NCERT Exemplar Solutions Class 11 Maths Sets — Short Answer Type Q1–22

NCERT Exemplar Problems, Class 11 Mathematics — Sets, Short Answer Type questions 1 to 22. All 22 questions solved.

This band covers three skills: writing a set in roster form from a description (Questions 1, 2, 3, 7, 8, 9), deciding whether a statement about a set is true or false (4, 13 to 17), and proving set identities by element chasing or by the algebra of sets (5, 6, 12, 18 to 21). The identities run on De Morgan’s laws, $A – B = A \cap B’$, and the distributive laws.

The printed key is wrong or incomplete for Questions 3, 4(iv) and 7, and Question 12 is misprinted in the book. Each is explained where it occurs.

Key insight. Turn every difference into an intersection with a complement, $A – B = A \cap B’$. After that, set identities become ordinary algebra with the laws $(A \cap B)’ = A’ \cup B’$ and $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

Question 1

Write the following sets in the roster form.

(i) $A = \{x : x \in \mathbf{R},\ 2x + 11 = 15\}$ (ii) $B = \{x \mid x^2 = x,\ x \in \mathbf{R}\}$ (iii) $C = \{x \mid x$ is a positive factor of a prime number $p\}$

Solution. Roster form means listing the elements, so first find which values satisfy the defining condition.

(i) $2x + 11 = 15$ gives $2x = 4$, so $x = 2$.

(ii) $x^2 = x$ means $x(x – 1) = 0$, so $x = 0$ or $x = 1$. Do not divide by $x$, which would lose the root $0$.

(iii) A prime $p$ has exactly two positive factors, $1$ and $p$.

Question 2

Write the following sets in the roster form.

(i) $D = \{t \mid t^3 = t,\ t \in \mathbf{R}\}$ (ii) $E = \left\{w \,\middle|\, \dfrac{w – 2}{w + 3} = 3,\ w \in \mathbf{R}\right\}$ (iii) $F = \{x \mid x^4 – 5x^2 + 6 = 0,\ x \in \mathbf{R}\}$

Solution. (i) $t^3 – t = t(t – 1)(t + 1) = 0$, so $t \in \{-1, 0, 1\}$.

(ii) Multiply out, remembering $w \ne -3$: $w – 2 = 3w + 9$, so $-2w = 11$ and $w = -\frac{11}{2}$. This is not $-3$, so it is admissible.

(iii) Put $y = x^2$: $y^2 – 5y + 6 = (y – 2)(y – 3) = 0$, so $x^2 = 2$ or $x^2 = 3$. All four roots $\pm\sqrt2, \pm\sqrt3$ are real.

Question 3

If $Y = \{x \mid x$ is a positive factor of the number $2^{p-1}(2^p – 1)$, where $2^p – 1$ is a prime number$\}$, write $Y$ in the roster form.

Solution. Let $q = 2^p – 1$, which is an odd prime. The number is $2^{p-1}q$, and since $q$ is a prime different from $2$, its positive factors are exactly the numbers $2^k$ and $2^k q$ for $k = 0, 1, \dots, p – 1$. Check with $p = 2$: the number is $2 \cdot 3 = 6$, with factors $1, 2, 3, 6$, which agrees.

The printed key lists only $\{1, 2, 2^2, \dots, 2^{p-1}, (2^p – 1)\}$, which stops after one multiple of $q$. That list is incomplete: for $p = 2$ it gives only $\{1, 2, 3\}$ and misses the factor $6$. The full set has $2p$ elements.

Question 4

State which of the following statements are true and which are false. Justify your answer.

(i) $35 \in \{x \mid x$ has exactly four positive factors$\}$. (ii) $128 \in \{y \mid$ the sum of all the positive factors of $y$ is $2y\}$. (iii) $3 \notin \{x \mid x^4 – 5x^3 + 2x^2 – 112x + 6 = 0\}$. (iv) $496 \notin \{y \mid$ the sum of all the positive factors of $y$ is $2y\}$.

Solution. Each statement asks whether a number meets a membership condition, so test the number directly.

(i) $35 = 5 \times 7$ has positive factors $1, 5, 7, 35$, exactly four. True.

(ii) $128 = 2^7$ has factor sum $1 + 2 + \dots + 128 = 255$. For membership we would need $2 \times 128 = 256$. Since $255 \ne 256$, $128$ is not in the set. False.

