The mapping f: R to R defined as f(x) = cos x | vedclass

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(A) Let $x_1, x_2 \in R$. Then $f(x_1) = \cos x_1$ and $f(x_2) = \cos x_2$. For the function to be one-one,$f(x_1) = f(x_2)$ must imply $x_1 = x_2$. H

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Neither one-one nor onto

Vedclass · Answer

(A) Let $x_1, x_2 \in R$. Then $f(x_1) = \cos x_1$ and $f(x_2) = \cos x_2$. For the function to be one-one,$f(x_1) = f(x_2)$ must imply $x_1 = x_2$. However,$\cos x_1 = \cos x_2$ implies $x_1 = 2n\pi \pm x_2$ for some integer $n$. Since $x_1$ does not necessarily equal $x_2$ (e.g.,$\cos(0) = \cos(2\pi) = 1$),the function is not one-one. For the function to be onto,the range must equal the codomain $R$. The range of $f(x) = \cos x$ is $[-1, 1]$,which is a proper subset of $R$. Since the range $
eq$ codomain,the function is not onto. Therefore,the mapping is neither one-one nor onto.

The mapping $f: R \to R$ defined as $f(x) = \cos x, x \in R$ is:

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