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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Application of Derivatives

For the function f(x)=( x)-x+1, x ∈ ℝ , between the following two...

For the function f(x)=(cos⁡x)−x+1,x∈Rf(x)=(\cos x)-x+1, x \in \mathbb{R}, between the following two statements (S1) f(x)=0f(x)=0 for only one value of xin[0,π]x in [0, \pi]. (S2) f(x)f(x) is decreasing in \left[0, \frac{\pi}{2}\right]andincreasingin[π2,π]and increasing in \left[\frac{\pi}{2}, \pi\right].

Options

  1. ABoth (S1) and (S2) are correct.
  2. BBoth (S1) and (S2) are incorrect.
  3. COnly (S2) is correct.
  4. DOnly (S1) is correct.

Correct answer

D. Only (S1) is correct.

Step-by-step solution

f′(x)=cos⁡x−x+1f′(x)=−sin⁡x−1\begin{aligned} & f^{\prime}(x)=\cos x-x+1 \\ & f^{\prime}(x)=-\sin x-1\end{aligned}\mathrm{f}isdecreasingis decreasing\forall \mathrm{x} \in \mathrm{R} f(x)=0f(0)=2,f(π)=−π\begin{aligned} & \mathrm{f}(\mathrm{x})=0 \\ & \mathrm{f}(0)=2, \mathrm{f}(\pi)=-\pi\end{aligned} \mathrm{f}isstrictlydecreasinginis strictly decreasing in[0, \pi] and \mathrm{f}(0) . \mathrm{f}(\pi) < 0 \Rightarrowonlyonesolutionofonly one solution of\mathrm{f}(\mathrm{x})=0 \mathrm{S} 1iscorrectandis correct and\mathrm{S} 2$ is incorrect.
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