Q.Draw appropriate Venn diagram for each of the following :

Concept: Set Difference / Complement — the region outside a given set within the universal set $U$.

(i) $(A \cup B)'$

Shade everything outside the union of $A$ and $B$.

(ii) $A' \cap B'$

Shade the region that is outside $A$ and outside $B$ — same as (i) by De Morgan's law.

(iii) $(A \cap B)'$

Shade everything outside the intersection of $A$ and $B$.

(iv) $A' \cup B'$

Shade the region that is outside $A$ or outside $B$ — same as (iii) by De Morgan's law. …

The Concept: Set Difference and Complement

When we talk about the complement of a set, say A′, we mean everything outside A but inside the universal set U. Similarly, (A ∪ B)′ means everything outside the union of A and B — that is, elements that belong to neither A nor B.

The trick here is that De Morgan’s laws give us two pairs of identical expressions:

• (A ∪ B)′ = A′ ∩ B′
• (A ∩ B)′ = A′ ∪ B′

So you only need to understand how to shade one of each pair, and the other follows. Let’s draw them step by step.

Step-by-Step Drawing

We’ll use a standard Venn diagram with two overlapping circles (A and B) inside a rectangle (the universal set U). Shading will show the required region.

1. (i) (A ∪ B)′

First, think: A ∪ B is everything in A or B (or both). Its complement is everything not in A ∪ B — that is, the region outside both circles.

• Shade the part of U that lies outside both A and B.
• This is the area in the rectangle but not inside either circle.

2. (ii) A′ ∩ B′

Now, A′ is everything outside A, and B′ is everything outside B. Their intersection is the region that is outside A and outside B simultaneously.

• That’s exactly the same region as in (i): the area outside both circles.
• So the diagram for (ii) is identical to (i).

3. (iii) (A ∩ B)′

A ∩ B is the overlapping region (the “middle” part where A and B meet). Its complement is everything except that overlap.

• Shade all of U except the tiny lens-shaped region where A and B intersect.
• This includes: the part of A that is not in B, the part of B that is not in A, and the area outside both circles.

4. (iv) A′ ∪ B′

A′ is outside A, B′ is outside B. Their union is everything that is outside A or outside B (or both).

• That means: any point that is not inside A, or not inside B, or both. …

Concept First — The Idea Being Tested

This question tests Set Intersection and Complement — specifically, De Morgan’s Laws.

Intuition:

• The complement of a set (like $A'$) means “everything not in $A$” (within a universal set $U$).
• The union $A \cup B$ means “everything in $A$ or $B$ (or both)”.
• The intersection $A \cap B$ means “everything in both $A$ and $B$”.

De Morgan’s Laws tell us:

• $(A \cup B)' = A' \cap B'$
• $(A \cap B)' = A' \cup B'$

So parts (i) and (ii) are equal, and parts (iii) and (iv) are equal.

We’ll draw each separately to see why.

Step-by-Step — Drawing Each Venn Diagram

We assume a universal set $U$ (the rectangle) and two overlapping circles $A$ and $B$ inside it.

(i) $(A \cup B)'$

Reasoning:

• First, shade $A \cup B$ — that’s all of $A$ and all of $B$ (the two circles together).
• Then, the complement $(A \cup B)'$ is everything outside both circles — the part of the rectangle not covered by $A$ or $B$.

Diagram:

• Shade the region outside both circles (the “frame” of the rectangle).

Shaded area = everything not in A or B.

(ii) $A' \cap B'$

Reasoning:

• $A'$ = everything not in $A$ (shade the rectangle except circle $A$).
• $B'$ = everything not in $B$ (shade the rectangle except circle $B$).
• $A' \cap B'$ = the overlap of those two shaded regions — i.e., the part that is neither in $A$ nor in $B$.

That’s exactly the same region as in (i) — the area outside both circles.

Diagram:

• Shade the region outside both circles (same as (i)).

✓ Observation: $(A \cup B)' = A' \cap B'$ — De Morgan’s first law.

(iii) $(A \cap B)'$

Reasoning:

• First, shade $A \cap B$ — that’s the overlap region where both circles meet (the “lens” in the middle).
• Then, the complement $(A \cap B)'$ is everything except that overlap — i.e., all of $A$ except the overlap, all of $B$ except the overlap, and the outside region.

Diagram:

• Shade everything except the small central lens where $A$ and $B$ overlap.

The only unshaded part is the overlap of A and B.

(iv) $A' \cup B'$

Reasoning:

• $A'$ = everything not in $A$ (shade outside $A$).
• $B'$ = everything not in $B$ (shade outside $B$).
• $A' \cup B'$ = all of those shaded parts combined — i.e., everything that is either outside $A$ or outside $B$ (or both).

What’s missing from this union? Only the part that is inside both $A$ and $B$ — the overlap. So the shaded region is everything except the overlap.

Diagram:

• Shade everything except the central lens (same as (iii)). …

Here are the common mistakes students make with set intersection problems, specifically using the sets you provided, and how to avoid each.

Mistake 1: Forgetting the Definition of "Natural Number" ($\mathbb{N}$)

The Error: Students often assume natural numbers start from 0 or include negative numbers. In the Indian curriculum (NCERT/CBSE), natural numbers are defined as $\{1, 2, 3, 4, ...\}$.

How it affects the answer: If you include 0, then $A \cap B$ might incorrectly include 0. If you include negatives, the intersection with primes ($D$) becomes confusing.

How to Avoid: Memorize the standard definition. For Class 11 NCERT, $\mathbb{N} = \{1, 2, 3, ...\}$. Always write this set down before solving.

Mistake 2: Confusing "Even" and "Odd" with "Prime"

The Error: Students think that because a number is prime, it cannot be even (or odd). They forget that 2 is the only even prime number.

The Consequence: For question (v) $B \cap D$, students often write $\phi$ (empty set) instead of $\{2\}$.

How to Avoid: List the first few elements of each set.

• $B = \{2, 4, 6, 8, 10, ...\}$
• $D = \{2, 3, 5, 7, 11, ...\}$ Now, visually scan for common elements. The only common element is 2.

Mistake 3: Assuming "Odd" and "Prime" are Mutually Exclusive

The Error: Students think that since most primes are odd, the intersection $C \cap D$ must be all odd primes. They forget that 2 is prime but not odd.

The Consequence: For question (vi) $C \cap D$, students write $\{3, 5, 7, 11, ...\}$ (all odd primes) but forget to explicitly exclude 2. While the set of odd primes is correct, the reasoning is flawed if they don't mention that 2 is excluded.

How to Avoid: Always check the boundary case (the number 2).

• $C = \{1, 3, 5, 7, 9, ...\}$
• $D = \{2, 3, 5, 7, 11, ...\}$ The intersection is $\{3, 5, 7, 11, ...\}$ (all odd primes). This is correct, but be explicit: "All prime numbers except 2."

Mistake 4: Writing the Answer in Roster Form Incorrectly

The Error: For infinite sets like $A \cap B$ (which is just $B$), students try to list all elements or write an incomplete roster like $\{2, 4, 6\}$.

The Consequence: Marks are deducted for not showing the pattern or using the wrong notation.

How to Avoid: Use set-builder notation for infinite answers, or use roster form with an ellipsis ($...$).

• Correct: $A \cap B = \{2, 4, 6, 8, ...\}$ or $A \cap B = \{x : x \text{ is an even natural number}\}$.
• Incorrect: $A \cap B = \{2, 4, 6\}$ (this implies the set stops at 6).

Mistake 5: Misinterpreting the Intersection Symbol ($\cap$) …