Download App

Relations and Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsRelations and Functions1 Mark
Question
Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x, y) : y = x + 1 }. Write down the domain, codomain and range of R
✓

Answer

We have, R={(x,y):y=x+1}
Since A={1,2,3,4,5,6}
Therefore, R={(1,2),(2,3),(3,4),(4,5),(5,6)}
Therefore, domain ={1, 2, 3, 4, 5,}
Similarly, the range = {2, 3, 4, 5, 6} and the co-domain = {1, 2, 3, 4, 5, 6}

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

Explore more

More "1 Marks Question" from Relations and Functions→Relations and Functions — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

Find the value of expression $\left\{\cos A \cdot \cos \left(90^{\circ}- A \right)\right\}$.
→
Write the quadration the arithmetic and geometric means of whose roots are A and G respectively.
→
Write $-1 + \text{i}\sqrt{3}$ in polar form.
→
There is a complex number which is not a real number.
→
A relation $R$ is defined on a set $A=\{1,2,3,4,5,6\}$ such that $x R y \Leftrightarrow x+2 y=8$, then write the domain of R.
→
If f is a real function satisfying $\text{f}\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{x}^2+\frac{1}{\text{x}^2}$ for all $\text{x}\in\text{R}-\{0\},$ then write the expression for f(x).
→
Write the component statements of the following compound statements and check whether the compound statement is true or false:
To enter into a public library children need an identity card from the school or a letter from the school authorities.
→
Evaluate the following:
$\cos80^\circ\cos20^\circ+\sin80^\circ\sin20^\circ$
→
Write the equation of hyperbola whose transverse axis and conjugate axis are 4 and 5 respectively.
→
Write the coordinates of the ends of the latus rectum of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
→