A) \[3\]
B) \[0\]
C) \[-3\]
D) \[-1\]
Correct Answer: A
Solution :
Key Idea: For function \[f(x)\] to be continuous at\[x=a\] \[=\underset{x\to a}{\mathop{\lim }}\,f(x)=\underset{x\to a}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)\] Given,\[f(x)=\left\{ \begin{matrix} \frac{\sin 3x}{\sin x}, & x\ne 0 \\ k, & x=0 \\ \end{matrix} \right.\] \[\therefore \]For continuity \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=k\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin 3x}{\sin x}=k\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin 3x}{3x}\cdot \frac{3x}{\sin x}=k\] \[\Rightarrow \] \[3=k\] \[\therefore \]Value of\[k=3\].
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