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KVPY Sample Paper KVPY Stream-SX Model Paper-9

  • question_answer
    The equation \[2\,{{\cos }^{2}}\frac{x}{2}{{\sin }^{2}}x={{x}^{2}}+{{x}^{-2}}\]\[0<x\le \frac{\pi }{2}\] has

    A) No real solution

    B) One real solution

    C) More than one solution

    D) None of these

    Correct Answer: A

    Solution :

    The given equation is\[2{{\cos }^{2}}\left( \frac{x}{2} \right){{\sin }^{2}}x={{x}^{2}}+\frac{1}{{{x}^{2}}}\]
    Where \[0<x\le \frac{\pi }{2}\]
    LHS=\[2{{\cos }^{2}}\frac{x}{2}{{\sin }^{2}}x=\left( 1+\cos \,x \right){{\sin }^{2}}x\]\[\because 1+\cos x<2\,and\,{{\sin }^{2}}x\le for\,0<x\le \frac{\pi }{2}\]
    \[\therefore \left( 1+\cos x \right){{\sin }^{2}}x<2\]and
    R.H.S=\[{{x}^{2}}+\frac{1}{{{x}^{2}}}\ge 2\]
    \[\therefore for\,0<x\le \frac{\pi }{2},\]
    Given equation is not possible for any real value of x.


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