A) 2
B) 1
C) -1
D) 0
Correct Answer: D
Solution :
Given, \[f(x)=\left\{ \begin{align} & \frac{2\sin x-\sin 2x}{2x\,\cos x},\,\text{if}\,x\ne 0 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{if}\,x=0 \\ \end{align} \right.\] Now, \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x-\sin 2x}{2x\,\cos x}\] \[\left( \frac{0}{0}\text{form} \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\cos x-2\cos 2x}{2(\cos x-x\sin x)}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2-2}{2(1-0)}=0\] Since, \[f(x)\]is continuous at x = 0 \[\therefore \] \[f(0)=\underset{x\to 0}{\mathop{lim}}\,f(x)\,\,\,\,\,\Rightarrow \,\,\,\,\,a=0\]
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