Question

Let R be relation defined on Z (set of all integers) such that R = {(x, y): x2020 + y2 = 2y, x, y ∈ Z}, thenDomain of R is {0, 1}


Domain of R is {–1, 0}


Range of R is {0, 1}


Range of R is {0, 1, 2}

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

Let R be relation defined on Z (set of all integers) such that

$\sf{R = \{ (x, y) : {x}^{2020} + {y}^{2} = 2y \: , \: x,y \in Z} \: \}$

Then

Domain of R is {0, 1}

Domain of R is {–1, 0}

Range of R is {0, 1}

Range of R is {0, 1, 2}

EVALUATION

Here the given relation is

$\sf{R = \{ (x, y) : {x}^{2020} + {y}^{2} = 2y \: , \: x,y \in Z} \: \}$

Now

$\sf{ {x}^{2020} + {y}^{2} = 2y}$

$\sf{ \implies \: {x}^{2020} + {y}^{2} - 2y = 0}$

$\sf{ \implies \: {x}^{2020} + {y}^{2} - 2y + 1= 1}$

$\sf{ \implies \: {x}^{2020} + {(y - 1)}^{2} = 1}$

Now above holds when x = - 1 , 0 , 1

When x = - 1 we have y = 1

When x = 0 we have y = 0 & 2

When x = 1 we have y = 1

Hence

Domain of R = { - 1 , 0 , 1 }

Range of R = { 0 , 1 , 2 }

FINAL ANSWER

Hence the correct option is

Range of R is { 0, 1, 2 }

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