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(B) To check for one-one: Let $f(x_1) = f(x_2)$ for $x_1, x_2 \in [0, \infty)$.$\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$$x_1(1+x_2) = x_2(1+x_1)$$x_1 +

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One-one but not onto

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(B) To check for one-one: Let $f(x_1) = f(x_2)$ for $x_1, x_2 \in [0, \infty)$.$\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$$x_1(1+x_2) = x_2(1+x_1)$$x_1 + x_1x_2 = x_2 + x_1x_2$$x_1 = x_2$.Thus,$f$ is one-one.To check for onto: Let $y = \frac{x}{1+x}$.$y(1+x) = x \implies y + xy = x \implies y = x(1-y) \implies x = \frac{y}{1-y}$.For $x \in [0, \infty)$,we need $y \in [0, 1)$.Since the codomain is $[0, \infty)$,values like $y = 2$ have no pre-image in the domain.Thus,$f$ is not onto.Therefore,the function is one-one but not onto.

The function $f:[0, \infty) \rightarrow [0, \infty)$ defined by $f(x) = \frac{x}{1+x}$ is

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