A) \[3\]
B) \[0\]
C) \[-6\]
D) \[1/6\]
Correct Answer: B
Solution :
We have \[f(x)\left\{ \begin{matrix} \frac{{{x}^{2}}-9}{x-3}, & if\,\,x\ne 3 \\ 2x+k, & if\,\,x=3 \\ \end{matrix} \right.\] Now, \[\underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}\] \[=\underset{x\to 3}{\mathop{\lim }}\,\frac{(x-3)(x+3)}{(x-3)}=\underset{x\to 3}{\mathop{\lim }}\,(x+3)=6\] Since,\[f(x)\]is continuous at\[x=3\] \[\therefore \] \[\underset{x\to 3}{\mathop{\lim }}\,f(x)=f(3)\] \[\Rightarrow \] \[6=6+k\] \[\Rightarrow \] \[k=0\]
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