A) \[k=0\]
B) \[k=1\]
C) \[k=-1\]
D) None of these
Correct Answer: B
Solution :
\[f(x)=\left\{ \begin{align} & \frac{1-\cos 4x}{8{{x}^{2}}},x\ne 0 \\ & k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,x=0 \\ \end{align} \right.\] If \[f(x)\] is continous function at point \[x=0\] then \[\underset{x\to 0+}{\mathop{\lim }}\,[f(x)=\underset{x\to 0-}{\mathop{\lim }}\,[f(x)]\]\[\underset{x\to 0}{\mathop{\lim }}\,[f(x)]=\underset{h\to 0-}{\mathop{\lim }}\,[f(0+h)]\] \[=\underset{h\to 0}{\mathop{\lim }}\,[f(h)]=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-\cos 4h}{8{{h}^{2}}}=\underset{h\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}2h}{8{{h}^{2}}}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}2h}{4{{h}^{2}}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,{{\left( \frac{\sin 2h}{2h} \right)}^{2}}={{(1)}^{2}}=1\] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\underset{\,\,\,\,\,h\to 0}{\mathop{=\lim }}\,[f(0-h)\]\[=\underset{h\to 0}{\mathop{\lim }}\,[f(-h)]=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-\cos 4(-h)}{8{{(-h)}^{2}}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-\cos 4h}{8{{h}^{2}}}\]\[=1\] \[f(0)=1\Rightarrow k=1\].
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