Let f : R to R be defined by f(x) = frac{ax^2 | vedclass

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(A) Given $f(x) = \frac{ax^2 + ax + b}{ax + b}$.For $f$ to be defined for all $x \in R$,the denominator $ax + b$ must not be zero for any $x \in R$. T

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$f$ is many-one

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(A) Given $f(x) = \frac{ax^2 + ax + b}{ax + b}$.For $f$ to be defined for all $x \in R$,the denominator $ax + b$ must not be zero for any $x \in R$. This implies $a = 0$ and $b 
eq 0$.Substituting $a = 0$ into the function,we get $f(x) = \frac{0 \cdot x^2 + 0 \cdot x + b}{0 \cdot x + b} = \frac{b}{b} = 1$.Since $f(x) = 1$ for all $x \in R$,the function is a constant function.$A$ constant function maps every element of the domain to the same element in the codomain,therefore it is many-one.Thus,$f$ is many-one.

Let $f : R \to R$ be defined by $f(x) = \frac{ax^2 + ax + b}{ax + b}$. Then:

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