For the function f(x)= x+3 x-(2)/(π)(x^2+x), where x ∈[0, (π)/(2)], consider the following two statements : (I) f is increasing in (0, (π)/(2)) . (II) f^ is decreasing in (0, (π)/(2)) .Between the above two statements,

Correct answer: D. both (I) and (II) are true aligned & f(x)= x+3 x-(2)/(π)(x^2+x) x ∈[0, (π)/(2)] & f^ (x)= x+3-(2)/(π)(2 x+1)>0 f(x) & f^ (x)=- x+0-(π)/(2)(2) & =- x-(4)/(π) (-2)/(π)( +3 2 x+1)>-(2)/(π)( +3 π+1) & +ve 3-(2)/(π)>3-(2)/(π)(2 x+1)>3-(2)/(π)( +ve π+1) & aligned

JEE Main 20245 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Application of Derivatives

For the function f(x)= x+3 x-(2)/(π)(x^2+x), where x ∈[0, (π)/(2)],...

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,

Options

Aonly (II) is true.
Bonly (I) is true.
Cneither (I) nor (II) is true.
Dboth (I) and (II) are true

Correct answer

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

Step-by-step 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}

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