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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Mathematics
Chapter
Area Under Curves

JEE Advanced 2013 · Paper 1 · Mathematics · Area Under Curves

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

The area enclosed by the curves y=sin⁡x+cos⁡xandy = \sin x + \cos x and\text{y} = \left|\text{cos x} - \text{sin x}\right|overtheinterval[0,π2]over the interval \left[0 \text{,} \dfrac{\pi}{2}\right] is

Options

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

Correct answer

B. 22(2−1)2 \sqrt{2} \left(\sqrt{2} - 1\right)

Step-by-step solution

The given curves are y=sin⁡x+cos⁡x and y=∣cos⁡x−sin⁡x∣y = \sin x + \cos x \text{ and } y = \left|\cos x - \sin x\right|

{\text{y}}_{1} = \text{sin x} + \text{cos x} = \sqrt{2} \text\sin \left(\text{x} + \dfrac{\pi}{4}\right)
{\text{y}}_{2} = \sqrt{2} | \text\sin \left(\dfrac{\pi}{4} - \text{x}\right) |
So, Required Area=∫0π4((sin x+cos x)−(cos x−sin x))dx+∫π4π2((sin x+cos x)−(sin x−cos x))dx= \mathop{∫}_{0}^{\dfrac{\pi}{4}}\left(\left(\text{sin x} + \text{cos x}\right) - \left(\text{cos x} - \text{sin x}\right)\right)\text{dx} + \mathop{∫}_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}\left(\left(\text{sin x} + \text{cos x}\right) - \left(\text{sin x} - \text{cos x}\right)\right)\text{dx}
2∫0π4sinxdx+2∫π4π2cosxdx2{∫}_{0}^{\dfrac{\pi}{4}}sinxdx + 2{∫}_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}cosxdx
=4−22=22(2−1)= 4 - 2\sqrt{2} = 2\sqrt{2}\left(\sqrt{2} - 1\right).

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