KVPY Sample Paper KVPY Stream-SX Model Paper-5

  • 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 \[y=(gof)(x)\] and the lines \[x=\alpha ,\]\[x=\beta \] and \[y=0\] is

    A) \[\frac{\sqrt{3}+1}{2}\]            

    B) \[\frac{\sqrt{3}-\sqrt{2}}{2}\]

    C) \[\frac{\sqrt{2}-1}{2}\]              

    D) \[\frac{\sqrt{3}-1}{2}\]

    Correct Answer: D

    Solution :

    \[18{{x}^{2}}-9\pi x+{{\pi }^{2}}=0\]
    \[(3x-\pi )(6x-\pi )=0\]
    \[\Rightarrow \]\[\alpha =\frac{\pi }{6},\beta =\frac{\pi }{3}\]
    \[f(x)=\sqrt{x},g(x)=\cos {{x}^{2}}\] \[(gof)(x)=\cos x\]
    Area between \[\cos x,x=\frac{\pi }{6},x=\frac{\pi }{3}\] and \[y=0\] is \[A=\int_{\pi /6}^{\pi /3}{\cos xdx=[\sin x]_{\pi /6}^{\pi /3}}\]
    \[A=\sin \frac{\pi }{3}-\sin \frac{\pi }{6}=\frac{\sqrt{3}-1}{2}\]

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