QJEEAdv20Ma7c7b: JEE Advanced 2020 Mathematics - Class 12 JEEAdv20Ma7c7b Solution: The value of the limit is ...........

Question

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The value of the limit

\mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 (\sin 3x + \sin x)} \over {\left( {2\sin 2x\sin {{3x} \over 2} + \cos {{5x} \over 2}} \right) - \left( {\sqrt 2 + \sqrt 2 \cos 2x + \cos {{3x} \over 2}} \right)}}

is ...........

Accepted Answer

The limit

\mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 (\sin 3x + \sin x)} \over {\left( {2\sin 2x\sin {{3x} \over 2} + \cos {{5x} \over 2}} \right) - \left( {\sqrt 2 + \sqrt 2 \cos 2x + \cos {{3x} \over 2}} \right)}}

= \mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 (2\sin 2x\cos x)} \over {2\sin 2x\sin {{3x} \over 2} + \left( {\cos {{5x} \over 2} - \cos {{3x} \over 2}} \right) - \sqrt 2 (1 + cos2x)}}

= \mathop {\lim }\limits_{x \to {\pi \over 2}} {{8\sqrt 2 \sin 2x\cos x} \over {2\sin 2x\sin {{3x} \over 2} - 2\sin 2x\sin {x \over 2} - \sqrt 2 (2{{\cos }^2}x)}}

= \mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 \sin 2x\cos x} \over {\sin 2x\left( {\sin {{3x} \over 2} - \sin {x \over 2}} \right) - \sqrt 2 {{\cos }^2}x}}

= \mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 \sin 2x\cos x} \over {2\sin 2x\cos x\sin {x \over 2} - \sqrt 2 {{\cos }^2}x}}

= \mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 \sin 2x} \over {2\sin 2x\sin {x \over 2} - \sqrt 2 \cos x}}

= \mathop {\lim }\limits_{x \to {\pi \over 2}} {{8\sqrt 2 \sin x} \over {4\sin x\sin {x \over 2} - \sqrt 2 }}

= {{8\sqrt 2 } \over {{4 \over {\sqrt 2 }} - \sqrt 2 }} = {{16} \over {4 - 2}} = 8

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