RBSE Solutions Class 11 Maths Chapter 1 Sets Exercise 1.1

Question 1. Fill the symbols \( \in \) or \( \notin \) in the blanks to make the following statements correct:
(i) 3.... {1, 2, 3, 4, 5}
(ii) 2.5...N
(iii) 0.........Q
Answer:
(i) \( \in \)
(ii) \( \notin \)
(iii) \( \in \)
In simple words: The symbol \( \in \) means "is an element of," showing that something belongs to a set. The symbol \( \notin \) means "is not an element of," showing that something does not belong to a set. We use these symbols to describe if a number or item is inside a group of numbers or items.

🎯 Exam Tip: Remember that "N" stands for Natural Numbers (1, 2, 3...) and "Q" stands for Rational Numbers (numbers that can be written as a fraction, like 0 which is 0/1).

Question 2. Fill the symbols \( \subset \) or \( \not\subset \) in the blanks to make following statements correct:
(i) {2, 3, 4}.........{1, 2, 3, 4, 5}
(ii) {a, e, o}........{a, b, c}
(iii) {x: x is an equilateral triangle in a plane)........{x : x is a triangle in a plane)
(iv) {x: x is a natural number}.......{x : x is an odd whole number}

🎯 Exam Tip: The symbol \( \subset \) means "is a subset of," implying all elements of the first set are also in the second set. The symbol \( \not\subset \) means "is not a subset of." Carefully check each element to see if it belongs to the second set. Since the solution for this question is not available in the provided content, it is crucial to understand the definition of subsets for such problems.

Question 3. Examine the following statements :
(i) {a, b} \( \subset \) {b, a, c]
(ii) {a, e} \( \subset \) {x : x is vowel of English alphabet}
(iii) {1, 2, 3} \( \not\subset \) {1, 3, 2, 5}
(iv) {x: x is an even natural number less than 6} \( \not\subset \) {x : x is a natural number which divide 36}
Answer:
(i) {a, b} \( \subset \) {b, a, c}
The elements 'a' and 'b' from the first set are both present in the second set. This means that the first set is a subset of the second. Therefore, the statement is true.
(ii) {a, e} \( \subset \) {x : x is vowel of English alphabet}
The set of vowels in the English alphabet is {a, e, i, o, u}. Since 'a' and 'e' are both in this set, the first set is a subset of the second. So, the statement is true.
(iii) {1, 2, 3} \( \not\subset \) {1, 3, 2, 5}
All elements of the first set {1, 2, 3} are present in the second set {1, 3, 2, 5} because the order does not matter in sets. So, {1, 2, 3} *is* a subset of {1, 3, 2, 5}. Therefore, the statement that it is *not* a subset is false.
(iv) {x: x is an even natural number less than 6} \( \not\subset \) {x : x is a natural number which divide 36}
The first set is {2, 4}. The second set, which divides 36, is {1, 2, 3, 4, 6, 9, 12, 18, 36}. Both 2 and 4 are present in the second set. This means {2, 4} *is* a subset of the second set. Therefore, the statement that it is *not* a subset is false.
In simple words: We check if every item in the first group is also in the second group. If yes, it's a subset. If no, it's not a subset. When checking for subsets, the order of items does not matter.

🎯 Exam Tip: Pay close attention to the symbols \( \subset \) (subset) and \( \not\subset \) (not a subset). Listing out the elements of sets described by properties (like "vowels" or "divisors") helps in comparing them accurately.

