If $ f(x)=\cos \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x $, then

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

JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

f(x)=\cos \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x

, then

View answer and explanation

Correct option: D

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

Option D (

f(\frac{\pi}{2})=-1

) is correct because substituting

x = \frac{\pi}{2}

into the simplified function yields -1.

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