Let a function f: (0, infty) to (0, infty) be | vedclass

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(C) Given $f(x) = |1 - \frac{1}{x}|$ for $x \in (0, \infty)$.To check for injectivity (one-one): Consider $f(x) = y$. For $y \in (0, 1)$,there exist t

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neither injective nor surjective

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(C) Given $f(x) = |1 - \frac{1}{x}|$ for $x \in (0, \infty)$.To check for injectivity (one-one): Consider $f(x) = y$. For $y \in (0, 1)$,there exist two values of $x$. For example,if $y = 0.5$,then $|1 - \frac{1}{x}| = 0.5$,which gives $1 - \frac{1}{x} = 0.5 \implies \frac{1}{x} = 0.5 \implies x = 2$,and $1 - \frac{1}{x} = -0.5 \implies \frac{1}{x} = 1.5 \implies x = \frac{2}{3}$. Since $f(2) = f(\frac{2}{3}) = 0.5$,the function is not injective.To check for surjectivity (onto): The codomain is $(0, \infty)$. The range of $f(x) = |1 - \frac{1}{x}|$ for $x \in (0, \infty)$ is $[0, \infty)$. Since the range $[0, \infty)$ is not equal to the codomain $(0, \infty)$ (as $0$ is in the range but not in the codomain),the function is not surjective.Thus,$f$ is neither injective nor surjective.

Let a function $f: (0, \infty) \to (0, \infty)$ be defined by $f(x) = |1 - \frac{1}{x}|$. Then $f$ is

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