GSEB Class 11 Maths Solutions Chapter 1 Sets Exercise 1.3

Question 1. Make correct statements by filling in the symbols \( \subset \) or \( \not\subset \) in the blank spaces:
1. {2, 3, 4,} {1, 2, 3, 4, 5}
2. {a, b, c} {b, c, d}
3. {x : x is a student of Class XI of your school} {x : x is a student of your school}
4. {x: x is a circle in the plane} {x: x is a rectangle in the plane}
5. {x: x is a triangle in the plane} {x: x is a rectangle in the plane}
6. {x: x is an equilateral triangle in the plane} {x: x is a triangle in the plane}
7. {x: x is an even natural number} {x: x is an integer}.
Answer:
1. \( \subset \)
2. \( \not\subset \)
3. \( \subset \)
4. \( \not\subset \)
5. \( \not\subset \)
6. \( \subset \)
7. \( \subset \)
In simple words: Look at each pair of sets. If all the things in the first set are also in the second set, use the 'is a subset of' symbol \( \subset \). If even one thing from the first set is missing from the second set, use the 'is not a subset of' symbol \( \not\subset \).

Exam Tip: Remember that the subset symbol (\( \subset \)) indicates that every element of the first set is also an element of the second set. The symbol \( \not\subset \) means at least one element is missing.

Question 2. Examine whether the following statements are true or false:
1. {a, b} \( \not\subset \) {b, c, a}
2. [a, e} \( \subset \) {x : x is a vowel in the English alphabet}
3. {1, 3, 5} \( \subset \) {1, 3, 5}
4. {a} \( \subset \) {a, b, c}
5. {a} \( \in \) {a, b, c}
6. {x: x is an even natural number which divides 6} \( \subset \) {x: x is a natural number which divides 36}.
Answer:
1. This statement is **False**. The elements 'a' and 'b' from the set {a, b} are both present in the set {b, c, a}. Therefore, {a, b} is a subset of {b, c, a}, making the original statement incorrect.
2. This statement is **True**. The vowels in the English alphabet are a, e, i, o, u. The set {a, e} contains only vowels, so it is a subset of the set of all English vowels.
3. This statement is **True**. The set {1, 3, 5} is a subset of itself because all its elements are contained within itself. Equal sets are always subsets of each other.
4. This statement is **True**. The element 'a' in the set {a} is also an element of the set {a, b, c}. Therefore, {a} is a subset of {a, b, c}.
5. This statement is **False**. The element 'a' belongs to the set {a, b, c}, so 'a \( \in \) {a, b, c}' is true. However, the set {a} is a subset, not an element, of {a, b, c}. The 'belongs to' symbol (\( \in \)) is used for elements, not sets.
6. This statement is **True**. The set {x : x is an even natural number which divides 6} simplifies to {2, 4}. The set {x : x is a natural number which divides 36} becomes {1, 2, 3, 4, 6, 9, 12, 18, 36}. Since all elements of {2, 4} (which are 2 and 4) are also present in the second set, the statement is correct.
In simple words: Check each statement carefully. For subsets, every item in the first group must be in the second. For elements, the exact item must be directly inside the group. Equal sets are always subsets of each other.

Exam Tip: Distinguish carefully between the 'subset' symbol (\( \subset \)) and the 'element of' symbol (\( \in \)). A set containing elements is a subset of another set if all its elements are included. An element is a single item belonging to a set.

Question 3. Let A = {1, 2, {3, 4}, 5}. Which of the following statements are correct and why?
1. {3, 4} \( \subset \) A
2. {3, 4} \( \in \) A
3. 1 \( \in \) A
4. 1 \( \subset \) A
5. {1, 2, 5} \( \subset \) A
6. {1, 2, 5} \( \in \) A
7. {1, 2, 3} \( \subset \) A
8. \( \phi \) \( \subset \) A
9. { \( \phi \) } \( \subset \) A
Answer: Given set A = {1, 2, {3, 4}, 5}.
1. **Incorrect**: The item {3, 4} is an element of set A, not a subset of A. A subset would be formed by taking existing elements of A and placing them in a new set.
2. **Correct**: The set {3, 4} is directly listed as an element within set A.
3. **Correct**: The number 1 is directly listed as an element within set A.
4. **Incorrect**: The number 1 is an element, not a set, so it cannot be a subset of A. A subset must itself be a set.
5. **Correct**: The elements 1, 2, and 5 are all found within set A. Therefore, the set {1, 2, 5} is a subset of A.
6. **Incorrect**: The set {1, 2, 5} itself is not listed as an element within set A.
7. **Incorrect**: The number 3 is not an individual element of A; rather, {3, 4} is an element. Thus, {1, 2, 3} cannot be a subset of A.
8. **Correct**: The empty set (\( \phi \)) is always a subset of every set, including A.
9. **Incorrect**: For { \( \phi \) } to be a subset of A, the empty set \( \phi \) would need to be an element of A. However, \( \phi \) is not listed as an element within set A.
In simple words: For the set A, remember that {3, 4} is treated as one whole item, not two separate items (3 and 4). If an item is inside the main set, use the 'is an element' symbol (\( \in \)). If a small group of items are all found in the main set, use the 'is a subset' symbol (\( \subset \)). The empty set is always a subset, but it's only an element if it's explicitly written inside the set.