Question 4. Write power sets of the following:
(i) {a}
(ii) {1, 2, 3}
(iii) {a, b}
(iv) \( \Phi \)
Answer:
(i) For the set {a}, the subsets are \( \Phi \) (the empty set) and {a}.
So, the power set P({a}) = { \( \Phi \), {a} }.
(ii) For the set {1, 2, 3}, the subsets are \( \Phi \), {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and {1, 2, 3}. A set with 'n' elements has \( 2^n \) subsets. For 3 elements, it has \( 2^3 = 8 \) subsets.
So, the power set P({1, 2, 3}) = { \( \Phi \), {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} }.
(iii) For the set {a, b}, the subsets are \( \Phi \), {a}, {b}, and {a, b}.
So, the power set P({a, b}) = { \( \Phi \), {a}, {b}, {a, b} }.
(iv) For \( \Phi \) (the null set), there is only one subset, which is \( \Phi \) itself.
So, the power set P(\( \Phi \)) = { \( \Phi \) }.
In simple words: A power set is a set that contains all possible subsets of the original set, including the empty set and the set itself. For a set with 'n' items, its power set will always have \( 2^n \) items.

🎯 Exam Tip: Always remember to include the empty set ( \( \Phi \) ) and the set itself when listing all subsets for a power set. The total number of subsets is always \( 2^n \), where 'n' is the number of elements in the original set.

Question 5. Write the following as intervals :
(i) {x: x \( \in \) R, -3 < x < 6}
(ii) {x: x \( \in \) R, -4 \( \le \) x \( \le \) 8}
(iii) {x : x \( \in \) R, 4 < x \( \le \) 9}
(iv) {x: x \( \in \) R, -6 \( \le \) x < -1}
Answer:
(i) {x: x \( \in \) R, -3 < x < 6} is an open interval. This means that x is greater than -3 but less than 6, and -3 and 6 are not included. Open intervals are shown with parentheses.
So, the interval is (-3, 6).
(ii) {x: x \( \in \) R, -4 \( \le \) x \( \le \) 8} is a closed interval. This means that x is greater than or equal to -4 and less than or equal to 8, and both -4 and 8 are included. Closed intervals are shown with square brackets.
So, the interval is [-4, 8].
(iii) {x : x \( \in \) R, 4 < x \( \le \) 9} is a semi-closed interval. This means x is greater than 4 (not included) but less than or equal to 9 (included).
So, the interval is (4, 9].
(iv) {x: x \( \in \) R, -6 \( \le \) x < -1} is also a semi-closed interval. This means x is greater than or equal to -6 (included) but less than -1 (not included).
So, the interval is [-6, -1).
In simple words: We write sets of numbers that are between two points using intervals. Curved brackets '()' mean the numbers at the ends are not included, like 'open doors'. Square brackets '[]' mean the numbers at the ends are included, like 'closed doors'.

🎯 Exam Tip: Remember that '<' and '>' denote open intervals (using parentheses), while '\( \le \)' and '\( \ge \)' denote closed intervals (using square brackets). Mixed use creates semi-open or semi-closed intervals.

Question 7. A = {1, 3, 5}, B = {1, 4, 6} and C = {2, 4, 6, 8}, then which of the following may be considered as in universal set:
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) {1, 2, 3, 4, 5, 6, 7, 8}
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) \( \Phi \)
Answer:
A universal set (U) must contain all the elements of every other set being considered. For sets A = {1, 3, 5}, B = {1, 4, 6}, and C = {2, 4, 6, 8}, we need a set that has at least all these unique numbers: 1, 2, 3, 4, 5, 6, 8. Any other elements in the universal set are fine.
Let's check the given options:
(i) {0, 1, 2, 3, 4, 5, 6}: This set is missing the element 8 from set C.
(ii) {1, 2, 3, 4, 5, 6, 7, 8}: This set contains all elements from A, B, and C.
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}: This set also contains all elements from A, B, and C.
(iv) \( \Phi \): The empty set cannot be a universal set because it contains no elements.
So, both option (ii) and option (iii) can be considered as universal sets because they both contain all the required elements from sets A, B, and C. It is possible to have more than one universal set depending on the context.
In simple words: A universal set is like a big container that holds all the items from all the smaller sets we are looking at. We need to pick an option that has every single number found in sets A, B, and C.

🎯 Exam Tip: A universal set is typically denoted by U. To identify a universal set, list all unique elements from all given sets, and then find an option that includes all those unique elements (and possibly more). The empty set can never be a universal set unless all other sets are empty.