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Area Under the Curves question

2018 · Shift 0 · Q41
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  5. /2018 · Shift 0 · Q41

Area Under the Curves question

2018 · Shift 0 · Q41

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
Let g(x) = cosx2, f(x) = x\sqrt xx​ and α,β(α<β)\alpha ,\beta \left( {\alpha \lt \beta } \right)α,β(α<β) be the roots of the quadratic equation 18x2 - 9 π\piπ x + π2{\pi ^2}π2= 0. Then the area (in sq. units) bounded by the curve y = (gof)(x) and the lines x=αx = \alphax=α, x=βx = \betax=β and y = 0 is :
  1. A
    12(2−1){1 \over 2}\left( {\sqrt 2 - 1} \right)21​(2​−1)
  2. B
    12(3−1){1 \over 2}\left( {\sqrt 3 - 1} \right)21​(3​−1)
  3. C
    12(3+1){1 \over 2}\left( {\sqrt 3 + 1} \right)21​(3​+1)
  4. D
    12(3−2){1 \over 2}\left( {\sqrt 3 - \sqrt 2 } \right)21​(3​−2​)
View written solutionFree

Correct answer: B

  1. Interpret the functions

Given:

  • g(x)=cos⁡(x2)g(x)=\cos(x^2)g(x)=cos(x2)
  • f(x)=xf(x)=\sqrt{x}f(x)=x​

So,

(g∘f)(x)=g(f(x))=cos⁡((x)2)=cos⁡x(g\circ f)(x)=g(f(x))=\cos\left((\sqrt{x})^2\right)=\cos x(g∘f)(x)=g(f(x))=cos((x​)2)=cosx

Hence the required curve is:

y=cos⁡xy=\cos xy=cosx
  1. Find α\alphaα and β\betaβ

They are the roots of:

18x2−9πx+π2=018x^2-9\pi x+\pi^2=018x2−9πx+π2=0

Using the quadratic formula,

x=9π±(9π)2−4⋅18⋅π22⋅18x=\frac{9\pi\pm\sqrt{(9\pi)^2-4\cdot 18\cdot \pi^2}}{2\cdot 18}x=2⋅189π±(9π)2−4⋅18⋅π2​​ =9π±81π2−72π236=\frac{9\pi\pm\sqrt{81\pi^2-72\pi^2}}{36}=369π±81π2−72π2​​ =9π±3π36=\frac{9\pi\pm 3\pi}{36}=369π±3π​

So the roots are:

α=6π36=π6,β=12π36=π3\alpha=\frac{6\pi}{36}=\frac{\pi}{6}, \qquad \beta=\frac{12\pi}{36}=\frac{\pi}{3}α=366π​=6π​,β=3612π​=3π​

since α<β\alpha<\betaα<β.

  1. Set up the required area

The area bounded by y=cos⁡xy=\cos xy=cosx, x=αx=\alphax=α, x=βx=\betax=β, and y=0y=0y=0 is

∫αβcos⁡x dx\int_{\alpha}^{\beta} \cos x\,dx∫αβ​cosxdx

Since on [π6,π3]\left[\frac{\pi}{6},\frac{\pi}{3}\right][6π​,3π​], cos⁡x>0\cos x>0cosx>0, this integral directly gives the area.

Thus,

Area=∫π/6π/3cos⁡x dx\text{Area}=\int_{\pi/6}^{\pi/3} \cos x\,dxArea=∫π/6π/3​cosxdx =sin⁡x∣π/6π/3=\sin x\Big|_{\pi/6}^{\pi/3}=sinx​π/6π/3​ =sin⁡π3−sin⁡π6=\sin\frac{\pi}{3}-\sin\frac{\pi}{6}=sin3π​−sin6π​ =32−12=\frac{\sqrt{3}}{2}-\frac{1}{2}=23​​−21​ =12(3−1)=\frac{1}{2}(\sqrt{3}-1)=21​(3​−1)
  1. Match with the options
12(3−1)\frac{1}{2}(\sqrt{3}-1)21​(3​−1)

corresponds to Option B.

  1. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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