(iii) Substitute $x = 3$: $81 – 135 + 18 – 336 + 6 = -366 \ne 0$. So $3$ is not a root and $3 \notin$ the set. True.

(iv) $496 = 2^4 \times 31$, so its factor sum is $(1 + 2 + 4 + 8 + 16)(1 + 31) = 31 \times 32 = 992 = 2 \times 496$. So $496$ does belong to the set (it is a perfect number), and the claim “$496 \notin$” is False.

The printed key says True for (iv). That is wrong: the factor sum of $496$ is exactly $992 = 2 \cdot 496$, so $496$ is in the set.

Question 5

Given $L = \{1, 2, 3, 4\}$, $M = \{3, 4, 5, 6\}$ and $N = \{1, 3, 5\}$. Verify that $L – (M \cup N) = (L – M) \cap (L – N)$.

Solution. Compute each side separately and compare.

$M \cup N = \{1, 3, 4, 5, 6\}$, so $L – (M \cup N) = \{2\}$.

$L – M = \{1, 2\}$ and $L – N = \{2, 4\}$, so $(L – M) \cap (L – N) = \{2\}$.

Both sides equal $\{2\}$.

Question 6

If $A$ and $B$ are subsets of the universal set $U$, show that (i) $A \subset A \cup B$ (ii) $A \subset B \Leftrightarrow A \cup B = B$ (iii) $(A \cap B) \subset A$.

Solution. To show $P \subset Q$, take an arbitrary $x \in P$ and prove $x \in Q$.

(i) If $x \in A$, then $x \in A$ or $x \in B$, which is the definition of $x \in A \cup B$. So $A \subset A \cup B$.

(ii) ($\Rightarrow$) Suppose $A \subset B$. By (i), $B \subset A \cup B$. Conversely, if $x \in A \cup B$ then either $x \in B$, or $x \in A \subset B$, so in both cases $x \in B$; hence $A \cup B \subset B$. The two inclusions give $A \cup B = B$. ($\Leftarrow$) Suppose $A \cup B = B$. By (i), $A \subset A \cup B = B$.

(iii) If $x \in A \cap B$, then $x \in A$ and $x \in B$; in particular $x \in A$. So $(A \cap B) \subset A$.

Question 7

Given that $N = \{1, 2, 3, \dots, 100\}$, write (i) the subset of $N$ whose elements are even numbers, (ii) the subset of $N$ whose elements are perfect square numbers.

Solution. (i) The even numbers from $1$ to $100$ run from $2$ to $100$, fifty of them. (ii) The perfect squares up to $100$ are $1^2, 2^2, \dots, 10^2$; the largest is $100$ itself, since $10^2 = 100 \in N$.

The printed key stops at $98$ in (i) and at $81$ in (ii). Both lists omit $100$, which belongs to $N$ and is both even and a perfect square.

Question 8

If $X = \{1, 2, 3\}$ and $n$ represents any member of $X$, write the sets containing all numbers represented by (i) $4n$ (ii) $n + 6$ (iii) $\dfrac n2$ (iv) $n – 1$.

Solution. Substitute $n = 1, 2, 3$ in each expression.

Question 9

If $Y = \{1, 2, 3, \dots, 10\}$ and $a$ represents any element of $Y$, write the sets containing all the elements satisfying the given conditions. (i) $a \in Y$ but $a^2 \notin Y$ (ii) $a + 1 = 6,\ a \in Y$ (iii) $a$ is less than $6$ and $a \in Y$.

Solution. (i) $a^2 \le 10$ exactly when $a \le 3$, so the squares of $1, 2, 3$ stay inside $Y$ ($1, 4, 9$) and $a^2 \notin Y$ for $a \ge 4$. (ii) $a = 5$. (iii) The elements of $Y$ below $6$.

Question 10

$A$, $B$ and $C$ are subsets of the universal set $U$. If $A = \{2, 4, 6, 8, 12, 20\}$, $B = \{3, 6, 9, 12, 15\}$, $C = \{5, 10, 15, 20\}$ and $U$ is the set of all whole numbers, draw a Venn diagram showing the relation of $U$, $A$, $B$ and $C$.

