NCERT Solutions & Concepts for Class 11 Maths Chapter 1 Exercise 1.5
Welcome, Class 11 students! In this guide, we dive deep into CBSE Class 11 Maths Chapter 1 Exercise 1.5 (Sets), which focuses primarily on the Complement of a Set and the mathematical beauty of De Morgan's Laws. Understanding how to find complements and work with universal sets is highly critical, not just for clearing your school exams, but also for mastering advanced chapters like Probability and Mathematical Reasoning. By the end of this module, you will confidently solve every problem in sets ex 1 5 class 11 ncert with speed and conceptual accuracy. Let's pick up our virtual sketchpad and begin!
Understanding the Complement of a Set
To understand the complement of a set, we must first establish a reference boundary known as the Universal Set (denoted by $U$). Let $A$ be a subset of $U$. The complement of $A$ (denoted by $A'$ or $A^c$) is the set of all elements belonging to $U$ that do not belong to $A$.
Mathematically, we write this as:
$A' = \{x \in U : x \notin A\}$
In simple terms, you can calculate the complement using the set difference operation: $A' = U - A$. Visually, if you picture a Venn diagram where the rectangle represents $U$ and a circle inside represents $A$, the shaded region outside the circle $A$ but inside the rectangle is $A'$. This basic concept extends to complex relations, especially when verifying De Morgan's Laws.
Properties of Complement Sets & De Morgan's Laws
- Complement Laws
- 1. $A \cup A' = U$ (The union of a set and its complement equals the Universal Set). 2. $A \cap A' = \phi$ (A set and its complement share no common elements; they are disjoint).
- De Morgan's Laws
- These are highly crucial equations for boards: 1. $(A \cup B)' = A' \cap B'$ (The complement of union is the intersection of complements). 2. $(A \cap B)' = A' \cup B'$ (The complement of intersection is the union of complements).
- Law of Double Complementation
- $(A')' = A$. Taking the complement of a complement set returns you back to the original set.
- Laws of Empty Set & Universal Set
- $\phi' = U$ (Complement of nothing is everything) and $U' = \phi$ (Complement of everything is nothing).
How to Find and Verify Complement Set Operations
- Identify the Universal Set — Always determine the scope of elements listed in the Universal Set ($U$). All operations are confined to this set.
- Perform the Inner Operation First — If asked to find $(A \cup B)'$, first calculate the union set $A \cup B$ by listing all unique elements in both $A$ and $B$.
- Subtract from Universal Set — Remove the elements of the inner operation from $U$. The remaining elements in $U$ form the final complement set.
Fully Worked CBSE Examples
- Example 1: Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $A = \{2, 4, 6, 8\}$, and $B = \{2, 3, 5, 7\}$. Find $(A \cup B)'$. Step 1: First find $A \cup B$. Combining elements from both sets gives $A \cup B = \{2, 3, 4, 5, 6, 7, 8\}$. Step 2: Now, find the complement of $(A \cup B)$ with respect to $U$. Subtract these elements from $U$. $(A \cup B)' = U - \{2, 3, 4, 5, 6, 7, 8\} = \{1, 9\}$.
- Example 2: Using the same sets, verify De Morgan's Law: $(A \cap B)' = A' \cup B'$. Step 1: Find the Left Hand Side (LHS): $A \cap B = \{2\}$ (since only 2 is common). $(A \cap B)' = U - \{2\} = \{1, 3, 4, 5, 6, 7, 8, 9\}$. Step 2: Find the Right Hand Side (RHS): $A' = U - A = \{1, 3, 5, 7, 9\}$ $B' = U - B = \{1, 4, 6, 8, 9\}$ $A' \cup B' = \{1, 3, 4, 5, 6, 7, 8, 9\}$. Since LHS = RHS, the law is verified!
Board Exam Trap: Watch out for parentheses and operators!
A very common error Class 11 students commit during exams is writing $(A \cup B)' = A' \cup B'$. Remember, the union symbol flips to intersection, and intersection flips to union when distributing the complement operator.
Always verify your operations using a quick Venn Diagram sketch in your rough work section to ensure you haven't made silly computational slips!
Practice Questions with Solutions
- Q: Let $U = \{x : x \in \mathbb{N}, x \le 10\}$ and $A = \{x : x \text{ is a prime number} \}$. Find $A'$. A: Step 1: Write down the sets in roster form: $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ $A = \{2, 3, 5, 7\}$ Step 2: Calculate $A' = U - A$ by removing the elements of $A$ from $U$. Final answer: $A' = \{1, 4, 6, 8, 9, 10\}$
- Q: Let $U$ be the set of all triangles in a plane. If $A$ is the set of all triangles with at least one angle different from $60^\circ$, what is $A'$? A: Step 1: Analyze the definition of set $A$. If a triangle has at least one angle different from $60^\circ$, it means it is NOT equilateral (since equilateral triangles have all angles equal to $60^\circ$). Step 2: The complement set $A'$ contains triangles from $U$ that do not fall under $A$. Therefore, $A'$ must contain triangles where NO angle is different from $60^\circ$. Final answer: $A'$ is the set of all equilateral triangles.
- Q: Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, $A = \{1, 3, 5, 7, 9\}$, and $B = \{2, 4, 6, 8, 10\}$. Find $(A \cup B)'$. A: Step 1: Calculate $A \cup B$. Since $A$ is odd natural numbers up to 10 and $B$ is even natural numbers up to 10: $A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} = U$ Step 2: Find the complement: $(A \cup B)' = U - U = \phi$ Final answer: $\phi$ (Empty set)
- Q: Simplify the expression: $(A \cup B)' \cup (A' \cap B)$ using set laws. A: Step 1: Apply De Morgan's Law to the first term: $(A \cup B)' = A' \cap B'$ Step 2: Substitute this back into the expression: $(A' \cap B') \cup (A' \cap B)$ Step 3: Apply the Distributive Law in reverse (take $A' \cap$ common): $= A' \cap (B' \cup B)$ Step 4: Use Complement Law ($B' \cup B = U$): $= A' \cap U$ Step 5: Intersection with universal set yields the set itself: $= A'$ Final answer: $A'$
Frequently Asked Questions
What is De Morgan's Law in Sets?
De Morgan's Laws state that the complement of the union of two sets is equal to the intersection of their complements, $(A \cup B)' = A' \cap B'$, and vice versa.
What is the complement of a universal set?
The complement of a universal set is an empty set (i.e., $U' = \phi$), because there are no elements left outside the universal set within our domain of discourse.
How do you represent the complement of a set on a Venn diagram?
To represent the complement $A'$ on a Venn diagram, you draw a rectangle for $U$ containing a circle for $A$. Shading everything inside the rectangle except the circle $A$ depicts $A'$.