Check whether following function is strictly decreasing on $(0, \frac{π}{2})$ or not.

cos 2x

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Solution

Let $f (x) = c o s 2 x \Rightarrow f^{'} (x) = - 2 s i n 2 x$ . In interval $(0, \frac{π}{2}) . f^{'} (x) < 0$
Because sin2x will either lie in the first or second quadrant which will give a positive value. Therefore, f(x) is strictly decreasing on $(0, \frac{π}{2})$ .

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