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

For the function $f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right), \text { where } x \in\left[0, \frac{\pi}{2}\right],$ consider the following two statements : (I) $\mathrm{f}$ is increasing in $\left(0, \frac{\pi}{2}\right)$. (II) $f^{\prime}$ is decreasing in $\left(0, \frac{\pi}{2}\right)$.
Between the above two statements,
  1. only (II) is true.
  2. only (I) is true.
  3. neither (I) nor (II) is true.
  4. both (I) and (II) are true

Solution

$\begin{aligned} & \mathrm{f}(\mathrm{x})=\sin \mathrm{x}+3 \mathrm{x}-\frac{2}{\pi}\left(\mathrm{x}^2+\mathrm{x}\right) \quad \mathrm{x} \in\left[0, \frac{\pi}{2}\right] \\ & \mathrm{f}^{\prime}(\mathrm{x})=\cos \mathrm{x}+3-\frac{2}{\pi}(2 \mathrm{x}+1)>0 \mathrm{f}(\mathrm{x}) \uparrow \\ & \mathrm{f}^{\prime}(\mathrm{x})=-\sin \mathrm{x}+0-\frac{\pi}{2}(2) \\ & =-\sin \mathrm{x}-\frac{4}{\pi} < 0 \quad \mathrm{f}^{\prime}(\mathrm{x}) \downarrow \\ & 0 < \mathrm{x} < \frac{\pi}{2}\end{aligned}$ $\Rightarrow-\frac{2}{\pi}(\underset{+1}{0}{ < \underset{+1}2 \mathrm{x}} < \underset{+1}\pi)$ $\begin{aligned} & \underset{+3}{-\frac{2}{\pi}}>\frac{-2}{\pi}(\underset{+3}2 \mathrm{x}+1)>-\frac{2}{\pi}(\underset{+3}\pi+1) \\ & \underset{+ve} 3-\frac{2}{\pi}>3-\frac{2}{\pi}(2 x+1)>3-\frac{2}{\pi}(\underset{+ve}\pi+1) \\ & \end{aligned}$

Asked in: JEE Main 2024 (05 Apr Shift 1)

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