Let $P = \left\{ {\left( {x,y} \right){\rm{|}}{x^2} + {y^2} = 1,x,y \in R} \right\}.$ Then, $P$ is

JEE MATHEMATICS MCQ: RELATIONS

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Understanding +4 | -1

Let

P = \left\{ {\left( {x,y} \right)\mathrm{|}{x^2} + {y^2} = 1,x,y \in R} \right\}.

Then,

P

View answer and explanation

Correct option: B

Symmetric

Option B (Symmetric) is correct because if (x, y) is in P, then (y, x) is also in P, since

x^2 + y^2 = y^2 + x^2 = 1

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\mathrm{Z}

R

\mathrm{N}

\mathrm{R}

\mathrm{N} \times \mathrm{N}

(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})

a d(b+c)=b c(a+d)

\mathrm{R}

W

R

R=\{(x, y) \in W \times W \mid

x

y

\}

S

R = \{ \left( {a,\;b} \right):1 + ab > 0\}

S

X

R

X

A

B

R

R

N

R = \left\{ {\left( {x,\;y} \right):4x + 3y = 20} \right\}

R