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Area Under the Curves question

2021 · 1 Sep · Shift 2 · Q31
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  5. /2021 · 1 Sep · Shift 2 · Q31

Area Under the Curves question

2021 · 1 Sep · Shift 2 · Q31

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area, enclosed by the curves y=sin⁡x+cos⁡xy = \sin x + \cos xy=sinx+cosx and y=∣cos⁡x−sin⁡x∣y = \left| {\cos x - \sin x} \right|y=∣cosx−sinx∣ and the lines x=0,x=π2x = 0,x = {\pi \over 2}x=0,x=2π​, is :
  1. A
    22(2−1)2\sqrt 2 (\sqrt 2 - 1)22​(2​−1)
  2. B
    2(2+1)2(\sqrt 2 + 1)2(2​+1)
  3. C
    4(2−1)4(\sqrt 2 - 1)4(2​−1)
  4. D
    22(2+1)2\sqrt 2 (\sqrt 2 + 1)22​(2​+1)
View written solutionFree

Correct answer: A

  1. Given curves

We need the area enclosed between y1=sin⁡x+cos⁡xy_1=\sin x+\cos xy1​=sinx+cosx and y2=∣cos⁡x−sin⁡x∣y_2=|\cos x-\sin x|y2​=∣cosx−sinx∣ for 0≤x≤π2.0\le x\le \frac{\pi}{2}.0≤x≤2π​.

So the required area is A=∫0π/2∣y1−y2∣ dx.A=\int_0^{\pi/2} |y_1-y_2|\,dx.A=∫0π/2​∣y1​−y2​∣dx.

  1. Handle the modulus carefully

On [0,π/2][0,\pi/2][0,π/2], the expression cos⁡x−sin⁡x\cos x-\sin xcosx−sinx changes sign at cos⁡x=sin⁡x  ⟹  x=π4.\cos x=\sin x \implies x=\frac{\pi}{4}.cosx=sinx⟹x=4π​.

Hence,

  • for 0≤x≤π40\le x\le \frac{\pi}{4}0≤x≤4π​, ∣cos⁡x−sin⁡x∣=cos⁡x−sin⁡x,|\cos x-\sin x|=\cos x-\sin x,∣cosx−sinx∣=cosx−sinx,
  • for π4≤x≤π2\frac{\pi}{4}\le x\le \frac{\pi}{2}4π​≤x≤2π​, ∣cos⁡x−sin⁡x∣=sin⁡x−cos⁡x.|\cos x-\sin x|=\sin x-\cos x.∣cosx−sinx∣=sinx−cosx.

Thus,

+\int_{\pi/4}^{\pi/2}\left[(\sin x+\cos x)-(\sin x-\cos x)\right]dx.$$ 3. **Simplify the integrands** For $0\le x\le \pi/4$: $$ (\sin x+\cos x)-(\cos x-\sin x)=2\sin x.$$ For $\pi/4\le x\le \pi/2$: $$ (\sin x+\cos x)-(\sin x-\cos x)=2\cos x.$$ So, $$A=\int_0^{\pi/4}2\sin x\,dx+\int_{\pi/4}^{\pi/2}2\cos x\,dx.$$ 4. **Evaluate the integrals** First integral: $$\int_0^{\pi/4}2\sin x\,dx=2\left[-\cos x\right]_0^{\pi/4} =2\left(1-\frac{1}{\sqrt2}\right).$$ Second integral: $$\int_{\pi/4}^{\pi/2}2\cos x\,dx=2\left[\sin x\right]_{\pi/4}^{\pi/2} =2\left(1-\frac{1}{\sqrt2}\right).$$ Therefore, $$A=4\left(1-\frac{1}{\sqrt2}\right).$$ Now simplify: $$4\left(1-\frac{1}{\sqrt2}\right)=4\left(\frac{\sqrt2-1}{\sqrt2}\right) =\frac{4(\sqrt2-1)}{\sqrt2}=2\sqrt2(\sqrt2-1).$$ 5. **Match with the options** Thus the required area is $$\boxed{2\sqrt2(\sqrt2-1)}.$$ This corresponds to **Option A**.
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