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A mapping $f: R \rightarrow R$ which is defined as $[2009-I I]$
$f(x)=\cos x ; x \in R$ is
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$f(x)=\cos x ; x \in R$ is
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The correct answer is:
Neither one-one nor onto
Let $x_{1}, x_{2} \in \mathrm{R}$
Then, $f\left(x_{1}\right)=f\left(x_{2}\right)$
$\Rightarrow \cos x_{1}=\cos x_{2}$
$\Rightarrow x_{1}=2 n \pi \pm x_{2}$
So, $x_{1} \neq x_{2}$
Hence, $\cos \mathrm{x}$ is not one-one function. Now, let $\mathrm{y}=\cos \mathrm{x}$
We know, $-1 \leq \cos x \leq 1$
$\therefore \mathrm{y} \in[-1,1]$
$[-1,1] \subset \mathrm{R}$. So, $\cos \mathrm{x}$ is into function, not onto. Hence, $\mathrm{f}(\mathrm{x})=\cos \mathrm{x}$ is neither one-one nor onto.
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Then, $f\left(x_{1}\right)=f\left(x_{2}\right)$
$\Rightarrow \cos x_{1}=\cos x_{2}$
$\Rightarrow x_{1}=2 n \pi \pm x_{2}$
So, $x_{1} \neq x_{2}$
Hence, $\cos \mathrm{x}$ is not one-one function. Now, let $\mathrm{y}=\cos \mathrm{x}$
We know, $-1 \leq \cos x \leq 1$
$\therefore \mathrm{y} \in[-1,1]$
$[-1,1] \subset \mathrm{R}$. So, $\cos \mathrm{x}$ is into function, not onto. Hence, $\mathrm{f}(\mathrm{x})=\cos \mathrm{x}$ is neither one-one nor onto.
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