For the function f(x)= x+3 x-2 (x^2+x ) , where x [0, 2 ] , consi… - Vidyadip

MCQ

For the function

$f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right) \text {, where } x \in\left[0, \frac{\pi}{2}\right] \text {, }$

consider the following two statements :

($I$) $\mathrm{f}$ is increasing in $\left(0, \frac{\pi}{2}\right)$.

($II$) $\mathrm{f}^{\prime}$ is decreasing in $\left(0, \frac{\pi}{2}\right)$.

Between the above two statements,

A
only ($I$) is true.
B
only ($II$) is true.
C
neither ($I$) nor ($II$) is true .
✓
both ($I$) and ($II$) are true.

✓

Answer

Correct option: D.

both ($I$) and ($II$) are true.

d
$ f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right) \quad x \in\left[0, \frac{\pi}{2}\right] $

$ \mathrm{f}^{\prime}(\mathrm{x})=\cos \mathrm{x}+3-\frac{2}{\pi}(2 \mathrm{x}+1)>0 \mathrm{f}(\mathrm{x}) \uparrow $

$ f^{\prime}(x)=-\sin x+0-\frac{\pi}{2}(2) $

$ =-\sin x-\frac{4}{\pi}<0 \quad f^{\prime}(x) \downarrow $

$ 0<\mathrm{x}<\frac{\pi}{2} $

$ \Rightarrow-\frac{2}{\pi}(\underset{+1}{0} \underset{+1}{0} \underset{+1}{2 x}<\pi) $

$ -\frac{2}{\pi}>\frac{-2}{\pi}(2 \mathrm{x}+1)>-\frac{2}{\pi}(\pi+1) $

$ 3-\frac{2}{\pi}>3-\frac{2}{\pi}(2 x+1)>3-\frac{2}{\pi}(\pi+1) $

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