A) 1/2
B) 1
C) 0
D) -1/2
Correct Answer: D
Solution :
Let\[\int\limits_{e}^{x}{t\,f(t)dt}=\sin x-x\cos x-\frac{{{x}^{2}}}{2}\] By using Leibnitsz rule, we get \[\frac{d}{dx}\left[ \int\limits_{e}^{x}{t\,f(t)}dt \right]=\frac{d}{dx}\left[ \sin x-x\cos x-\frac{{{x}^{2}}}{2} \right]\] \[\Rightarrow \]\[xf(x)-ef(e).0=x\sin x-x\] Now, put\[x=\frac{\pi }{6},\]we get \[\frac{\pi }{6}.f\left( \frac{\pi }{6} \right)=\frac{\pi }{6}.\sin \frac{\pi }{6}-\frac{\pi }{6}\] \[\Rightarrow \]\[f\left( \frac{\pi }{6} \right)=\sin \frac{\pi }{6}-1=\frac{1}{2}-1=-\frac{1}{2}\]
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