NCERT Exemplar Solutions Class 11 Maths Sets — Fill in the Blanks and True or False Q44–58

NCERT Exemplar Problems, Class 11 Mathematics — Sets, Fill in the Blanks and True or False questions 44 to 58. All 15 questions solved.

Short statements, each tested against the same few facts: a set with $n$ elements has $2^n$ subsets, $A – B = A \cap B’$, $A \subset B \Rightarrow A \cup B = B$, and the distributive and De Morgan laws. Question 52 is a six-part matching question, which the book prints at the end of the fill-in-the-blanks questions.

The printed key is wrong for Question 44 (the bracket) and Question 51. Both are explained where they occur.

Key insight. In every question here, translate the English into a plain statement about elements, then check one concrete element or small example. For anything stated “for all sets”, a single counterexample settles falsity, and an element-chasing argument settles truth.

Fill in the blanks in each of the Exercises from 44 to 51.

Question 44

The set $\{x \in \mathbf{R} : 1 \le x < 2\}$ can be written as ________.

Solution. The condition has a closed left end ($\le$, so $1$ is included) and an open right end ($<$, so $2$ is excluded). In interval notation, a square bracket includes the end point and a round bracket excludes it.

The printed key shows $[1, 2]$, which includes $2$ and is therefore wrong.

Question 45

When $A = \phi$, then the number of elements in $P(A)$ is ________.

Solution. The power set of the empty set has exactly one element, the empty set itself: $P(\phi) = \{\phi\}$. This agrees with $2^0 = 1$.

Question 46

If $A$ and $B$ are finite sets such that $A \subset B$, then $n(A \cup B) = $ ________.

Solution. If $A \subset B$, then $A \cup B = B$, so $n(A \cup B) = n(B)$. (Equivalently, $n(A \cup B) = n(A) + n(B) – n(A \cap B)$ and $A \cap B = A$.)

Question 47

If $A$ and $B$ are any two sets, then $A – B$ is equal to ________.

Solution. $A – B$ is the set of elements in $A$ that are not in $B$, that is, in $A$ and in the complement of $B$.

Question 48

The power set of the set $A = \{1, 2\}$ is ________.

Solution. List every subset, by size: the empty set, the two singletons, and the whole set. There are $2^2 = 4$ of them.

Question 49

Given the sets $A = \{1, 3, 5\}$, $B = \{2, 4, 6\}$ and $C = \{0, 2, 4, 6, 8\}$. Then the universal set of all the three sets $A$, $B$ and $C$ can be ________.

Solution. A universal set must contain every element of all three sets, so it must contain $A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$. The smallest choice is exactly that union; any larger set, such as $\mathbf{N} \cup \{0\}$ or $\mathbf{Z}$, would also do, which is why the question says “can be”.

Question 50

If $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, $A = \{1, 2, 3, 5\}$, $B = \{2, 4, 6, 7\}$ and $C = \{2, 3, 4, 8\}$. Then (i) $(B \cup C)’$ is ________. (ii) $(C – A)’$ is ________.

Solution. (i) $B \cup C = \{2, 3, 4, 6, 7, 8\}$. Its complement in $U$ is what is left: $\{1, 5, 9, 10\}$.

(ii) $C – A$ removes from $C$ the elements that are in $A$ ($2$ and $3$), leaving $\{4, 8\}$. The complement of that is $U$ without $4$ and $8$.

Question 51

For all sets $A$ and $B$, $A – (A \cap B)$ is equal to ________.

Solution. $A – (A \cap B) = A \cap (A \cap B)’ = A \cap (A’ \cup B’) = (A \cap A’) \cup (A \cap B’) = A \cap B’$, which is $A – B$. In words: taking away from $A$ only the part that is inside $B$ leaves the part of $A$ outside $B$.

The printed key shows $A \cup B’$, which is wrong. Test it: with $U = \{1, 2\}$, $A = \phi$ and $B = \{1\}$, the left side is $\phi – \phi = \phi$, but $A \cup B’ = \{2\}$. The key’s answer also contradicts its own answer to Question 47, since $A – B = A \cap B’$.

Question 52

Match the following sets for all sets $A$, $B$ and $C$.

Solution. Simplify each left side with the laws of sets.

(i) $A’ \cup B’ = (A \cap B)’$, and subtracting $A$ means intersecting with $A’$: $(A \cap B)’ \cap A’$. Since $A’ \subset (A \cap B)’$, this is just $A’$. Its complement is $A$. So (i) matches (b).

