MCQ
A relation R is defined as $R =\{(1,2)\}$ on set $A =$ $\{1,2,3\}$, then R is :
  • A
    Reflexive
  • B
    Symmetric
  • Transitive
  • D
    None of these.

Answer

Correct option: C.
Transitive
(C) - Transitive.
$\because(1,1),(2,2),(3,3) \notin R \Rightarrow R$ is not reflexive.
$\because(1,2) \in R$ but $(2,1) \notin R \Rightarrow R$ is not symmetric. $R$ is transitive because $(1,2) \notin R$ there is not any ordered pair starting from 2 .

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If the system of equations, $x + 2y -3z = 1, (k + 3) z = 3, (2k + 1)x + z = 0$ is inconsistent, then the value of $k$ is :-
The positive integral solution of the equation$\tan^{-1}\text{x}+\cos^{-1}\frac{\text{y}}{\sqrt{1+\text{y}^2}}=\sin^{-1}\frac{3}{\sqrt{10}}$ is:
If $\text{f}(\text{x})=\tan^{-1}(\text{g}(\text{x})),$ where g(x) is monotonically increasing for $0<\text{x}<\frac{\pi}{2}.$ Then, f(x) is:
If $\text{A}=\displaystyle \begin{vmatrix} 1 &\text{amp; } 0 \\ 1 &\text{amp; } 0 \end{vmatrix}$ And $\text{B}=\displaystyle \begin{vmatrix} 1 &\text{amp; } 0 \\ 0 &\text{amp; } 1 \end{vmatrix}$ then $\text{A+B}=$
Number of positive solution of the equation,$\int\limits_0^x {\,\,{{\left( {t\,\, - \,\,\left\{ t \right\}} \right)}^2}}$ $dt = 2 (x - 1)$ where $\{ \}$ denotes the fractional part function is :
Choose the correct option from given four options:
If $\int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C},$ then:
$\mathop {Lim}\limits_{n \to \infty } $$\frac{\pi }{{2\,n}}\,\,\left( {1\,\, + \,\,\cos \,\frac{\pi }{{2\,n}}\,\, + \,\,\cos \,\frac{{2\,\pi }}{{2\,n}}\,\, + \,\,.....\,\, + \,\,\cos \,\frac{{(n\, - \,1)\,\pi }}{{2\,n}}} \right)$ equal to
The area of the shorter region bounded by $|y| = 4\, -\, x^2$ and $|y| = 3x$ is given by $\left( {3K + \frac{1}{3}} \right)$ sq-unit where $K$ is equal to
Let a solution $y=y(x)$ of the differential equation $x \sqrt{x^2-1} d y-y \sqrt{y^2-1} d x=0 \quad$ satisfy $y(2)=\frac{2}{\sqrt{3}}$.

$STATEMENT-1$: $y(x)=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$ and

$STATEMENT-2$ : $\mathrm{y}(\mathrm{x})$ is given by $\frac{1}{\mathrm{y}}=\frac{2 \sqrt{3}}{\mathrm{x}}-\sqrt{1-\frac{1}{\mathrm{x}^2}}$

${\cos ^{ - 1}}\frac{1}{2} + 2{\sin ^{ - 1}}\frac{1}{2}$ is equal to