The relation $R$ on the set of natural numbers is defined by $R = \{(a, b) : a = 2b\}$. Then ${R^{-1}}$ is:
- A$\{(2, 1), (4, 2), (6, 3), \dots\}$
- B$\{(1, 2), (2, 4), (3, 6), \dots\}$
- C${R^{-1}}$ is not defined
- DNone of these
Explore More
Similar Questions
If $g$ is the inverse of $f$ and $f^{\prime}(x)=\frac{1}{1+x^3}$,then $g^{\prime}(x)$ is
Let $f: R - \{-\frac{4}{3}\} \rightarrow R$ be a function defined as $f(x) = \frac{4x}{3x+4}$. The inverse of $f$ is the map $g: \text{Range } f \rightarrow R - \{-\frac{4}{3}\}$ given by
Medium
View SolutionIf the function $f(x) = -4e^{\left(\frac{1-x}{2}\right)} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3}$ and $g(x) = f^{-1}(x)$,then the value of $g'(-\frac{7}{6})$ equals
If $f : R \to R$ is defined by $f(x) = x^2 + 1$,then $f^{-1}(17)$ and $f^{-1}(-3)$ are
Medium
View SolutionLet $f(x)=(x+1)^2-1$ for $x \geq -1$.
Statement-$1$: $S=\{x:f(x)=f^{-1}(x)\}=\{0, -1\}$
Statement-$2$: $f$ is a bijection.
Statement-$1$: $S=\{x:f(x)=f^{-1}(x)\}=\{0, -1\}$
Statement-$2$: $f$ is a bijection.
MediumIIT 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