RAJASTHAN ­ PET Rajasthan PET Solved Paper-2007

  • question_answer
    Let a non-negative continuous function\[f(x)\]is such that the area of the region bounded by the curve\[y=f(x),\]\[x-\]axis and the lines\[x=\frac{\pi }{4}\]ands\[x=\beta >\frac{\pi }{4}\]is\[\left( \beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta  \right),\]then the value of \[f\left( \frac{\pi }{2} \right)\]is

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

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

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

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

    Correct Answer: D

    Solution :

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


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