Question

If $f(x)=\cos \,x, g(x)=\cos\, 2 x, h(x)=\cos \,3 x$ and $I(x)=\tan \,x$ then which of the following option is correct?

• A. $f ( x )$ and $g ( x )$ are strictly decreasing in $(0, \pi / 2)$
• B. $h ( x )$ is neither increasing nor decreasing in $(0, \pi / 2)$
• C. $I ( x )$ is strictly increasing in $(0, \pi / 2)$
• D. All are correct

Answer 1

Answer: (D)

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