A) \[f\left( \frac{\pi }{4} \right)=2\]
B) \[f(-\pi )=2\]
C) \[f(\pi )=1\]
D) \[f\left( \frac{\pi }{2} \right)=-1\]
Correct Answer: D
Solution :
\[f(x)=\cos \,[{{\pi }^{2}}]x+\cos \,[-{{\pi }^{2}}]\,x\] \[f(x)=\cos (9x)+\cos (-10x)\]\[=\cos (9x)+\cos (10x)\] \[=2\cos \left( \frac{19x}{2} \right)\cos \left( \frac{x}{2} \right)\] \[f\left( \frac{\pi }{2} \right)=2\cos \left( \frac{19\pi }{4} \right)\cos \left( \frac{\pi }{4} \right)\]; \[f\left( \frac{\pi }{2} \right)=2\times \frac{-1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=-1\].
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