Consider the function f(x) = cos(x^2). Then, | vedclass

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(C) We have the function $f(x) = \cos(x^2)$.For a function to be periodic with period $T > 0$, it must satisfy $f(x+T) = f(x)$ for all $x$ in its doma

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$f$ is not periodic

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(C) We have the function $f(x) = \cos(x^2)$.For a function to be periodic with period $T > 0$, it must satisfy $f(x+T) = f(x)$ for all $x$ in its domain.Substituting the function, we get $\cos((x+T)^2) = \cos(x^2)$.This implies $(x+T)^2 = x^2 + 2n\pi$ or $(x+T)^2 = -(x^2) + 2n\pi$ for some integer $n$.Expanding the left side, $x^2 + 2xT + T^2 = x^2 + 2n\pi$, which simplifies to $2xT + T^2 = 2n\pi$.For this to hold for all $x$, the coefficient of $x$ must be zero, which implies $T=0$. However, the period $T$ must be a positive constant.Since no such $T > 0$ exists, the function $f(x) = \cos(x^2)$ is not periodic.

Consider the function $f(x) = \cos(x^2)$. Then,

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