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QJEEM23Ma79dc: JEE Main 2023 Mathematics - Class 12 JEEM23Ma79dc Solution: The area of the region enclosed by the curve and the -axis is

The area of the region enclosed by the curve f(x)=max⁡{sin⁡x,cos⁡x},−π≤x≤πf(x)=\max \{\sin x, \cos x\},-\pi \leq x \leq \pi and the xx -axis is

Multiple Choice Options

  • A 22(2+1)2 \sqrt{2}(\sqrt{2}+1)
  • B 4
  • C 2(2+1)2(\sqrt{2}+1)
  • D 4(2)4(\sqrt{2})

Step-by-Step Verified Solution

JEE Main 2023 (Online) 13th April Morning Shift Mathematics - Area Under The Curves Question 42 English Explanation

Area =

= amp;∣∫−π−3π/4sin⁡xdx∣+∣∫π/4πsin⁡xdx∣+∣∫−3π/4−π/2cos⁡xdx∣+∣∫−π/2π/4cos⁡xdx∣\begin{aligned}&\left|\int_{-\pi}^{-3 \pi / 4} \sin x d x\right|+\left|\int_{\pi / 4}^\pi \sin x d x\right|+\left|\int_{-3 \pi / 4}^{-\pi / 2} \cos x d x\right| +\left|\int_{-\pi / 2}^{\pi / 4} \cos x d x\right|\end{aligned}

amp;=∣12−1∣+∣1+12∣+∣−1+12∣+∣12+1∣amp;=2+22+2−22=4\begin{aligned}& =\left|\frac{1}{\sqrt{2}}-1\right|+\left|1+\frac{1}{\sqrt{2}}\right|+\left|-1+\frac{1}{\sqrt{2}}\right|+\left|\frac{1}{\sqrt{2}}+1\right| \\\\& = 2+\frac{2}{\sqrt{2}}+2-\frac{2}{\sqrt{2}}=4\end{aligned}

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