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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Mathematics
Chapter
Area Under Curves

BITSAT 2022 · Mathematics · Area Under Curves

The area enclosed by the curves y= x+ x and y=| x- x| over the...

The area enclosed by the curves y=sin⁡x+cos⁡xy=\sin x+\cos x and y=∣cos⁡x−sin⁡x∣y=|\cos x-\sin x| over the interval [0,π2]\left[0, \frac{\pi}{2}\right] isThe area enclosed by the curves y= x+ x and y=| x- x| over the interval [0, (π)/(2)] isThe area enclosed by the curves y= x+ x and y=| x- x| over the interval [0, (π)/(2)] is

Options

  1. A4(2−1)4(\sqrt{2}-1)
  2. B22(2−1)2 \sqrt{2}(\sqrt{2}-1)
  3. C2(2+1)2(\sqrt{2}+1)
  4. D22(2+1)2 \sqrt{2}(\sqrt{2}+1)

Correct answer

B. 22(2−1)2 \sqrt{2}(\sqrt{2}-1)

Step-by-step solution

Given, y=sin⁡x+cos⁡x,x∈[0,π/2]dydx=cos⁡x−sin⁡xy=\sin x+\cos x, x \in[0, \pi / 2]\frac{d y}{d x}=\cos x-\sin xy=∣cos⁡x−sin⁡x∣=[cos⁡x−sin⁡xx∈[0,π/4]sin⁡x−cos⁡xx∈[π/4,π/2]]\begin{aligned} y & =|\cos x-\sin x| \\ & =\left[\begin{array}{cc}\cos x-\sin x & x \in[0, \pi / 4] \\ \sin x-\cos x & x \in[\pi / 4, \pi / 2]\end{array}\right]\end{aligned}<imgsrc="/api/proxy−image?url=https<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fbitsat%252FBqoq95Rpm-j5ycIe8E3qOccP5zZI1leuKUG20Kyp-38.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2FBqoq95Rpm-j5ycIe8E3qOccP5zZI1leuKUG20Kyp-38.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2FBqoq95Rpm-j5ycIe8E3qOccP5zZI1leuKUG20Kyp-38.original.fullsize.png?alt=media"><br>\thereforeRequiredareaRequired area=∫0π/4∣(sin⁡x+cos⁡x)−(cos⁡x−sin⁡x)∣dx+∫π/4π/2∣2cos⁡x∣dx\begin{aligned} &=\int_0^{\pi / 4}|(\sin x+\cos x)-(\cos x-\sin x)| d x+\int_{\pi / 4}^{\pi / 2}|2 \cos x| d x \end{aligned}=∫0π/4∣2sin⁡x∣dx+∫π/4π/2∣2cos⁡x∣dx=2(−cos⁡x)0π/4+2(sin⁡x)π/4π/2=2[−12+1+1−12]<imgsrc="/api/proxy−image?url=https=\int_0^{\pi / 4}|2 \sin x| d x+\int_{\pi / 4}^{\pi / 2}|2 \cos x| d x=2(-\cos x)_0^{\pi / 4}+2(\sin x)_{\pi / 4}^{\pi / 2}=2 \left[-\frac{1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}}\right]<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fbitsat%252Fh0i2UrjkiqoD78hh_1S802WjWVfCXjP7KBP5OpVWwrA.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2Fh0i2UrjkiqoD78hh_1S802WjWVfCXjP7KBP5OpVWwrA.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2Fh0i2UrjkiqoD78hh_1S802WjWVfCXjP7KBP5OpVWwrA.original.fullsize.png?alt=media"><br>=2(2−22)=2(2−2)=4−22=22(2−1)\begin{aligned}=& 2\left(2-\frac{2}{\sqrt{2}}\right)\\ = & 2(2-\sqrt{2})\\ = & 4-2 \sqrt{2}\\ = & 2 \sqrt{2}(\sqrt{2}-1)\end{aligned}$
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