Question

If $f(x)=\cos x,g(x)=\cos 2x,h(x)=\cos 3x$ 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 $\left(0,\frac{\pi }{2}\right),f^{\prime }(x)<0$
Therefore, $f(x)$ is strictly decreasing on $\left(0,\frac{\pi }{2}\right)$
(b) Let $f(x)=\cos 2x\Rightarrow f^{\prime }(x)=-2\sin x2x$
In interval $\left(0,\frac{\pi }{2}\right),f^{\prime }(x)<0$
Because $\sin 2x$ 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 $f^{\prime }(x)=\cos 3x$
$\Rightarrow f^{\prime }(x)=-3\sin 3x.$
In Interval $\left(0,\frac{\pi }{3}\right),f^{\prime }(x)<0$
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 $\left(0,\frac{\pi }{3}\right)$.
When $x\in \left(\frac{\pi }{3},\frac{\pi }{2}\right)$, then $f^{\prime }(x)>0$
Because $\sin 3x$ will lie in the third quadrant.
Therefore, $f(x)$ is not strictly decreasing on $\left(0,\frac{\pi }{2}\right)$
(d) Let $f(x)=\tan x\Rightarrow f^{\prime }(x)={\sec }^{2}x$.
In Interval $x\in \left(0,\frac{\pi }{2}\right),f^{\prime }(x)>0$
Therefore, $f(x)$ is not strictly decreasing on $\left(0,\frac{\pi }{2}\right)$