Examples (Revised) - Chapter 1 -Sets - Ncert Solutions class 11 - Maths

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Chapter 1 - Sets | NCERT Solutions for Class 11 Maths

Example 1

Write the solution set of the equation $x^2+x-2=0$ in roster form.
Solution

The given equation can be written as
$
(x-1)(x+2)=0 \text {, i. e., } \quad x=1,-2
$

Therefore, the solution set of the given equation can be written in roster form as $\{1,-2\}$. Example 2 Write the set $\left\{x: x\right.$ is a positive integer and $\left.x^2<40\right\}$ in the roster form.

Solution

The required numbers are $1,2,3,4,5,6$. So, the given set in the roster form is $\{1,2,3,4,5,6\}$.

Example 3

Write the set $\mathrm{A}=\{1,4,9,16,25, \ldots\}$ in set-builder form.
Solution

We may write the set $\mathrm{A}$ as
$
\mathrm{A}=\{x: x \text { is the square of a natural number }\}
$

Alternatively, we can write
$
\mathrm{A}=\left\{x: x=n^2 \text {, where } n \in \mathbf{N}\right\}
$

Example 4

Write the set $\left\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}\right\}$ in the set-builder form.
Solution

We see that each member in the given set has the numerator one less than the denominator. Also, the numerator begin from 1 and do not exceed 6. Hence, in the set-builder form the given set is
$
\left\{x: x=\frac{n}{n+1} \text {, where } n \text { is a natural number and } 1 \leq n \leq 6\right\}
$

Example 5

Match each of the set on the left described in the roster form with the same set on the right described in the set-builder form :
(i) $\{\mathrm{P}, \mathrm{R}, \mathrm{I}, \mathrm{N}, \mathrm{C}, \mathrm{A}, \mathrm{L}\}$
(a) $\{x: x$ is a positive integer and is a divisor of 18$\}$
(ii) $\{0\}$
(b) $\left\{x: x\right.$ is an integer and $\left.x^2-9=0\right\}$
(iii) $\{1,2,3,6,9,18\}$
(c) $\{x: x$ is an integer and $x+1=1\}$
(iv) $\{3,-3\}$
(d) $\{x: x$ is a letter of the word PRINCIPAL $\}$

Solution

Since in (d), there are 9 letters in the word PRINCIPAL and two letters P and I are repeated, so (i) matches (d). Similarly, (ii) matches (c) as $x+1=1$ implies $x=0$. Also, $1,2,3,6,9,18$ are all divisors of 18 and so (iii) matches (a). Finally, $x^2-9=0$ implies $x=3,-3$ and so (iv) matches (b).

Example 6

State which of the following sets are finite or infinite :
(i) $\{x: x \in \mathrm{N}$ and $(x-1)(x-2)=0\}$
(ii) $\left\{x: x \in \mathrm{N}\right.$ and $\left.x^2=4\right\}$
(iii) $\{x: x \in \mathrm{N}$ and $2 x-1=0\}$
(iv) $\{x: x \in \mathrm{N}$ and $x$ is prime $\}$
(v) $\{x: x \in \mathrm{N}$ and $x$ is odd $\}$

Solution

(i) Given set $=\{1,2\}$. Hence, it is finite.
(ii) Given set $=\{2\}$. Hence, it is finite.
(iii) Given set $=\phi$. Hence, it is finite.
(iv) The given set is the set of all prime numbers and since set of prime numbers is infinite. Hence the given set is infinite
(v) Since there are infinite number of odd numbers, hence, the given set is infinite.

Example 7

Find the pairs of equal sets, if any, give reasons:
$
\begin{array}{ll}
\mathrm{A}=\{0\}, & \mathrm{B}=\{x: x>15 \text { and } x<5\}, \\
\mathrm{C}=\{x: x-5=0\}, & \mathrm{D}=\left\{x: x^2=25\right\}, \\
\mathrm{E}=\left\{x: x \text { is an integral positive root of the equation } x^2-2 x-15=0\right\} .
\end{array}
$

Solution

Since $0 \in \mathrm{A}$ and 0 does not belong to any of the sets B, C, D and E, it follows that, $\mathrm{A} \neq \mathrm{B}, \mathrm{A} \neq \mathrm{C}, \mathrm{A} \neq \mathrm{D}, \mathrm{A} \neq \mathrm{E}$.

