The function $f : N \to N$ defined by $f(x) = x - 5[\frac{x}{5}]$,where $N$ is the set of natural numbers and $[x]$ denotes the greatest integer less than or equal to $x$,is
- Aone-one and onto.
- Bone-one but not onto.
- Conto but not one-one.
- Dneither one-one nor onto.
Explore More
Similar Questions
Statement-$I$: Let $f : R \rightarrow R$ be a function such that $f(x) = x^3 + x^2 + 3x + \sin x$. Then $f$ is a one-one function.
Statement-$II$: $f(x)$ is a decreasing function.
Statement-$II$: $f(x)$ is a decreasing function.
Difficult
View SolutionLet $f:[0,10] \rightarrow [1,20]$ be a function defined as $f(x) = \begin{cases} \frac{60-5x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3x, & 7 \leq x \leq 10 \end{cases}$. The function $f$ is:
If a function $f: R-\{l\} \to R-\{m\}$ defined by $f(x) = \frac{x+3}{x-2}$ is a bijection,then $3l - 2m =$
Let $X$ be a set with exactly $5$ elements and $Y$ be a set with exactly $7$ elements. If $\alpha$ is the number of one-one functions from $X$ to $Y$ and $\beta$ is the number of onto functions from $Y$ to $X$,then the value of $\frac{1}{5!}(\beta-\alpha)$ is.
DifficultJEE MAIN 2018
View SolutionIf the function $f:[-1,1] \rightarrow R$ is defined by $f(x) = \begin{cases} 2^x+1, & \text{for } x \in [-1,0) \\ 1, & \text{for } x=0 \\ 2^x-1, & \text{for } x \in (0,1] \end{cases}$,then in $[-1,1]$,$f(x)$ has
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo