Question

Which of the following functions are strictly decreasing on $(0,\frac{\pi }{2})$.

i) cosx

ii) cos 2x

iii) cos 3x

iv) tan x

Answer 1

Let $f\left(x\right)=cosx,then{f}^{\prime }\left(x\right)=-sinx$, In interval $(0,\frac{\pi }{2}),f^{\prime }\left(x\right)<0$
Therefore, f(x) is strictly decreasing on $(0,\frac{\pi }{2})$.

Let $f\left(x\right)=cos2x\Rightarrow f^{\prime }\left(x\right)=-2sin2x$. In interval $(0,\frac{\pi }{2}).f^{\prime }\left(x\right)<0$
Because sin2x will either lie in the first or second quadrant which will give a positive value. Therefore, f(x) is strictly decreasing on $(0,\frac{\pi }{2})$.

Let f(x) =cos3x , f'(x)=-3sin3x. In interval $(0,\frac{\pi }{2}),f^{\prime }\left(x\right)<0$
Because sin 3x will either lie in the first or second quadrant which will give a positive value. Therefore, f(x) is strictly decreasing on $(0,\frac{\pi }{2})$.
When $xϵ(\frac{\pi }{3},\frac{\pi }{2})$, then $f^{\prime }\left(x\right)>0$
Because $sin3x$ will lie in the third quadrant.
Therefore, f(x) is not strictly decreasing on $(0,\frac{\pi }{2})$

Let $f\left(x\right)=tanx\Rightarrow f^{\prime }\left(x\right)=se{c}^{2}x$. In interval $xϵ(0,\frac{\pi }{2}),f^{\prime }\left(x\right)>0$
Therefore, f(x) is not strictly decreasing on $(0,\frac{\pi }{2})$