Since $\mathrm{B}=\phi$ but none of the other sets are empty. Therefore $\mathrm{B} \neq \mathrm{C}, \mathrm{B} \neq \mathrm{D}$ and $\mathrm{B} \neq \mathrm{E}$. Also $\mathrm{C}=\{5\}$ but $-5 \in \mathrm{D}$, hence $\mathrm{C} \neq \mathrm{D}$.
Since $E=\{5\}, C=E$. Further, $D=\{-5,5\}$ and $E=\{5\}$, we find that, $D \neq E$.

Thus, the only pair of equal sets is $\mathrm{C}$ and $\mathrm{E}$.
Example 8

Which of the following pairs of sets are equal? Justify your answer.
(i) $\mathrm{X}$, the set of letters in "ALLOY" and B, the set of letters in "LOYAL".
(ii) $\mathrm{A}=\left\{n: n \in \mathrm{Z}\right.$ and $\left.n^2 \leq 4\right\}$ and $\mathrm{B}=\left\{x: x \in \mathrm{R}\right.$ and $\left.x^2-3 x+2=0\right\}$.

Solution

(i) We have, $\mathrm{X}=\{\mathrm{A}, \mathrm{L}, \mathrm{L}, \mathrm{O}, \mathrm{Y}\}, \mathrm{B}=\{\mathrm{L}, \mathrm{O}, \mathrm{Y}, \mathrm{A}, \mathrm{L}\}$. Then $\mathrm{X}$ and $\mathrm{B}$ are equal sets as repetition of elements in a set do not change a set. Thus,
$
\mathrm{X}=\{\mathrm{A}, \mathrm{L}, \mathrm{O}, \mathrm{Y}\}=\mathrm{B}
$
(ii) $A=\{-2,-1,0,1,2\}, B=\{1,2\}$. Since $0 \in A$ and $0 \notin B, A$ and $B$ are not equal sets.

Example 9

Consider the sets
$
\phi, \mathrm{A}=\{1,3\}, \mathrm{B}=\{1,5,9\}, \mathrm{C}=\{1,3,5,7,9\} .
$

Insert the symbol $\subset$ or $\not \subset$ between each of the following pair of sets:
(i) $\phi \ldots$ B
(ii) A . . B
(iii) A . . C
(iv) $\mathrm{B}$. . . C

Solution

(i) $\phi \subset \mathrm{B}$ as $\phi$ is a subset of every set.
(ii) $\mathrm{A} \not \subset \mathrm{B}$ as $3 \in \mathrm{A}$ and $3 \notin \mathrm{B}$
(iii) $\mathrm{A} \subset \mathrm{C}$ as $1,3 \in \mathrm{A}$ also belongs to $\mathrm{C}$
(iv) $\mathrm{B} \subset \mathrm{C}$ as each element of $\mathrm{B}$ is also an element of $\mathrm{C}$.

Example 10

Let $\mathrm{A}=\{a, e, i, o, u\}$ and $\mathrm{B}=\{a, b, c, d\}$. Is A a subset of $\mathrm{B}$ ? No. (Why?). Is B a subset of A? No. (Why?)
Example 11

Let $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ be three sets. If $\mathrm{A} \in \mathrm{B}$ and $\mathrm{B} \subset \mathrm{C}$, is it true that $\mathrm{A} \subset \mathrm{C}$ ? . If not, give an example.
Solution

No. Let $\mathrm{A}=\{1\}, \mathrm{B}=\{\{1\}, 2\}$ and $\mathrm{C}=\{\{1\}, 2,3\}$. Here $\mathrm{A} \in \mathrm{B}$ as $\mathrm{A}=\{1\}$ and $\mathrm{B} \subset \mathrm{C}$. But $\mathrm{A} \not \subset \mathrm{C}$ as $1 \in \mathrm{A}$ and $1 \notin \mathrm{C}$.
Note that an element of a set can never be a subset of itself.

