A) \[k=0\]
B) \[k=1\]
C) \[k=-1\]
D) None of the above
Correct Answer: B
Solution :
Given, \[f(x)=\left\{ \begin{matrix} \frac{1-\cos 4x}{8{{x}^{2}}} & x\ne 0 \\ k, & x=0 \\ \end{matrix} \right.\] \[LHL=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)=\underset{h\to 0}{\mathop{\lim }}\,f(h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-\cos 4h}{8{{h}^{2}}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{4\sin 4h}{16h}\]? (using L Hospital?s rule) \[=1\times 1=1\] Since,/(x) is continuous at x = 0. \[\therefore f(0)=LHL\Rightarrow k=1\]
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