State with reason whether the following function has an inverse: $g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\}$ with $g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$.
(NO) function $g$ has an inverse if and only if it is a bijection (both one-one and onto).
Given $g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$.
We observe that $g(5) = 4$ and $g(7) = 4$.
Since two distinct elements in the domain,$5$ and $7$,map to the same element $4$ in the codomain,the function $g$ is not one-one (it is many-one).
Because $g$ is not one-one,it is not a bijection.
Therefore,the function $g$ does not have an inverse.
Given $g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$.
We observe that $g(5) = 4$ and $g(7) = 4$.
Since two distinct elements in the domain,$5$ and $7$,map to the same element $4$ in the codomain,the function $g$ is not one-one (it is many-one).
Because $g$ is not one-one,it is not a bijection.
Therefore,the function $g$ does not have an inverse.
Explore More
Similar Questions
Let $f(x) = (x + 1)^2 - 1$ for $x \ge -1$. Then the set $S = \{ x : f(x) = f^{-1}(x) \}$ is
DifficultIIT 1995
View SolutionLet $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x},$ where $b$ is a constant such that $0 < b < 1$. Then
If $f: R \rightarrow R$ is defined by $f(x)=x^{3}$,then $f^{-1}(8)$ is equal to
Let $f:(2, 3) \to (0, 1)$ be defined by $f(x) = x - [x]$. Then ${f^{ - 1}}(x)$ equals:
Easy
View SolutionIf $f: R \rightarrow R$ is a mapping defined by $f(x)=x^{3}+5$,then $f^{-1}(x)$ is equal to
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