Example 12

Let $\mathrm{A}=\{2,4,6,8\}$ and $\mathrm{B}=\{6,8,10,12\}$. Find $\mathrm{A} \cup \mathrm{B}$.
Solution

We have $\mathrm{A} \cup \mathrm{B}=\{2,4,6,8,10,12\}$
Note that the common elements 6 and 8 have been taken only once while writing $\mathrm{A} \cup \mathrm{B}$.
Example 13

Let $\mathrm{A}=\{a, e, i, o, u\}$ and $\mathrm{B}=\{a, i, u\}$. Show that $\mathrm{A} \cup \mathrm{B}=\mathrm{A}$
Solution

We have, $\mathrm{A} \cup \mathrm{B}=\{a, e, i, o, u\}=\mathrm{A}$.
This example illustrates that union of sets $\mathrm{A}$ and its subset $\mathrm{B}$ is the set $\mathrm{A}$ itself, i.e., if $\mathrm{B} \subset \mathrm{A}$, then $\mathrm{A} \cup \mathrm{B}=\mathrm{A}$.

Example 14

Let $\mathrm{X}=\{$ Ram, Geeta, Akbar $\}$ be the set of students of Class XI, who are in school hockey team. Let $\mathrm{Y}=\{$ Geeta, David, Ashok $\}$ be the set of students from Class XI who are in the school football team. Find $\mathrm{X} \cup \mathrm{Y}$ and interpret the set.
Solution

We have, $\mathrm{X} \cup \mathrm{Y}=\{$ Ram, Geeta, Akbar, David, Ashok $\}$. This is the set of students from Class XI who are in the hockey team or the football team or both.

Example 15

Consider the sets $\mathrm{A}$ and $\mathrm{B}$ of Example 12. Find $\mathrm{A} \cap \mathrm{B}$.
Solution

We see that 6,8 are the only elements which are common to both $A$ and $B$. Hence $\mathrm{A} \cap \mathrm{B}=\{6,8\}$.
Example 16

Consider the sets $\mathrm{X}$ and $\mathrm{Y}$ of Example 14. Find $\mathrm{X} \cap \mathrm{Y}$.
Solution

We see that element 'Geeta' is the only element common to both. Hence, $\mathrm{X} \cap \mathrm{Y}=\{$ Geeta $\}$.
Example 17

Let $\mathrm{A}=\{1,2,3,4,5,6,7,8,9,10\}$ and $\mathrm{B}=\{2,3,5,7\}$. Find $\mathrm{A} \cap \mathrm{B}$ and hence show that $\mathrm{A} \cap \mathrm{B}=\mathrm{B}$.
Solution

We have $\mathrm{A} \cap \mathrm{B}=\{2,3,5,7\}=\mathrm{B}$. We note that $\mathrm{B} \subset \mathrm{A}$ and that $\mathrm{A} \cap \mathrm{B}=\mathrm{B}$.

Example 18

Let $\mathrm{A}=\{1,2,3,4,5,6\}, \mathrm{B}=\{2,4,6,8\}$. Find $\mathrm{A}-\mathrm{B}$ and $\mathrm{B}-\mathrm{A}$.
Solution

We have, $\mathrm{A}-\mathrm{B}=\{1,3,5\}$, since the elements $1,3,5$ belong to $\mathrm{A}$ but not to $\mathrm{B}$ and $\mathrm{B}-\mathrm{A}=\{8\}$, since the element 8 belongs to $\mathrm{B}$ and not to $\mathrm{A}$.
We note that $\mathrm{A}-\mathrm{B} \neq \mathrm{B}-\mathrm{A}$.

Example 19

Let $\mathrm{V}=\{a, e, i, o, u\}$ and $\mathrm{B}=\{a, i, k, u\}$. Find $\mathrm{V}-\mathrm{B}$ and $\mathrm{B}-\mathrm{V}$

Solution

We have, $\mathrm{V}-\mathrm{B}=\{e, o\}$, since the elements $e$, $o$ belong to $\mathrm{V}$ but not to $\mathrm{B}$ and $\mathrm{B}-\mathrm{V}=\{k\}$, since the element $k$ belongs to $\mathrm{B}$ but not to $\mathrm{V}$.

