#### Which of the following functions is decreasing on $\left ( 0,\frac{\pi}{2} \right )$. A. sin2x B. tan x C. cos x D. cos 3x

(i) Let f(x)=sin 2x

Apply first derivative and get

f’(x)=2cos 2x

Put f’(x)=0, and get

2cos 2x =0

⇒ cos 2x=0

It is possible when

0≤x≤2π

Thus, sin 2x does not decrease or increase on $x\in \left ( 0,\frac{\pi}{2} \right )$

(ii) Let f(x)=tan x

Apply first derivative and get

f’(x)= $sec^2 x$

Now. square of every number is always positive,

So, tan x is increasing function in $x\in \left ( 0,\frac{\pi}{2} \right )$

(iii) Let f(x)=cos x

Apply first derivative and get

f’(x)=-sin x

But, sin x>0 for $x\in \left ( 0,\frac{\pi}{2} \right )$

And -sin x<0 for $x\in \left ( 0,\frac{\pi}{2} \right )$

Hence f’(x)<0 for $x\in \left ( 0,\frac{\pi}{2} \right )$

⇒ cos x is strictly decreasing on$x\in \left ( 0,\frac{\pi}{2} \right )$

(iv) Let f(x)=cos 3x

Apply first derivative and get

f’(x)=-3sin 3x

Put f’(x)=0, we get

-3sin 3x=0

⇒ sin 3x=0

Because sin θ=0 if θ=0, π, 2π, 3π

⇒ 3x=0,π, 2π, 3π

\begin{aligned} &\Rightarrow x=0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi\\ &x \in\left(0, \frac{\pi}{2}\right)\\ &\Rightarrow x=\frac{\pi}{3}\\ &\text { since }\\ &\mathrm{x} \in\left(0, \frac{\pi}{2}\right) \end{aligned}

so we write it on number line as

Now, this point $x=\frac{\pi}{3} divides the interval \left(0, \frac{\pi}{2}\right)$ into 2 disjoint intervals.
i.e. $\left(0, \frac{\pi}{3}\right) and \left(\frac{\pi}{3}, \frac{\pi}{2}\right)$
case 1 : for $\mathrm{x} \in\left(0, \frac{\pi}{3}\right)$
$\\0
So wher $\mathrm{x} \in\left(0, \frac{\pi}{3}\right), 3 \mathrm{x} \in(0, \pi) \ldots . .(\mathrm{a})$

Also,
$\\ \sin \theta>0 for \theta \in(0, \pi) \\\sin 3 x>0 for 3 x \in(0, \pi)$
From equation (a), we get
$\sin 3 x>0 for x \in\left(0, \frac{\pi}{3}\right)$
sin 3x <0 for $x \in\left(0, \frac{\pi}{3}\right)$

$\\\Rightarrow f^{\prime}(x)<0 for \\x \in\left(0, \frac{\pi}{3}\right) \\\Rightarrow f(x) is strictly decreasing \left(0, \frac{\pi}{3}\right)$
case 2: for $x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$
Now $\frac{\pi}{3}
$\\\Rightarrow 3 \times \frac{\pi}{3}<3 x<\frac{\pi}{2} \times 3 \\\Rightarrow \pi<3 x<\frac{3 \pi}{2} \\\mathrm{x} \in\left(\frac{\pi}{3}, \frac{\pi}{3}\right), 3 \mathrm{x} \in\left(\pi, \frac{3 \pi}{2}\right) \ldots(b)$

Also,
$\\ \sin \theta<0 in 3^{\text {rd }} quadrant \\\sin \theta<0 for \theta \in(0, \pi) \\\sin \theta<0 for \\\theta \in\left(\pi, \frac{3 \pi}{2}\right) \\\sin 3 x<0 for \\3 \mathrm{x} \in\left(\pi, \frac{3 \pi}{2}\right)$
Equation (b) gives
$\\ \sin 3 x<0 for x \in\left(\frac{\pi}{3}, \frac{3 \pi}{2 \times 3}\right) \\-\sin 3 x>0 for x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right) \\\Rightarrow f^{\prime}(x)<0 for x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right) \\\Rightarrow f(x) is strictly increasing on \left(\frac{\pi}{3}, \frac{\pi}{2}\right)$
Hence, cos 3 x does not decrease or increase on $\mathrm{x} \in\left(0, \frac{\pi}{2}\right)$
So, the correct answer is option C i.e., $\cos \mathrm{x}$ is decreasing in $\left(0, \frac{\pi}{2}\right)$