Verified Solution • 2025–26 Session moderate

QJEEM24Maf16e: JEE Main 2024 Mathematics - Class 12 JEEM24Maf16e Solution: For the function consider the following two statements : (I) is …

For the function

f(x)=sin⁡x+3x−2π(x2+x), where x∈[0,π2],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) ff is increasing in (0,π2)\left(0, \frac{\pi}{2}\right) .

(II) f′f^{\prime} is decreasing in (0,π2)\left(0, \frac{\pi}{2}\right) .

Between the above two statements,

Multiple Choice Options

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

Step-by-Step Verified Solution

amp;f(x)=sin⁡x+3x−2π(x2+x), where x∈[0,π2]amp;f′(x)=cos⁡x+3−2π(2x+1)amp;=cos⁡x−4xπ−2π+3amp; as x∈[0,π2]amp;4xπ∈[0,2]\begin{aligned}& f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right), \text { where } x \in\left[0, \frac{\pi}{2}\right] \\& f^{\prime}(x)=\cos x+3-\frac{2}{\pi}(2 x+1) \\& =\cos x-\frac{4 x}{\pi}-\frac{2}{\pi}+3 \\& \text { as } x \in\left[0, \frac{\pi}{2}\right] \\& \frac{4 x}{\pi} \in[0,2]\end{aligned}

\Rightarrow 3-\frac{2}{\pi}-\frac{4 x}{\pi}>0

and also \cos x>0 when x∈[0,π2]x \in\left[0, \frac{\pi}{2}\right]

\Rightarrow f^{\prime}(x)>0

⇒f(x)\Rightarrow f(x) is increasing

Now, f^{\prime \prime}(x)=-\sin x-\frac{4}{\pi}<0 \forall x \in\left[0, \frac{\pi}{2}\right]

Hence, f′(x)f^{\prime}(x) is decreasing

∴\therefore \quad Both statements (I) and (II) are true

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring