If \int\limits_0^{{\pi \over 3}} {{{ an heta } \over {\sqrt {2k\,\sec heta } }}} \,d heta = 1 - {1 \over {\sqrt 2 }},\left( {k 0} ight), then value of k is :

Question

If \int\limits_0^{{\pi \over 3}} {{{	an 	heta } \over {\sqrt {2k\,\sec 	heta } }}} \,d	heta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} ight), then value of k is :

Accepted Answer

Answer: D. 2 \int\limits_0^{{\pi \over 3}} {{{	an 	heta } \over {\sqrt {2k\,\sec 	heta } }}} \,d	heta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} ight), \Rightarrow \,\,\int\limits_0^{\pi /3} {{{	an 	heta } \over {\sqrt {2k\sec 	heta } }}} \,d	heta = 1 - {1 \over {\sqrt 2 }}\left( {k > 0} ight) \Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - 2\sqrt {\cos 	heta } } ight)_0^{\pi /3} = 1 - {1 \over {\sqrt 2 }} \Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - \sqrt 2 + 2} ight) = 1 - {1 \over {\sqrt 2 }} \Rightarrow \,\,{{\sqrt 2 \left( {\sqrt 2 - 1} ight)} \over {\sqrt {2k} }} = {{\sqrt 2 - 1} \over {\sqrt 2 }} \Rightarrow \,\,k = 2

If $$\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),$$ then value of k is :

Answer

Explanation

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