(ii) $B’ – A = B’ \cap A’ \subset B’$, so $B’ \cup (B’ – A) = B’$. Its complement is $B$. So (ii) matches (c).

(iii) $(A – B) – (B – C) = (A \cap B’) \cap (B \cap C’)’ = A \cap B’ \cap (B’ \cup C) = A \cap B’$, because $A \cap B’$ is already inside $B’$. This is $A – B$. So (iii) matches (a).

(iv) $(A \cap B’) \cap (C \cap B’) = A \cap C \cap B’ = (A \cap C) – B$. So (iv) matches (f).

(v) A pair $(a, y)$ is in $A \times (B \cap C)$ exactly when $a \in A$ and $y$ is in both $B$ and $C$, which is the same as being in both $A \times B$ and $A \times C$. So (v) matches (d).

(vi) Likewise, $y \in B \cup C$ means $y \in B$ or $y \in C$, which gives $(a, y) \in A \times B$ or $(a, y) \in A \times C$. So (vi) matches (e).

State True or False for the following statements in each of the Exercises from 53 to 58.

Question 53

If $A$ is any set, then $A \subset A$.

Solution. Every element of $A$ is an element of $A$, so $A$ is a subset of itself (not a proper subset, but “$\subset$” in this book allows equality).

Question 54

Given that $M = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and if $B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, then $B \not\subset M$.

Solution. The two sets have exactly the same elements, so $B = M$ and in particular $B \subset M$. The statement $B \not\subset M$ is false.

Question 55

The sets $\{1, 2, 3, 4\}$ and $\{3, 4, 5, 6\}$ are equal.

Solution. Equal sets have the same elements. The first contains $1$, which the second does not.

Question 56

$\mathbf{Q} \cup \mathbf{Z} = \mathbf{Q}$, where $\mathbf{Q}$ is the set of rational numbers and $\mathbf{Z}$ is the set of integers.

Solution. Every integer $n$ is rational ($n = \frac n1$), so $\mathbf{Z} \subset \mathbf{Q}$. When $\mathbf{Z} \subset \mathbf{Q}$, the union is $\mathbf{Q}$.

Question 57

Let sets $R$ and $T$ be defined as $R = \{x \in \mathbf{Z} \mid x$ is divisible by $2\}$ and $T = \{x \in \mathbf{Z} \mid x$ is divisible by $6\}$. Then $T \subset R$.

Solution. If $x$ is divisible by $6$, then $x = 6k = 2(3k)$ for an integer $k$, so $x$ is divisible by $2$. Every element of $T$ is in $R$.

Question 58

Given $A = \{0, 1, 2\}$ and $B = \{x \in \mathbf{R} \mid 0 \le x \le 2\}$. Then $A = B$.

Solution. $A$ has three elements, while $B$ is the whole interval $[0, 2]$ of real numbers, which contains infinitely many numbers such as $\frac12$ and $\sqrt2$ that are not in $A$.

Common mistakes

• Question 44. Writing $[1, 2]$ (as the printed key does) because the interval “ends at $2$”. The strict inequality $x < 2$ needs the round bracket at $2$.
• Question 45. Answering $0$ because $A$ has no elements. The power set of $\phi$ still has one element, $\phi$ itself, since $2^0 = 1$.
• Questions 47 and 51. Mixing up $A \cap B’$ with $A \cup B’$. The difference $A – B$ keeps elements of $A$ that satisfy a second condition, so it needs an intersection; the key’s $A \cup B’$ in Question 51 adds elements outside $A$.
• Question 49. Writing only $A \cup B$ or $C$, forgetting that a universal set must contain all three sets.
• Question 52(i) and (ii). Dropping a complement at the end. In (i) the inner set simplifies to $A’$, and the outer prime then gives $A$, not $A’$.
• Questions 54 and 58. In 54, reading $\not\subset$ as “is not a proper subset”, when equal sets are subsets of each other; in 58, treating $\{0, 1, 2\}$ and the interval $[0, 2]$ as the same because they have the same end points.

Practise next

• The other pages of this unit: Short Answer Type, Q1–22 and Long Answer and Objective Type, Q23–43.
• NCERT textbook practice on the same ideas: Exercise 1.2 (subsets, power sets, equal sets), Exercise 1.3 and the Miscellaneous Exercise.