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+.....-1 \right)}^{2}}}{\frac{x}{a}.\left( \frac{\sin \left( \frac{x}{a} \right)}{\frac{x}{a}} \right).\frac{x}{4}\left( \frac{\log \left( 1+\frac{x}{4} \right)}{\frac{x}{4}} \right)}=12\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\left( 1+\frac{x}{2!}+\frac{{{x}^{2}}}{3!}+\frac{{{x}^{3}}}{4!}+.... \right)}^{2}}}{\frac{1}{4a}.\left( \frac{\sin \left( \frac{x}{a} \right)}{\frac{x}{a}} \right)\left( \frac{\log \left( 1+\frac{x}{4} \right)}{\frac{x}{4}} \right)}=12\] \[\Rightarrow \] \[4a=12\] \[\Rightarrow \] \[a=3\]
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