Let f(x) = cos sqrt{x},then which of the foll | vedclass

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(D) function $f(x)$ is periodic if there exists a constant $T > 0$ such that $f(x + T) = f(x)$ for all $x$ in the domain.For $f(x) = \cos \sqrt{x}$,we

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$f(x)$ is not a periodic function

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(D) function $f(x)$ is periodic if there exists a constant $T > 0$ such that $f(x + T) = f(x)$ for all $x$ in the domain.For $f(x) = \cos \sqrt{x}$,we check if $\cos \sqrt{x + T} = \cos \sqrt{x}$.This implies $\sqrt{x + T} = \sqrt{x} + 2n\pi$ or $\sqrt{x + T} = 2n\pi - \sqrt{x}$.Squaring both sides,we get $x + T = x + 4n\pi\sqrt{x} + 4n^2\pi^2$.Since $T$ must be a constant independent of $x$,and the expression contains $\sqrt{x}$,no such constant $T$ exists.Therefore,$f(x) = \cos \sqrt{x}$ is not a periodic function.

Let $f(x) = \cos \sqrt{x}$,then which of the following is true?

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