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JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    If \[\int\limits_{\sin x}^{l}{{{t}^{2}}f(t)dt]=1-\sin x,x\in \left( 0,\frac{\pi }{2} \right)}\]then \[f\left( \frac{1}{\sqrt{2}} \right)\] is equal to

    A)  1                    

    B)  2

    C)  3                                

    D)  4

    Correct Answer: B

    Solution :

    On differentiating both sides with respect to x, we get \[0-{{\sin }^{2}}\,x.f(\sin x)\,\cos x=-\cos x\,\] \[\Rightarrow \,\,f(\sin x)=\frac{1}{{{\sin }^{2}}x}\] \[\therefore \,\,f\left( \frac{1}{\sqrt{2}} \right)\,={{(\sqrt{2})}^{2}}=2\]


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