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JEE Main & Advanced Sample Paper JEE Main Sample Paper-24

  • question_answer
    Let \[g(x)\] be a differentiable function on R and\[\int\limits_{\sin \,t}^{1}{{{x}^{2}}g(x)dx=(1-\sin \,t)}\], where \[t\in \left( 0,\frac{\pi }{2} \right)\]. Then the value of \[g\left( \frac{1}{\sqrt{2}} \right)\] equals

    A)  \[\frac{1}{2}\]                         

    B)  \[\frac{1}{\sqrt{2}}\]

    C)  2                                

    D)  1          

    Correct Answer: C

    Solution :

    \[\int\limits_{\sin t}^{1}{{{x}^{2}}g(x)\,dx=(1-\sin t)}\]c Differentiate w.r.t. t \[0-{{\sin }^{2}}\,t.g(\sin t)\,\cos t=-\cos t\] \[\therefore \,\,g(\sin t)\,=\frac{1}{{{\sin }^{2}}t}\] Put \[t=\frac{\pi }{4}\] \[\therefore \,\,g\left( \frac{1}{\sqrt{2}} \right)=2\]


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