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Explain solution rd sharma class 12 chapter 16 Increasing and decreasing function exercise multiple choice question, question 34

Answers (1)

$\cos x$ , Option (c)

Hints: Check all the options and choose is satisfies

Given: Function is decreasing in $\left ( 0,\frac{\pi}{2} \right )$

Solution:

Option (A)

$\begin{aligned} &f(x)=\sin 2 x \\ &f^{\prime}(x)=2 \cos 2 x \end{aligned}$

$f(x)$ increases from ‘0’ to ‘1’ in $\left ( 0,\frac{\pi}{2} \right )$

Option (B)

$\begin{aligned} &f(x)=\tan x \\ &f^{\prime}(x)=\sec ^{2} x \end{aligned}$

In interval $\left ( 0,\frac{\pi}{2} \right )$ , $f^{\prime}(x)=-\sin x<0$

$f(x)=\cos x$ is strictly increasing in interval $\left ( 0,\frac{\pi}{2} \right )$

Option (C)

$\begin{aligned} &f(x)=\cos x \\ &f^{\prime}(x)=-\sin x \end{aligned}$

In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x)=-\sin x<0$

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

Option (D)

$\begin{aligned} &f(x)=\cos 3 x \\ &f^{\prime}(x)=-3 \sin 3 x \end{aligned}$

Now,

$\begin{aligned} &f^{\prime}(x)=0 \\ &\sin 3 x=0 \\ &3 x=\pi \end{aligned}$

As $x\epsilon \left ( 0,\frac{\pi}{2} \right )$

$x=\frac{\pi}{3}$

$f(x)=\cos 3 x$ is decreases only when $3 x \in\left(0, \frac{\pi}{2}\right)$

And $x \in\left(0, \frac{\pi}{6}\right)$

Therefore, Option (C) =cos x satisfies because $f(x)=\cos x$ is strictly decreasing in $\left ( 0,\frac{\pi}{2} \right )$

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