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JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    Let \[g(x)=\cos {{x}^{2}},\] \[f(x)=\sqrt{x},\] and \[\alpha ,\beta (\alpha <\beta )\] be the roots of the quadratic equation \[18{{x}^{2}}-9\pi x+{{\pi }^{2}}=0\]. Then the area (in sq. units) bounded by the curve \[y=(g\circ f)(x)\] and the lines \[x=\alpha ,\,\,x=\beta \] and\[y=0\], is: [JEE Main Online 08-04-2018]

    A)  \[\frac{1}{2}\left( \sqrt{3}-\sqrt{2} \right)\]              

    B)  \[\frac{1}{2}\left( \sqrt{2}-1 \right)\]

    C)  \[\frac{1}{2}\left( \sqrt{3}-1 \right)\]             

    D)  \[\frac{1}{2}\left( \sqrt{3}+1 \right)\]

    Correct Answer: C

    Solution :

    \[g(x)=\cos {{x}^{2}},f(x)=\sqrt{x}\] \[y=g(f(x))=\cos x\] \[18{{x}^{2}}-9\pi x+{{\pi }^{2}}=0\] \[18{{x}^{2}}-6\pi x-3\pi x+{{\pi }^{2}}=0\] \[(6x-\pi )(3x-\pi )=0\] \[x=\frac{\pi }{6},\frac{\pi }{3}\] \[\alpha =\frac{\pi }{6},\beta =\frac{\pi }{3}\] Req. Area\[=\int_{\pi /6}^{\pi /3}{\cos x.dx=[\sin x]_{\pi /6}^{\pi /3}=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}}\]


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