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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