Question

Let A = (1,8,27,64,125) and B= (1,2,3,4,5,6)and R be the relation ‘is cube of 'from A to B then domain of R is

Answer 1

$\textbf{Given:}$

$A=\{1,8,27,64,125\}$

$B=\{1,2,3,4,5,6\}$

$\textbf{To find:}$

$\text{Domain of R}$

$\textbf{Solution:}$

$\text{Here,}$

$_aR_b\;\iff\;a=b^3$

$1=1^3\;\implies\;_1R_1\;\implies\;(1,1)\;\in\;R$

$8=2^3\;\implies\;_8R_2\;\implies\;(8,2)\;\in\;R$

$27=3^3\;\implies\;_{27}R_3\;\implies\;(27,3)\;\in\;R$

$64=4^3\;\implies\;_{64}R_4\;\implies\;(64,4)\;\in\;R$

$125=5^3\;\implies\;_{125}R_5\;\implies\;(125,5)\;\in\;R$

$\implies\bf\,R=\{(1,1),(8,2),(27,3),(4,64),(125,5)\}$

$\therefore\textbf{Domain of R}=\{1,8,27,64,125\}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\sf{ \: A = \{1,8,27,64,125 \}\: }$

$\sf{B= \{1,2,3,4,5,6 \} \: }$

R be the relation ‘is cube of 'from A to B

TO FIND

The domain of R

CALCULATION

The relation R is defined as :

$\sf{aRb \: if \: and \: only \: if \: a = {b}^{3} \: for \: a \in \: A \: , b \in \: B}$

$\implies \sf{ R = \{(a, b) \in A \times B : a = {b}^{3} \}\: }$

By the above relation R

$\sf{If \: b = 1 \: then \: a = {(1)}^{3} = 1 }$

$\sf{If \: b = 2 \: then \: a = {(2)}^{3} = 8 }$

$\sf{If \: b = 3 \: then \: a = {(3)}^{3} = 27 }$

$\sf{If \: b = 4 \: then \: a = {(4)}^{3} = 64 }$

$\sf{If \: b = 5 \: then \: a = {(5)}^{3} = 125 }$

So

$\sf{R = \{(1,1), (8,2),(27,3),(64,4),(125,5) \}}$

Hence

$\sf{Domain \: of \: R = \{ 1,8,27,64,125 \} }$

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LEARN MORE FROM BRAINLY

Let A={1,2,0,-1} B={1,3-1,-3} and

The function f(x) =px+q

If f={ (1,1),(2,3),(0,-1),(-1,-3)}

Find the value of p & q

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