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Mathematics · Application of Derivatives

JEE Main 2024 — 5 April, Shift 1 — Question 5

For the function f(x)=sin⁡x+3x−2π(x2+x)\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+3 \mathrm{x}-\frac{2}{\pi}\left(\mathrm{x}^{2}+\mathrm{x}\right), where x∈[0,π2]\mathrm{x} \in\left[0, \frac{\pi}{2}\right], consider the following two statements :

Statement I : f is increasing in (0,π2)\left(0, \frac{\pi}{2}\right).

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

In the light of the above statements, choose the correct answer from the options given below :

  1. Option A:

    Both statement I and statement Il are correct.

  2. Option B:

    Statement I is correct and statement Il is incorrect.

    Correct
  3. Option C:

    Statement I is incorrect and statement Il is correct.

  4. Option D:

    Both statements 1 and statements ll are incorrect.

Answer: B

Step-by-step solution

f(x)=sin⁡x+3x−2π(x2+x)x∈[0,π2]f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^{2}+x\right) \quad x \in\left[0, \frac{\pi}{2}\right]

f′(x)=cos⁡x+3−2π(2x+1)>0;f(x)↑f^{\prime}(x)=\cos x+3-\frac{2}{\pi}(2 x+1)>0 ; f(x) \uparrow

f′′(x)=−sin⁡x+0−2π(2)f^{\prime\prime}(x)=-\sin x+0-\frac{2}{\pi}(2)

=−sin⁡x−4π<0;f′(x)↓=-\sin x-\frac{4}{\pi}<0 ; \quad f^{\prime}(x) \downarrow

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 1
Subject
Mathematics
Chapter
Application of Derivatives
Topic
Monotonicity

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