Solution. The diagram is fixed by the overlaps, so find them first: $A \cap B = \{6, 12\}$, $B \cap C = \{15\}$, $A \cap C = \{20\}$, and $A \cap B \cap C = \varnothing$. Three circles are therefore drawn so that each pair overlaps but there is no region common to all three, all inside a rectangle for $U$. Every element goes in exactly one region:

Region	Elements
$A$ only	$2, 4, 8$
$B$ only	$3, 9$
$C$ only	$5, 10$
$A \cap B$ only	$6, 12$
$B \cap C$ only	$15$
$A \cap C$ only	$20$
$A \cap B \cap C$	none
In $U$, outside all three	all other whole numbers, such as $0, 1, 7, 11$

The key’s diagram writes $20$ twice, once in $A$ and once in $C$; the single element $20$ belongs to the overlap of $A$ and $C$.

Question 11

Let $U$ be the set of all boys and girls in a school, $G$ the set of all girls, $B$ the set of all boys, and $S$ the set of all students who take swimming. Some, but not all, students take swimming. Draw a Venn diagram showing one of the possible interrelationships among $U$, $G$, $B$ and $S$.

Solution. $G$ and $B$ have no student in common, and between them they make up $U$, so draw $G$ and $B$ as two non-overlapping circles inside the rectangle $U$. “Some, but not all” means $S$ is neither empty nor all of $U$, so draw $S$ as a third circle that overlaps both $G$ and $B$ without covering either completely (swimmers of both sexes, and non-swimmers of both sexes).

The printed key’s diagram labels the left circle $A$ where the question calls it $B$; the structure is the same.

Question 12

For all sets $A$, $B$ and $C$, show that $(A – B) \cap (C – B) = A – (B \cup C)$.

Solution. As printed, this statement is false, so it cannot be shown. Take $A = C = \{1\}$ and $B = \varnothing$: the left side is $\{1\} \cap \{1\} = \{1\}$, while the right side is $\{1\} – \{1\} = \varnothing$. In general $$(A – B) \cap (C – B) = (A \cap B’) \cap (C \cap B’) = A \cap C \cap B’ = (A \cap C) – B,$$ whereas $$A – (B \cup C) = A \cap B’ \cap C’.$$ These differ whenever an element lies in $A$ and $C$ but not in $B$. The book appears to have run two correct identities together. Both are proved by the same expansion:

• $(A – B) \cap (C – B) = (A \cap C) – B$, as shown above;
• $(A – B) \cap (A – C) = (A \cap B’) \cap (A \cap C’) = A \cap (B’ \cap C’) = A \cap (B \cup C)’ = A – (B \cup C)$ by De Morgan’s law.

(The first identity is also the one matched in Question 52(iv) of this unit.) The key gives no answer for this question.

Determine whether each of the statements in Exercises 13 to 17 is true or false. Justify your answer.

Question 13

For all sets $A$ and $B$, $(A – B) \cup (A \cap B) = A$.

Solution. $A – B$ is the part of $A$ outside $B$ and $A \cap B$ is the part of $A$ inside $B$; together they are all of $A$. Algebraically, $(A \cap B’) \cup (A \cap B) = A \cap (B’ \cup B) = A \cap U = A$.

Question 14

For all sets $A$, $B$ and $C$, $A – (B – C) = (A – B) – C$.

Solution. Expand both sides: $A – (B – C) = A \cap (B \cap C’)’ = A \cap (B’ \cup C) = (A \cap B’) \cup (A \cap C)$, while $(A – B) – C = A \cap B’ \cap C’$. The left side has the extra piece $A \cap C$. Counterexample: $A = C = \{1\}$, $B = \varnothing$ gives $A – (B – C) = \{1\} – \varnothing = \{1\}$ but $(A – B) – C = \{1\} – \{1\} = \varnothing$.

Question 15

For all sets $A$, $B$ and $C$, if $A \subset B$, then $A \cap C \subset B \cap C$.

Solution. Let $x \in A \cap C$. Then $x \in A \subset B$, so $x \in B$, and also $x \in C$. Hence $x \in B \cap C$.

Question 16

For all sets $A$, $B$ and $C$, if $A \subset B$, then $A \cup C \subset B \cup C$.

Solution. Let $x \in A \cup C$. If $x \in A$ then $x \in B$ because $A \subset B$; if $x \in C$ it is already in $C$. Either way $x \in B \cup C$.

Question 17

For all sets $A$, $B$ and $C$, if $A \subset C$ and $B \subset C$, then $A \cup B \subset C$.

