Question

If the elements of the set A, B and C are given as A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}, then $\left( {A - C} \right) \times \left( {B - C} \right)$is equal to
(a){\text{ }}\left\\{ {\left( {1,4} \right)} \right\\} \\\ (b){\text{ }}\left\\{ {\left( {1,4} \right),\left( {4,4} \right)} \right\\} \\\ (c){\text{ }}\left\\{ {\left( {4,1} \right),\left( {4,4} \right)} \right\\} \\\ (d){\text{ none of these}} \\\

Answer 1

Hint – In this question use the concept that (A – C) will be the set of elements which is in (A) but not in (C), similarly (B – C) will be the set of elements which is (B) but not in (C). Multiplication of two sets means indulging them both together.

Complete step-by-step solution -
Given data
A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}
Now (A – C) is the set of elements which is (A) but not in (C).
So we see that the element which is in (A) but not in (C) is {1}.
\Rightarrow \left( {A - C} \right) = \left\\{ 1 \right\\}
Now (B – C) is the set of elements which is (B) but not in (C).
So we see that the element which is in (B) but not in (C) is {4}.
\Rightarrow \left( {B - C} \right) = \left\\{ 4 \right\\}
Now (A – C) $\times$(B – C) is a set of all elements in (A – C) and (B – C).
\Rightarrow \left( {A - C} \right) \times \left( {B - C} \right) = \left\\{ {\left( {1,4} \right)} \right\\}
So this is the required answer.
Hence option (A) is correct.

Note – A set is a collection of well-defined and distinct objects, the most basic properties of a set are that two sets are equal if and only if every element of the first set is also the element of the second set. Sets can be of different types like singleton set, finite set, infinite set, equal set, null set, proper set, null set etc.