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
If $g$ is the inverse of a function $f$ and $f'(x) = \frac{1}{1 + x^5}$,then $g'(x)$ is equal to:
Consider the function $y = \log_{a}(x + \sqrt{x^{2} + 1})$ where $a > 0$ and $a \neq 1$. The inverse of the function:
MediumWBJEE 2018
View SolutionIf $f: R \rightarrow R$ is given by $f(x)=7x+8$ and $f^{-1}(12)=\frac{k}{7}$,then the value of $k$ is
If $f(x) = \frac{x^2-x}{x^2+2x}$,then the value of $\frac{d}{dx}(f^{-1}(x))$ at $x = 2$ is:
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 SolutionVedclass 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