Answer
False
Work Step by Step
Set\[A\cap B\]contains all those elements which are common in both the sets A and B. Now, if set A is contained in set B then elements present in set\[A\cap B\]are the elements of set A. Thus, set\[A\cap B\]is exactly equal to the set A.
Mathematically, it can be written as:\[A\cap B=A\]
Therefore, the statement ‘If\[A\subseteq B\], then\[A\cap B=\varnothing \]’ is false.
The true statement is: If\[A\subseteq B\], then\[A\cap B=A\]
