Let f: R rightarrow R be defined as f(x)=3x. | vedclass

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(D) Given $f: R \rightarrow R$ defined by $f(x)=3x$.Step $1$: Check for one-one (injective).Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.$3x_1 = 3

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$f$ is one-one onto

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(D) Given $f: R \rightarrow R$ defined by $f(x)=3x$.Step $1$: Check for one-one (injective).Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.$3x_1 = 3x_2$$x_1 = x_2$Since $f(x_1) = f(x_2) \implies x_1 = x_2$,the function $f$ is one-one.Step $2$: Check for onto (surjective).Let $y \in R$ be any element in the codomain.We need to find $x \in R$ such that $f(x) = y$.$3x = y \implies x = \frac{y}{3}$.Since $y \in R$,$\frac{y}{3}$ is also a real number,so $x \in R$.For every $y \in R$,there exists $x = \frac{y}{3} \in R$ such that $f(x) = 3(\frac{y}{3}) = y$.Thus,the function $f$ is onto.Conclusion: The function $f$ is both one-one and onto (bijective).The correct answer is $D$.

Let $f: R \rightarrow R$ be defined as $f(x)=3x$. Choose the correct answer.

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