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Given $f(x) = ax + b$, $f(1) = 1$ and $f(2) = 3$. Substituting $x = 1$ in $f(x)$, we get $f(1) = a(1) + b = a + b = 1$. Substituting $x = 2$ in $f(x)$, we get $f(2) = a(2) + b = 2a + b = 3$. Now we have a system of two linear equations: $$a + b = 1$$ $$2a + b = 3$$ Subtracting the first equation from the second equation, we get: $$(2a + b) - (a + b) = 3 - 1$$ $$a = 2$$ Substituting $a = 2$ in the first equation, we get: $$2 + b = 1$$ $$b = 1 - 2 = -1$$ Thus, $a = 2$ and $b = -1$.
Since $a = 2$ and $b = -1$, the function $f(x)$ is given by: $$f(x) = 2x - 1$$
To check if $f(x)$ is one-one, we need to show that if $f(x_1) = f(x_2)$, then $x_1 = x_2$. Let $f(x_1) = f(x_2)$. Then, $$2x_1 - 1 = 2x_2 - 1$$ $$2x_1 = 2x_2$$ $$x_1 = x_2$$ Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f(x)$ is one-one.
To check if $f(x)$ is onto, we need to show that for every $y \in R$, there exists an $x \in R$ such that $f(x) = y$. Let $y \in R$. We want to find $x$ such that $f(x) = y$. $$2x - 1 = y$$ $$2x = y + 1$$ $$x = \frac{y + 1}{2}$$ Since $y \in R$, $\frac{y + 1}{2} \in R$. Thus, for every $y \in R$, there exists an $x = \frac{y + 1}{2} \in R$ such that $f(x) = y$. Therefore, the function $f(x)$ is onto.
The function $f(x) = 2x - 1$ is both one-one and onto.
Final Answer: f(x) = 2x - 1, one-one and onto
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