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Question
Show that f(x) = cos2 x is a decreasing function on (0, π/2) ?
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Solution
\[f\left( x \right) = \cos^2 x\]
\[f'\left( x \right) = 2 \cos x \left( - \sin x \right)\]
\[ \Rightarrow f'\left( x \right) = - \sin \left( 2x \right) . . . \left( 1 \right)\]
\[\text { Now,}\]
\[0 < x < \frac{\pi}{2}\]
\[ \Rightarrow 0 < 2x < \pi \]
\[ \Rightarrow \sin 2x > 0 \left[ \because \text { Sine fuction is positive in first and second quadrant } \right]\]
\[ \Rightarrow - \sin 2x < 0\]
\[ \Rightarrow f'\left( x \right) < 0 \left[ \text { From eq.} (1) \right]\]
\[\text { So,f(x)is decreasing on}\left( 0, \frac{\pi}{2} \right).\]
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