JEE Main & Advanced Mathematics Definite Integration Question Bank Area Bounded by Region, Volume and Surface Area of Solids of Revolution

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

    Solution :

               Given that, \[\int_{\pi /4}^{\beta }{f\ (x)dx}\]\[=\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta \]                    Differentiating w.r.t. b, we get                     \[\therefore \] \[f(\beta )=\sin \beta +\beta \cos \beta -\frac{\pi }{4}\sin \beta +\sqrt{2}\],                    Hence, \[f\ \left( \frac{\pi }{2} \right)=\left( 1-\frac{\pi }{4}+\sqrt{2} \right)\].

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