A) 1
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) None of these
Correct Answer: B
Solution :
\[\int_{0}^{k}{\frac{1}{2+8{{x}^{2}}}dx=\frac{1}{2}\int_{0}^{k}{\frac{dx}{1+{{(2x)}^{2}}}=\frac{1}{4}\int_{0}^{2k}{\frac{dt}{1+{{t}^{2}}}}}}\] \[=\frac{1}{4}|{{\tan }^{-1}}t|_{0}^{2k}=\frac{1}{4}{{\tan }^{-1}}2k\]. Comparing it with the given value, we get \[{{\tan }^{-1}}2k=\frac{\pi }{4}\Rightarrow 2k=1\Rightarrow k=\frac{1}{2}\].
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