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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

Answers (1)

(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


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 onx\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<x<\frac{\pi}{3}$\\ $\Rightarrow 3 \times 0<3 x<\frac{\pi}{3} \times 3$ \\$0<3 x<\pi$
So wher \mathrm{x} \in\left(0, \frac{\pi}{3}\right), 3 \mathrm{x} \in(0, \pi) \ldots . .(\mathrm{a})$

\\ \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}<x<\frac{\pi}{2}$
\\\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)

\\ \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)$

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