2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced AIEEE Solved Paper-2005

  • question_answer
    Let\[f(x)\]be a non-negative continuous function such that the area bounded by the curve\[y=f(x),\]X-axis and the ordinates\[x=\pi /4\] and \[x=\beta >\pi /4\] is\[\left( \beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta  \right)\].Then\[f\left( \frac{\pi }{2} \right)\],is     AIEEE  Solved  Paper-2005

    A) \[\left( 1-\frac{\pi }{4}+\sqrt{2} \right)\]              

    B) \[\left( 1-\frac{\pi }{4}-\sqrt{2} \right)\]

    C)        \[\left( \frac{\pi }{4}-\sqrt{2}+1 \right)\]              

    D)        \[\left( \frac{\pi }{4}+\sqrt{2}-1 \right)\]

    Correct Answer: A

    Solution :

    Since, \[\int_{\pi /4}^{\beta }{f(x)}dx=\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta \] On differentiating w.r.t.\[\beta \]both sides, we get \[f(\beta )=\sin \beta +\beta \cos \beta -\frac{\pi }{4}\sin \beta +\sqrt{2}\] So, \[f\left( \frac{\pi }{2} \right)=1+0-\frac{\pi }{4}+\sqrt{2}=1-\frac{\pi }{4}+\sqrt{2}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner