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1. Here, A = {2, 4, 6, 8, 10}.
The elements of set B are positive even integers less than 10. Therefore,
B = {2, 4, 6, 8}
Since 10 ∈ A and 10 ∉ B, A and B are not equal sets.
2. Clearly, A = {A, P, L, E} and B = {L, E, A, P}
Therefore, both the sets have same elements, so A = B.
3. Elements of set A are integers whose square is less than or equal to 4. Therefore,
A = {2, 1, 0, 1, 2}
Elements of set B are real numbers that satisfy the equation x2 – 3x + 2 = 0. To determine the elements of set B, we need to solve the equation.
x2 – 3x + 2 = 0
=> x2 – 2x  x + 2 = 0
=> x(x  2)  1(x  2) = 0
=> (x  1)(x  2) = 0
=> x = 1, 2
Therefore, B = {1, 2}
Clearly, A and B are not equal sets.
Trick: In case of objectives, after determining set A, we could easily have said A and B can not be equal sets. This is because set A has 5 distinct elements and since elements of B are real values that are solution of a quadratic equation, it can have maximum of 2 distinct elements.
4. A = { 4, 8, 12, 16 } B = { 8, 4, 16, 18}
Here 12 ∈ A and 12 ∉ B, hence A and B are not equal sets.
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