Let N be the set of natural numbers and a relation R on N be defined by R = {(x,y) ∈ N×N : x³ - 3x²y – xy² + 3y³ = 0} . Then the relation R is :
Let N be the set of natural numbers and a relation R on N be defined by R = {(x,y) ∈ N×N : x³ - 3x²y – xy² + 3y³ = 0} . Then the relation R is :
Option 1 -
Reflexive but neither symmetric nor transitive
Option 2 -
An equivalence relation
Option 3 -
Symmetric but neither reflexive nor transitive
Option 4 -
Reflexive and symmetric, but not transitive
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1 Answer
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Correct Option - 1
Detailed Solution:for reflexive (x, x)
x³ - 3x²x + 3x³ = 0
. reflexive
For symmetric
(x, y) ∈ R
x³ - 3x²y - xy² + 3y³ = 0
⇒ (x - 3y) (x² - y²) = 0
For (y, x)
(y - 3x) (y² - x²) = 0
⇒ (3x - y) (x² - y²) = 0
Not symmetric
Similar Questions for you
...(1)
–2α + β = 0 …(2)
Solving (1) and (2)
a = 1
b = 2
-> a + b = 3
|z| = 0 (not acceptable)
|z| = 1
|z|2 = 1
Given : x2 – 70x + l = 0
->Let roots be a and b
->b = 70 – a
->= a (70 – a)
l is not divisible by 2 and 3
->a = 5, b = 65
->
z1 + z2 = 5
⇒ 20 + 15i = 125 – 15z1z2
⇒ 3z1z2 = 25 – 4 – 3i
3z1z2 = 21– 3i
z1⋅z2 = 7 – i
(z1 + z2)2 = 25
= 11 + 2i
= 121 − 4 + 44i
⇒
⇒ = 117 + 44i − 2(49 −1−14i )
= 21 + 72i
⇒
a = 1 > 0 and D < 0
4 (3k – 1)2 – 4 (8k2 – 7) < 0
K = 3
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