If the sets A and B are defined as A = \{(x,y): x = 1/y², y ≠ 0, y {R}\} B = \{(x, y) : x = -3y, y {R}\} then

Correct Answer is : None of these

Correct Answer is : None of these

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Sets Practice Test 1

Question : 15 of 33

Marks: +1, -0

A = \{(x,y): x = \frac{1}{y^2}, y \neq 0, y \in \mathbb{R}\}

B = \{(x, y) : x = -3y, y \in \mathbb{R}\}

then

Solution: 👈: Video Solution

Concept:

The intersection

A \cap B

contains only those points that satisfy both set conditions.

Explanation:

For set

A

x = \frac{1}{y^2}

, where

y \neq 0

For set

B

x = -3y

At the intersection, equate the two expressions for

x

\frac{1}{y^2} = -3y

Multiply both sides by

y^2

1 = -3y^3

y^3 = -\frac{1}{3}

This gives the real solution

y = -3^{-1/3}

Then

x = -3y = 3^{2/3}

A \cap B

is a single point

\left(3^{2/3}, -3^{-1/3}\right)

This is not equal to

A

B

\phi

, or

A'

Hence, the correct choice is E.

Answer:

A \cap B = \left\{ \left(3^{2/3}, -3^{-1/3}\right) \right\}

, so option E is correct.

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