A) \[2\]
B) \[0\]
C) \[3\]
D) \[4\]
Correct Answer: C
Solution :
\[f(x)=\left\{ \begin{matrix} \frac{{{x}^{3}}-8}{{{x}^{2}}-4}, & if & x\ne 2 \\ k, & if & x=2 \\ \end{matrix} \right.\] Since, \[f(x)\] is continuous at \[x=2,\] then \[f(2-h)=f(2+h)=f(2),\] ie, LHL = RHL = value of function at \[x=2,\] Now, \[LHL=\underset{h\to 0}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(2-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{{{(2-h)}^{3}}-8}{{{(2-h)}^{2}}-4}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{3{{(2-h)}^{2}}}{2(2-h)}\] (By L Hospital rule) \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{3}{2}\,(2-h)\] \[=\frac{3}{2}\,.2=3\] \[=f(2)=k\] \[\therefore \] \[k=3\]You need to login to perform this action.
You will be redirected in
3 sec