NIMCET 2010 — Mathematics PYQ

The wrong statement is (a).

Explanation of each option:

• (a) If $A$ and $B$ are two sets, then $A-B=A\cap B$ (WRONG)

  By definition, the set difference $A-B$ consists of elements that are in $A$ but not in $B$. The correct identity is:

  $$A-B=A\cap \bar{B}$$

  The statement $A\cap B$ represents the intersection (elements common to both), which is entirely different from the difference.

• (b) If $A$, $B$ and $C$ are sets, then $(A-B)-C=(A-C)-(B-C)$ (RIGHT)

  We can verify this using the identity $X-Y=X\cap \bar{Y}$:

  LHS: $(A\cap \bar{B})\cap \bar{C}=A\cap \bar{B}\cap \bar{C}$

  RHS: $(A\cap \bar{C})\cap \overline{(B\cap \bar{C})}=(A\cap \bar{C})\cap (\bar{B}\cup C)$

  Distributing the intersection: $(A\cap \bar{C}\cap \bar{B})\cup (A\cap \bar{C}\cap C)$

  Since $\bar{C}\cap C=\varnothing$, the second part becomes empty.

  RHS: $A\cap \bar{C}\cap \bar{B}$, which matches the LHS.

• (c) If $A$ and $B$ are two sets, then $\bar{A}\cup \bar{B}=\overline{A\cap B}$ (RIGHT)

  This is one of De Morgan's Laws. It states that the union of the complements of two sets is equal to the complement of their intersection.

• (d) If $A$, $B$ and $C$ are sets, then $A\cap B\cap \bar{C}\subseteq A\cap B$ (RIGHT)

  The set on the left is the intersection of three sets: $A$, $B$, and the complement of $C$. By the property of intersections, any intersection of sets is a subset of the individual sets or smaller intersections involved. Since $A\cap B\cap \bar{C}$ is a further restricted version of $A\cap B$, the subset relation holds true.