Let f(x) = cos(pi(|x| + 2[x])),where [.] repr | vedclass

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(C) Given $f(x) = \cos(\pi(|x| + 2[x]))$.Since $\cos(2n\pi + 	heta) = \cos(	heta)$ for any integer $n$,and $2[x]$ is an integer,we have:$f(x) = \cos

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The range of $f(x)$ is $[-1, 1]$.

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(C) Given $f(x) = \cos(\pi(|x| + 2[x]))$.Since $\cos(2n\pi + 	heta) = \cos(	heta)$ for any integer $n$,and $2[x]$ is an integer,we have:$f(x) = \cos(2\pi[x] + \pi|x|) = \cos(\pi|x|)$.Now,check for even/odd property:$f(-x) = \cos(\pi|-x|) = \cos(\pi|x|) = f(x)$.Since $f(-x) = f(x)$,the function is an even function.Regarding periodicity,$f(x) = \cos(\pi|x|)$ is periodic with period $T = 2$.Regarding the range,since $\cos(	heta)$ oscillates between $-1$ and $1$,the range of $f(x) = \cos(\pi|x|)$ is $[-1, 1]$.Regarding $f(x) = |f(x)|$,this is only true when $f(x) \geq 0$,which is not true for all $x$ (e.g.,at $x = 1$,$f(1) = \cos(\pi) = -1$).Thus,the correct statement is that the range of $f(x)$ is $[-1, 1]$.

Let $f(x) = \cos(\pi(|x| + 2[x]))$,where $[.]$ represents the greatest integer function. Then:

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