Question

A function f: R→R defined by f(x)=3x+5 is :

• A. one-one but not onto
• B. onto but not one-one
• C. both one-one and onto
• D. neither one-one nor onto

Answer 1

Answer: (C)

both one-one and onto

Answer 2

Let the function be $f: R \to R$ defined by $f(x) = 3x + 5$.

One-one (injective): Assume $f(x_1) = f(x_2)$ for any $x_1, x_2 \in R$. Then $3x_1 + 5 = 3x_2 + 5$. Subtracting 5 from both sides gives $3x_1 = 3x_2$, and dividing by 3 yields $x_1 = x_2$. Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f(x)$ is one-one.

Onto (surjective): We need to show that for every $y$ in the codomain $R$, there exists at least one $x$ in the domain $R$ such that $f(x) = y$. Let $y \in R$. We want to find $x \in R$ such that $f(x) = y$. So, $3x + 5 = y$. Subtracting 5 from both sides gives $3x = y - 5$, and dividing by 3 yields $x = \frac{y - 5}{3}$. Since $y$ is a real number, $\frac{y-5}{3}$ is also a real number. Thus, for any $y \in R$ (codomain), we found an $x = \frac{y-5}{3} \in R$ (domain) such that $f(x) = f\left(\frac{y-5}{3}\right) = 3\left(\frac{y-5}{3}\right) + 5 = (y-5) + 5 = y$. Since for every $y$ in the codomain $R$, there exists an $x$ in the domain $R$ such that $f(x) = y$, the function $f(x)$ is onto.

Since the function $f(x)$ is both one-one and onto, it is a bijective function.