A) \[\frac{\pi }{2}\]
B) 0
C) -1
D) \[-\frac{\pi }{2}\]
Correct Answer: D
Solution :
Let \[I=\int\limits_{0}^{\pi }{[\cos x]dx}\] ?(1) \[I=\int\limits_{0}^{\pi }{[\cos (\pi -x)]dx}=\int\limits_{0}^{\pi }{[-\cos x]dx}\] ?.(2) On adding (1) and (2), we get \[2I=\int\limits_{0}^{\pi }{[\cos x]dx+}\int\limits_{0}^{\pi }{[-\cos x]dx}\] \[2I=\int\limits_{0}^{\pi }{[\cos x]+}[-\cos x]dx\] \[2I=\int\limits_{0}^{\pi }{-1dx}(\because [x]+[-x]=-1\,if\,x\notin Z)\] \[\left. 2I=-x \right|_{0}^{\pi }=-\pi \]\[\Rightarrow \]\[I=\frac{-\pi }{2}\]
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