• # 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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