Solution. Let $x \in A \cup B$. Then either $x \in A$, so $x \in C$; or $x \in B$, so $x \in C$. In both cases $x \in C$.

Using properties of sets, prove the statements given in Exercises 18 to 22.

Question 18

For all sets $A$ and $B$, $A \cup (B – A) = A \cup B$.

Solution. Write $B – A = B \cap A’$ and distribute: $$A \cup (B \cap A’) = (A \cup B) \cap (A \cup A’) = (A \cup B) \cap U = A \cup B.$$

Question 19

For all sets $A$ and $B$, $A – (A – B) = A \cap B$.

Solution. $A – B = A \cap B’$, so $A – (A \cap B’) = A \cap (A \cap B’)’$. De Morgan gives $(A \cap B’)’ = A’ \cup B$, so the set is $A \cap (A’ \cup B) = (A \cap A’) \cup (A \cap B) = \varnothing \cup (A \cap B) = A \cap B$.

Question 20

For all sets $A$ and $B$, $A – (A \cap B) = A – B$.

Solution. $A – (A \cap B) = A \cap (A \cap B)’ = A \cap (A’ \cup B’) = (A \cap A’) \cup (A \cap B’) = \varnothing \cup (A \cap B’) = A \cap B’ = A – B$.

Question 21

For all sets $A$ and $B$, $(A \cup B) – B = A – B$.

Solution. $(A \cup B) – B = (A \cup B) \cap B’ = (A \cap B’) \cup (B \cap B’) = (A \cap B’) \cup \varnothing = A \cap B’ = A – B$.

Question 22

Let $T = \left\{x \,\middle|\, \dfrac{x + 5}{x – 7} – 5 = \dfrac{4x – 40}{13 – x}\right\}$. Is $T$ an empty set? Justify your answer.

Solution. The equation needs $x \ne 7$ and $x \ne 13$. Combine the left side over $x – 7$: $$\frac{x + 5 – 5(x – 7)}{x – 7} = \frac{40 – 4x}{x – 7} = \frac{-4(x – 10)}{x – 7}.$$ The right side is $\dfrac{4(x – 10)}{13 – x}$. Bring everything to one side and factor out $4(x – 10)$: $$4(x – 10)\left[\frac{-1}{x – 7} – \frac{1}{13 – x}\right] = 0.$$ So either $x = 10$, or the bracket is zero. The bracket is zero only if $-(13 – x) = x – 7$, that is $-13 = -7$, which is impossible. Hence the only solution is $x = 10$, which is allowed (it is neither $7$ nor $13$). Check: left side $\frac{15}{3} – 5 = 0$, right side $\frac{0}{3} = 0$.

So $T = \{10\}$, which is not empty. The mistake to avoid is cancelling the factor $x – 10$ from both sides, which would lose this root and wrongly suggest $T = \varnothing$.

Common mistakes

• Questions 1(ii) and 2(i). Dividing $x^2 = x$ by $x$ gives $x = 1$ and loses the root $x = 0$. Always move everything to one side and factor.
• Question 3. Students list only the powers of $2$ and the number $2^p – 1$ and forget the products $2^k(2^p – 1)$; a one-case check with $p = 2$ exposes the gap.
• Question 4(ii) and (iv). The factor sum of $2^7$ is $2^8 – 1 = 255$, not $256$, which is why $128$ fails while $496$ passes. Reading “$\notin$” as “$\in$” in (iv) is the other common slip, and the printed key itself makes it.
• Questions 12 and 14. Treating $A – (B – C)$ as if it equalled $(A – B) – C$, as though the difference were associative. It is not; the counterexample $A = C = \{1\}$, $B = \varnothing$ settles it.
• Questions 18 to 21. Forgetting $A \cap A’ = \varnothing$ and $A \cup A’ = U$ when a proof stalls, or applying De Morgan without switching $\cap$ to $\cup$.
• Question 22. Cancelling $x – 10$ on both sides and concluding that the equation has no solution. Cancelling a factor that can be zero throws away a root.

Practise next

• The other pages of this unit: Long Answer and Objective Type, Q23–43 and Fill in the Blanks and True or False, Q44–58.
• NCERT textbook practice on the same ideas: Exercise 1.3 (union, intersection, difference), Exercise 1.4 (complements and laws of sets) and the Miscellaneous Exercise.