A) 2
B) 1
C) 4
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
\[\Rightarrow \,\,\,\,\int\limits_{0}^{\pi /3}{\frac{\tan \,\theta }{\sqrt{2k\,\sec \,\theta }\,}d\theta \,=\,1-\frac{1}{\sqrt{2}}}(k>0)\] \[\Rightarrow \,\,\,\,\int\limits_{0}^{\pi /3}{\frac{\sin \,\theta }{\sqrt{2k\,}\,\sqrt{\cos \,\theta }\,}\,=\,1-\frac{1}{\sqrt{2}}}\] \[\Rightarrow \,\,\,\,\frac{1}{\sqrt{2\,k}}{{(-2\sqrt{cos\,\theta })}_{0}}^{\pi /3}=1-\frac{1}{\sqrt{2}}\] \[\Rightarrow \,\,\,\,\frac{1}{\sqrt{2\,k}}(-\sqrt{2}+2)=1-\frac{1}{\sqrt{2}}\] \[\Rightarrow \,\,\,\,\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}\,k}\,=\,\,\frac{\sqrt{2}-1}{\sqrt{2}}\] \[\Rightarrow \,\,\,k=2\]
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