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Let $f(\mathrm{x}):\left\{\begin{array}{ll}\mathrm{x}, & \mathrm{x} \text { is rational } \\ 0, & \mathrm{x} \text { is irrational }\end{array}\right.$
and $g(x):\left\{\begin{array}{ll}0, & x \text { is rational } \\ x, & x \text { is irrational }\end{array}\right.$
If $f: \mathrm{R} \rightarrow \mathrm{R}$ and $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$, then $(f-\mathrm{g})$ is
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and $g(x):\left\{\begin{array}{ll}0, & x \text { is rational } \\ x, & x \text { is irrational }\end{array}\right.$
If $f: \mathrm{R} \rightarrow \mathrm{R}$ and $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$, then $(f-\mathrm{g})$ is
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Verified Answer
The correct answer is:
one-one and onto
$\quad(\mathrm{f}-\mathrm{g})(\mathrm{x})=\left\{\begin{array}{c}\mathrm{x}-0=\mathrm{x}, \mathrm{x} \text { is rational } \\ 0-\mathrm{x}=-\mathrm{x}, \mathrm{x} \text { is irrational }\end{array}\right.$
Clearly, $\mathrm{f}-\mathrm{g}$ is one $-$ one and onto.
Clearly, $\mathrm{f}-\mathrm{g}$ is one $-$ one and onto.
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