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The number of real solutions of equarion \sqrt{1+\cos x}=\sqrt{2}\cos^{-1}\left ( \cos x \right ) in \left [ \frac{\pi}{2},\pi \right ] is

A. 0

B. 1

C.  2

D. Infinite

Answers (1)


We have , \sqrt{1+\cos 2x}=\sqrt{2}\cos^{-1}\left ( \cos x \right ), x is in \left [ \frac{\pi}{2}, \pi \right ]


\sqrt{2}\cos^{-1}\left ( \cos x \right )=\sqrt{2}x

So, \sqrt{1+\cos 2x}=\sqrt{2}x

Squaring both side , we get,

\left ( 1+\cos 2 x \right )=2x^{2}

\Rightarrow \cos 2x=2x^{2}-1

Now plotting cos 2x and 2x2-1, we get,

As , there is no point of intersection in \left [ \frac{\pi}{2},\pi \right ], so therre is no

solution of the given equation in \left [ \frac{\pi}{2},\pi \right ]




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