Functions Question 616
If $ f(x)=\cos(x)
[{\pi }^{2}]x+\cos [{{\pi }^{2}}]x $ , then [Orissa JEE 2002]
Options:
A) $ f( \frac{\pi }{4} )=2 $
B) $ f(-\pi )=2 $
C) $ f(\pi )=1 $
D) $ f( \frac{\pi }{2} )=1 $
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Answer:
Correct Answer: D
Solution:
$ f(x)=\cos ,[{{\pi }^{2}}]x+\cos ,[-{{\pi }^{2}}],x $ $ f(x)=\cos (9x)+\cos (-10x) $ $ =\cos (9x)+\cos (10x) $ $ =2\cos ( \frac{19x}{2} )\cos ( \frac{x}{2} ) $ $ f( \frac{\pi }{2} )=2\cos ( \frac{19\pi }{4} )\cos ( \frac{\pi }{4} ) $ ; $ f( \frac{\pi }{2} )=2\times \frac{-1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=-1 $ .
BETA
