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  • Can someone explain - Sets, Relations and Functions - JEE Main

Let N denote the set of all natural numbers.
Define two binary relations on N as R_{1}= ( \left ( x,y \right ) \right \ \epsilon N\times N : 2x+y=10})

and R_{2}= ( \left ( x,y \right ) \right \ \epsilon N\times N : x+2y=10})
Then :

  • Option 1)

    Range of  R_{1}is {2, 4, 8}.

  • Option 2)

    Range of R_{2} is {1, 2, 3, 4}.

  • Option 3)

    Both R_{1} and R_{2} are symmetric
    relations

  • Option 4)

    Both R_{1} and R_{2} are transitive
    relations

 

Answers (2)

best_answer

As we have learned

Range -

The range of the relation R is the set of all second elements of the ordered pairs in a relation R.

- wherein

eg. R={(a,b),(c,d)}. Then Range is {b,d}

 

 

R_{1}:\left \{ \left ( x,y \right )\in N\times N:2x+y\right \}= 10

R_{2}=\left \{ \left ( x,y \right ) \right \ \epsilon N\times N : x+2y \right \}=10

 

So for R_{1}\: \: \: \: \: \: -x= \frac{10-y}{2}

i.e.y= 2,4,6,8

For \: R_{1},ordered\: pairs\: are\left ( 4,2 \right )\left ( 3,4 \right )\left ( 2,6 \right )\left ( 1,8 \right )

 

for \\*For \: R_{2},\: \: \: \: \: x+2y= 10;y=\frac{10-x}{2}\\* ordered\: pairs\: are\left (2,4 \right )\left ( 4,3 \right )\left ( 6,2 \right )\left ( 8,1 \right )\\* range \: of\: R_{2}\: is\left \{ 1,2,3,4 \right \}

 


Option 1)

Range of  R_{1}is {2, 4, 8}.

Option 2)

Range of R_{2} is {1, 2, 3, 4}.

Option 3)

Both R_{1} and R_{2} are symmetric
relations

Option 4)

Both R_{1} and R_{2} are transitive
relations

Posted by

Himanshu

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