The function $f: \mathbb{R} \rightarrow [-1, 1]$ defined by $f(x) = \cos x$ is:

A function is one-one (injective) if each output comes from exactly one input. A function is onto (surjective) if every value in the target set is achieved by some input.

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To check if $f(x) = \cos x$ is one-one, consider:

$f(0) = \cos(0) = 1$

$f(2\pi) = \cos(2\pi) = 1$

Since $0 \neq 2\pi$ but both give the same output, different inputs produce the same output.

Similarly:

$f(\pi/3) = \cos(\pi/3) = 0.5$

$f(5\pi/3) = \cos(5\pi/3) = 0.5$

The function is not one-one.

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To check if $f(x) = \cos x$ is onto, verify that every value in $[-1, 1]$ can be achieved.

The range of $\cos x$ is exactly $[-1, 1]$:

$\cos(0) = 1$

$\cos(\pi/3) = 0.5$

$\cos(\pi/2) = 0$

$\cos(2\pi/3) = -0.5$

$\cos(\pi) = -1$

For any value between $-1$ and $1$, there exists an $x$ value such that $\cos x$ equals that value.

The function is onto.

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The function is onto but not one-one.