Let a relation $R$ be defined by $R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\}$. Then ${R^{ - 1}}oR$ is

  • A
    $\{(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)\}$
  • B
    $\{(1, 1), (4, 4), (7, 7), (3, 3)\}$
  • C
    $\{(1, 5), (1, 6), (3, 6)\}$
  • D
    None of these

Explore More

Similar Questions

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are functions such that $g \circ f: A \rightarrow C$ is onto,then a necessary condition is:

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=2x+3$ and $g(x)=x^2+7$,then the values of $x$ such that $g(f(x))=8$ are

If $R$ is a relation from a set $A$ to a set $B$ and $S$ is a relation from $B$ to a set $C$,then the relation $S \circ R$ is:

If $f: R \rightarrow R$ is defined by $f(x)=e^{x}$ and $g: R \rightarrow R$ is defined by $g(x)=x^{2}$,then the mapping $(g \circ f): R \rightarrow R$ is defined by $(g \circ f)(x) = g(f(x))$ for all $x \in R$. Which of the following is true?

For a suitably chosen real constant $a$,let the function $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x}$. Further,suppose that for any real number $x \neq-a$ and $f(x) \neq-a$,$(f \circ f)(x)=x$. Then,$f\left(-\frac{1}{2}\right)$ 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 Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo