A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{-\pi }{2}\]
D) \[\frac{-\pi }{4}\]
Correct Answer: C
Solution :
\[f(x)=\int\limits_{0}^{\pi }{\cos t.\cos (x-t)dt}\] ?(1) \[\int\limits_{0}^{\pi }{-\cos t.\cos \,(x-\pi +t)\,}\] (using King) \[\therefore \,\,f(x)=\int\limits_{0}^{\pi }{\cos t.\,\cos (x+t)dt}\] ?(2) Now, (1) + (2) gives \[2f(x)\,=\int\limits_{0}^{\pi }{\cos t.(2\cos x.\cos t)dt}\] \[\therefore \,\,f(x)=\cos x\int\limits_{0}^{\pi }{{{\cos }^{2}}tdt=2\cos x\int\limits_{0}^{\pi /2}{{{\cos }^{2}}tdt}}\] \[\Rightarrow \,\,f(x)=\frac{\pi \cos x}{2}\] Hence, minimum value of f(x) is \[\frac{-\pi }{2}\].
You need to login to perform this action.
You will be redirected in
3 sec