We note that $\mathrm{V}-\mathrm{B} \neq \mathrm{B}-\mathrm{V}$. Using the setbuilder notation, we can rewrite the definition of difference as

$
\mathrm{A}-\mathrm{B}=\{x: x \in \mathrm{A} \text { and } x \notin \mathrm{B}\}
$

The difference of two sets $\mathrm{A}$ and $\mathrm{B}$ can be represented by Venn diagram as shown in Fig 1.8.

The shaded portion represents the difference of the two sets A and B.

Remark

The sets $\mathrm{A}-\mathrm{B}, \mathrm{A} \cap \mathrm{B}$ and $\mathrm{B}-\mathrm{A}$ are mutually disjoint sets, i.e., the intersection of any of these two sets is the null set as shown in Fig 1.9.

Example 20

Let $\mathrm{U}=\{1,2,3,4,5,6,7,8,9,10\}$ and $\mathrm{A}=\{1,3,5,7,9\}$. Find $\mathrm{A}^{\prime}$.
Solution

We note that $2,4,6,8,10$ are the only elements of $\mathrm{U}$ which do not belong to
A. Hence
$
\mathrm{A}^{\prime}=\{2,4,6,8,10\} \text {. }
$

Example 21

Let U be universal set of all the students of Class XI of a coeducational school and $\mathrm{A}$ be the set of all girls in Class XI. Find $\mathrm{A}^{\prime}$.
Solution

Since $\mathrm{A}$ is the set of all girls, $\mathrm{A}^{\prime}$ is clearly the set of all boys in the class.

If $\mathrm{A}$ is a subset of the universal set $\mathrm{U}$, then its complement $\mathrm{A}^{\prime}$ is also a subset of $U$.
Again in Example 20 above, we have $\mathrm{A}^{\prime}=\{2,4,6,8,10\}$
Hence
$
\begin{aligned}
\left(\mathrm{A}^{\prime}\right)^{\prime} & =\left\{x: x \in \mathrm{U} \text { and } x \notin \mathrm{A}^{\prime}\right\} \\
& =\{1,3,5,7,9\}=\mathrm{A}
\end{aligned}
$

It is clear from the definition of the complement that for any subset of the universal set $\mathrm{U}$, we have
$
\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}
$

Now, we want to find the results for $(\mathrm{A} \cup \mathrm{B})^{\prime}$ and $\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}$ in the followng example.
Example 22

Let $\mathrm{U}=\{1,2,3,4,5,6\}, \mathrm{A}=\{2,3\}$ and $\mathrm{B}=\{3,4,5\}$.
Find $\mathrm{A}^{\prime}, \mathrm{B}^{\prime}, \mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}, \mathrm{A} \cup \mathrm{B}$ and hence show that $(\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}$.
Solution

Clearly A' $=\{1,4,5,6\}, \mathrm{B}^{\prime}=\{1,2,6\}$. Hence $\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}=\{1,6\}$
Also $A \cup B=\{2,3,4,5\}$, so that $(A \cup B)^{\prime}=\{1,6\}$
$
(\mathrm{A} \cup \mathrm{B})^{\prime}=\{1,6\}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}
$

It can be shown that the above result is true in general. If $\mathrm{A}$ and $\mathrm{B}$ are any two subsets of the universal set $\mathrm{U}$, then
$(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$. Similarly, $(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}$. These two results are stated in words as follows :

The complement of the union of two sets is the intersection of their complements and the
$\mathrm{U}$
complement of the intersection of two sets is the union of their complements. These are called De Morgan's laws. These are named after the mathematician De Morgan.
The complement $\mathrm{A}^{\prime}$ of a set $\mathrm{A}$ can be represented by a Venn diagram as shown in Fig 1.10.
The shaded portion represents the complement of the set $\mathrm{A}$.