JEE Main & Advanced JEE Main Paper (Held On 12 April 2014)

  • question_answer
    If for a continuous function \[f(x),\int\limits_{-\pi }^{t}{\left( f\left( x \right)+x \right)dx}={{\pi }^{2}}-t2,\]for all\[t\ge -\pi ,\]then\[\left( -\frac{\pi }{3} \right)\]is equal to:   [JEE Main Online Paper ( Held On 12 Apirl  2014 )

    A) \[\pi \] 

    B) \[\frac{\pi }{2}\]

    C) \[\frac{\pi }{3}\]                                              

    D) \[\frac{\pi }{6}\]

    Correct Answer: A

    Solution :

                    Let\[\int_{-\pi }^{t}{(f(x)+)dx}={{\pi }^{2}}-{{t}^{2}}\] \[\Rightarrow \]\[\int_{-\pi }^{t}{f(x)dx+\int_{-\pi }^{t}{xdx}}={{\pi }^{2}}-{{t}^{2}}\] \[\Rightarrow \]\[\int_{-\pi }^{t}{f(x)dx+\left( \frac{{{t}^{2}}}{2}-\frac{{{\pi }^{2}}}{2} \right)}={{\pi }^{2}}-{{t}^{2}}\] \[\Rightarrow \]\[\int_{-\pi }^{t}{f(x)dx=\frac{3}{2}}({{\pi }^{2}}-{{t}^{2}})\] differentiating with respect to t \[\frac{d}{dt}\left[ \int_{-\pi }^{t}{f}(x)dx \right]=\frac{3}{2}\frac{d}{dt}({{\pi }^{2}}-{{t}^{2}})\] \[f(t).\frac{dt}{dt}-f(-\pi )\frac{d}{dt}(-\pi )=-3t\] \[f(t)=-3t\] \[f\left( -\frac{\pi }{3} \right)=-3\left( -\frac{\pi }{3} \right)=\pi \]                                

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