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(A) Let $h(x) = (f - g)(x)$.For $x \in \mathbb{Q}$,$h(x) = f(x) - g(x) = x - 0 = x$.For $x 
otin \mathbb{Q}$,$h(x) = f(x) - g(x) = 0 - x = -x$.Thus,$

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One-one onto

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(A) Let $h(x) = (f - g)(x)$.For $x \in \mathbb{Q}$,$h(x) = f(x) - g(x) = x - 0 = x$.For $x 
otin \mathbb{Q}$,$h(x) = f(x) - g(x) = 0 - x = -x$.Thus,$h(x) = \begin{cases} x, & x \in \mathbb{Q} \ -x, & x 
otin \mathbb{Q} \end{cases}$.To check if $h(x)$ is one-one: Let $h(x_1) = h(x_2)$. If $x_1, x_2 \in \mathbb{Q}$,then $x_1 = x_2$. If $x_1, x_2 
otin \mathbb{Q}$,then $-x_1 = -x_2 \implies x_1 = x_2$. If $x_1 \in \mathbb{Q}$ and $x_2 
otin \mathbb{Q}$,then $x_1 = -x_2$,which is possible (e.g.,$x_1 = 1, x_2 = -1$,but $-1$ is rational,so this case is restricted). Since $h(x)$ maps every real number to a unique value,it is one-one.To check if $h(x)$ is onto: For any $y \in \mathbb{R}$,if $y \in \mathbb{Q}$,$h(y) = y$. If $y 
otin \mathbb{Q}$,$h(-y) = -(-y) = y$. Thus,for every $y \in \mathbb{R}$,there exists an $x$ such that $h(x) = y$. Hence,it is onto.

If $f(x) = \begin{cases} x, & \text{when } x \text{ is rational} \\ 0, & \text{when } x \text{ is irrational} \end{cases}$ and $g(x) = \begin{cases} 0, & \text{when } x \text{ is rational} \\ x, & \text{when } x \text{ is irrational} \end{cases}$,then $(f - g)$ is:

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