If R = \ (x,y):x,y in Z,x^2 + 3y^2 <= 8 \ is a relation on the set of integers Z, then the domain of R^- 1 is

(d) : Given R = \ (x,y):x,y in Z,x^2 + 3y^2 R = \ (x,y):x,y in Z,fracx^28 + fracy^28/3 <= 1 \Now, domain of R^- 1 = Range of R = \ - 1,0,1\.

If R = { (x,y):x,y in Z,x^{2} + 3y^{2} ? 8 } is a ... | Relation (Mathematics)

JEE Mains ( $2^{nd\ }$ Sept $1^{st\ }$ Shift 2020) 2nd September Shift 1

Mathematics

R = \left\{ (x,y):x,y \in Z,x^{2} + 3y^{2} \leq 8 \right\}

is a relation on the set of integers

Z

, then the domain of

R^{- 1}

\{ - 2, - 1,1,2\}

\{ 0,1\}

\{ - 2, - 1,0,1,2\}

\{ - 1,0,1\}

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Correct Answer:

Explanation

(d) : Given $R = \left\{ (x,y):x,y \in Z,x^{2} + 3y^{2} \leq 8 \right\}$ $\\ \Rightarrow R = \left\{ (x,y):x,y \in Z,\frac{x^{2}}{8} + \frac{y^{2}}{8/3} \leq 1 \right\}$ Now, domain of $R^{- 1} =$ Range of $R = \{ - 1,0,1\}$ .

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Explanation

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