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