Exam Tip: When a set contains other sets as its elements (like {3, 4} in this case), carefully distinguish between the element relationship (\( \in \)) and the subset relationship (\( \subset \)). An element is directly inside the set; a subset is a collection of elements from the original set.

Question 4. Write down all the subsets of the following subsets;
1. {a}
2. {a, b}
3. {1, 2, 3}
4. \( \phi \)
Answer:
1. Subsets of {a}: \( \phi \), {a}
2. Subsets of {a, b}: \( \phi \), {a}, {b}, {a, b}
3. Subsets of {1, 2, 3}: \( \phi \), {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}
4. Subsets of \( \phi \): \( \phi \)
In simple words: To find all subsets, think about every possible group you can make using the items in the main set, including a group with nothing in it (the empty set) and a group with all the items (the set itself). If a set has 'n' items, there will be \( 2^n \) subsets.

Exam Tip: Remember that the empty set (\( \phi \)) is always a subset of any set, and every set is a subset of itself. For a set with 'n' elements, the total number of distinct subsets is \( 2^n \).

Question 5. How many elements has P(A), if A = \( \phi \)?
Answer: If A = \( \phi \), then by the definition of a power set, we have P(A) = P(\( \phi \)) = { \( \phi \) }. This is a set that contains 1 element.
In simple words: The power set of an empty set is a set that holds only the empty set itself. Because it holds one item (the empty set), it has one element.

Exam Tip: The power set P(A) of a set A is the set of all subsets of A. If a set A has 'n' elements, then its power set P(A) will have \( 2^n \) elements. For A=\( \phi \), n=0, so P(A) has \( 2^0 = 1 \) element.

Question 6. Write the following as intervals:
1. {x: x \( \in \) R, -4 < x \( \leq \) 6)
2. {x: x \( \in \) R, – 12 < x < – 10}
3. {x: x \( \in \) R, 0 \( \leq \) x < 7}
4. {x: x \( \in \) R, 3 \( \leq \) x \( \leq \) 4}
Answer: Given intervals are:
1. (-4, 6]
2. (-12, -10)
3. [0, 7)
4. [3, 4]
In simple words: If the number is not included, use a round bracket. If the number is included, use a square bracket. Read the symbols carefully to know if the ends are open or closed.

Exam Tip: Remember that '(' and ')' denote open intervals (exclusive of endpoints), while '[' and ']' denote closed intervals (inclusive of endpoints). Combine these based on the strict (< or >) or inclusive ($\leq$ or $\geq$) inequalities.

Question 7. Write the following intervals in the set builder form:
1. (-3, 0)
2. [6, 12]
3. (6, 12]
4. [-23, 5)
Answer:
1. (-3, 0) = {x: x \( \in \) R, – 3 < x < 0}
2. [6, 12] = {x : x \( \in \) R, 6 \( \leq \) x \( \leq \) 12}
3. (6, 12] = {x: x \( \in \) R, 6 < x \( \leq \) 12}
4. [- 23, 5) = {x : x \( \in \) R, – 23 \( \leq \) x < 5}.
In simple words: To change from interval form to set builder form, write 'x such that x is a real number' and then write the number rules. Square brackets mean 'greater than or equal to' or 'less than or equal to'. Round brackets mean strictly 'greater than' or 'less than'.

Exam Tip: The set builder notation {x: x \( \in \) R, condition} is standard. Ensure you use the correct inequality symbols (<, >, \( \leq \), \( \geq \)) corresponding to whether the interval is open or closed at each end.

Question 8. What universal set would you propose for each of the following:
(i) The set of right triangles.
(ii) The set of isosceles triangles.
Answer: The set of all possible triangles is the universal set for each of the given sets.
In simple words: For both right triangles and isosceles triangles, the best universal set to pick is simply the set of ALL kinds of triangles. Every right triangle is a triangle, and every isosceles triangle is also a triangle.

Exam Tip: A universal set (U) is a set that includes all the elements under consideration in a particular context. When thinking about different types of triangles, the most encompassing set is usually 'all triangles'.

Question 9. Given the sets A = {1, 3, 5}, B = {2, 4, 6}, C = {0, 2, 4, 6, 8}. Which of the following may be considered as universal set(s) for all the three sets A, B and C?
(i) {0,1, 2, 3, 4, 5, 6}
(ii) \( \phi \)
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) {1, 2, 3, 4, 5, 6, 7, 8}
Answer: The set (iii), i.e., {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the given sets A, B and C because it contains all elements from A, B, and C.
In simple words: To be a universal set, a set must include every single item from all the other sets you are thinking about. Set (iii) has all the numbers that are in A, B, and C.

Exam Tip: A universal set must contain all elements of the sets it covers. Check each option to ensure it includes every element from all given sets (A, B, and C in this case). The empty set (\( \phi \)) can never be a universal set for non-empty sets.