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Question: If \(f(x) = \cos\lbrack\pi^{2}\rbrack x + \cos\lbrack - \pi^{2}\rbrack x,\) then...

If f(x)=cos⁡[π2]x+cos⁡[−π2]x,f(x) = \cos\lbrack\pi^{2}\rbrack x + \cos\lbrack - \pi^{2}\rbrack x, then

A

f(π4)=2f\left( \frac{\pi}{4} \right) = 2

B

f(−π)=2f( - \pi) = 2

C

f(π)=1f(\pi) = 1

D

f(π2)=−1f\left( \frac{\pi}{2} \right) = - 1

Answer

f(π2)=−1f\left( \frac{\pi}{2} \right) = - 1

Explanation

Solution

f(x)=cos⁡[π2]x+cos⁡[−π2]xf(x) = \cos\lbrack\pi^{2}\rbrack x + \cos\lbrack - \pi^{2}\rbrack x

f(x)=cos⁡(9x)+cos⁡(−10x)=cos⁡(9x)+cos⁡(10x)f(x) = \cos(9x) + \cos( - 10x) = \cos(9x) + \cos(10x) =2cos⁡(19x2)cos⁡(x2)= 2\cos\left( \frac{19x}{2} \right)\cos\left( \frac{x}{2} \right)

f(π2)=2cos⁡(19π4)cos⁡(π4)f\left( \frac{\pi}{2} \right) = 2\cos\left( \frac{19\pi}{4} \right)\cos\left( \frac{\pi}{4} \right); f(π2)=2×−12×12=−1f\left( \frac{\pi}{2} \right) = 2 \times \frac{- 1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = - 1.