If f: R to R,then f(x) = |x| is | vedclass

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(D) Given the function $f: R 	o R$ defined by $f(x) = |x|$.$1$. Check for one-one: $A$ function is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$. H

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None of these

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(D) Given the function $f: R 	o R$ defined by $f(x) = |x|$.$1$. Check for one-one: $A$ function is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$. Here,$f(-1) = |-1| = 1$ and $f(1) = |1| = 1$. Since $f(-1) = f(1)$ but $-1 
eq 1$,the function is not one-one (it is many-one).$2$. Check for onto: $A$ function is onto if its range equals its codomain. The codomain is $R$. The range of $f(x) = |x|$ is $[0, \infty)$. Since the range is not equal to the codomain $([0, \infty) 
eq R)$,the function is not onto.Therefore,the function is neither one-one nor onto. The correct option is $(d)$.

If $f: R \to R$,then $f(x) = |x|$ is

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