The rule $f(x) = x^2$ is a bijection if the domain and the co-domain are given by: (a) $(0, \infty), R$ (b) $R, R$ (c) $[0, \infty), [0, \infty)$ (d) $R, (0, \infty)$

> > [!TLDR] > $f(x)=x^2$ is a bijection precisely when both domain and co-domain are $[0,\infty)$. > > Check each option by testing one-one and onto: > - $(0,\infty)\to\mathbb{R}$: values of $x^2$ for $x>0$ are only positive, so it never hits negative numbers in $\mathbb{R}$ — not onto. > - $\mathbb{R}\to\mathbb{R}$: $f(-2)=f(2)=4$, so it is not one-one; also never negative, so not onto. > - $[0,\infty)\to[0,\infty)$: for $x_1,x_2\ge 0$, $x_1^2=x_2^2\Rightarrow x_1=x_2$ (one-one), and every $y\ge 0$ has $x=\sqrt{y}\ge 0$ mapping to it (onto). This is a bijection. …

The rule f(x) = x² is a bijection if the domain and the co-domain are given by: (a) (0, ∞), R (b) R, R (c) [0,

Question 89 of 104

f(x) = x^2

(a)

(0, \infty), R

(b)

R, R

(c)

[0, \infty), [0, \infty)

(d)

R, (0, \infty)

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$f(x)=x^2$ is a bijection precisely when both domain and co-domain are $[0,\infty)$ .

Check each option by testing one-one and onto:

$(0,\infty)\to\mathbb{R}$ : values of $x^2$ for $x>0$ are only positive, so it never hits negative numbers in $\mathbb{R}$ — not onto.
$\mathbb{R}\to\mathbb{R}$ : $f(-2)=f(2)=4$ , so it is not one-one; also never negative, so not onto.
$[0,\infty)\to[0,\infty)$ : for $x_1,x_2\ge 0$ , $x_1^2=x_2^2\Rightarrow x_1=x_2$ (one-one), and every $y\ge 0$ has $x=\sqrt{y}\ge 0$ mapping to it (onto). This is a bijection. …

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Q.The rule $f(x) = x^2$ is a bijection if the domain and the co-domain are given by:

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