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JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    If\[y=\sqrt{{{a}^{2}}-{{b}^{2}}{{\cos }^{2}}(lnx)},\]then\[{{x}^{2}}{{y}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}\]is equal to

    A) \[{{a}^{2}}-{{y}^{2}}\]                   

    B) \[{{a}^{2}}+{{b}^{2}}-{{y}^{2}}\]

    C) \[({{a}^{2}}-{{y}^{2}})({{b}^{2}}-{{a}^{2}}+{{y}^{2}})\]

    D) \[({{b}^{2}}-{{y}^{2}})({{a}^{2}}-{{b}^{2}}+{{y}^{2}})\]

    Correct Answer: D

    Solution :

    \[{{y}^{2}}={{a}^{2}}-{{b}^{2}}{{\cos }^{2}}(lnx)\] \[\Rightarrow \]\[2y\frac{dy}{dx}=2{{b}^{2}}\cos (lnx)sin(lnx)\frac{1}{x}\] \[\therefore \]\[{{x}^{2}}{{y}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}={{b}^{4}}{{\cos }^{2}}(lnx)si{{n}^{2}}(lnx)\] \[={{b}^{4}}\left( \frac{{{a}^{2}}-{{y}^{2}}}{{{b}^{2}}} \right)\left( 1-\frac{{{a}^{2}}-{{y}^{2}}}{{{b}^{2}}} \right)\]


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