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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$f$ is one-one onto

Vedclass · Answer

(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.

Similar Questions

The function $f: R \rightarrow R$ defined by $f(x)=\frac{x}{\sqrt{1+x^2}}$ is

If a real-valued function $f$ is defined by $f(x) = \frac{ax + \sqrt{a^2 - x^2}}{bx}$,then $f$ is

The function $f(x) = \sin (\log (x + \sqrt {x^2 + 1}))$ is

If $f: Z \rightarrow N$ is defined by $f(n) = \begin{cases} 2n, & \text{if } n > 0 \\ 1, & \text{if } n = 0 \\ -2n-1, & \text{if } n < 0 \end{cases}$, then the function $f$ is:

The function $f: R \rightarrow R$ defined by $f(x) = e^x + e^{-x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module