A) 3
B) 0
C) ?6
D) 1/6
Correct Answer: B
Solution :
\[\underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}=\underset{x\to 3}{\mathop{\lim }}\,(x+3)=6\] and \[f(3)=2(3)+k=6+k\] \[\because f\] is continuous at \[x=3\]; \ \[6+k=6\Rightarrow k=0\].You need to login to perform this action.
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