The function f: N rightarrow N defined by f(x | vedclass

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

Question

(C) To check if $f$ is one-one:Let $f(x_1) = f(x_2)$.If $x_1$ is odd and $x_2$ is even,then $x_1+1 = x_2-1$,so $x_2 - x_1 = 2$. Since $x_1$ is odd,$x_

Vedclass · Accepted Answer

one-one and onto

Vedclass · Answer

(C) To check if $f$ is one-one:Let $f(x_1) = f(x_2)$.If $x_1$ is odd and $x_2$ is even,then $x_1+1 = x_2-1$,so $x_2 - x_1 = 2$. Since $x_1$ is odd,$x_1+1$ is even. Since $x_2$ is even,$x_2-1$ is odd. This contradicts $f(x_1) = f(x_2)$.If both are odd,$x_1+1 = x_2+1 \implies x_1 = x_2$.If both are even,$x_1-1 = x_2-1 \implies x_1 = x_2$.Thus,$f$ is one-one.To check if $f$ is onto:For any $y \in N$,if $y$ is odd,we can choose $x = y+1$ (which is even),then $f(y+1) = (y+1)-1 = y$.If $y$ is even,we can choose $x = y-1$ (which is odd),then $f(y-1) = (y-1)+1 = y$.Since for every $y \in N$,there exists an $x \in N$ such that $f(x) = y$,$f$ is onto.Therefore,$f$ is one-one and onto.

The function $f: N \rightarrow N$ defined by $f(x) = \begin{cases} x+1, & x \text{ is odd} \\ x-1, & x \text{ is even} \end{cases}$ is . . . . . . .

Similar Questions

Show that the function $f : R \rightarrow \{ x \in R : -1 < x < 1 \}$ defined by $f(x) = \frac{x}{1+|x|}$ for all $x \in R$ is a one-one and onto function.

The function $f(x) = \sqrt{3} \sin 2x - \cos 2x + 4$ is one-one in the interval

If $f: N \times N \rightarrow N$ is defined by $f(m, n) = 2^{m-1}(2n-1)$ for all $(m, n) \in N \times N$,then $f$ is

Let $f: X \rightarrow X$ be such that $f(f(x)) = x$ for all $x \in X$ and $X \subseteq \mathbb{R}$. Then:

Check the injectivity and surjectivity of the function $f: Z \rightarrow Z$ defined by $f(x) = x^{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module