Class 11 Maths Chapter 1 – Sets Examples

Examples

Eg 1. Write the solution set of the equation $x^2+x-2=0$ in roster form.

Sol. Factorising the given equation,

\begin{aligned} & x^2+x-2 =0 \\ & x^2+2x-x-2 =0 \\ & (x-1)(x+2) =0 \\ & x =1\text{ or }x=-2 \end{aligned}

∴ The solution set in roster form is $\{1,-2\}$ .

Eg 2. Write the set $\{x:x$ is a positive integer $\text{and }x^2<40\}$ in the roster form.

Sol. Squares of positive integers: $1^2=1$ , $2^2=4$ , $3^2=9$ , $4^2=16$ , $5^2=25$ , $6^2=36$ , $7^2=49$ .

So, $x^2<40$ holds for $x=1,2,3,4,5,6$ and fails from $x=7$ onwards.

∴ The given set in roster form is $\{1,2,3,4,5,6\}$ .

Eg 3. Write the set $A=\{1,4,9,16,25,\ldots\}$ in set-builder form.

Sol. Here, $1=1^2,\ 4=2^2,\ 9=3^2,\ 16=4^2,\ 25=5^2,\ \ldots$

So, every element of $A$ is the square of a natural number.

∴ $A\allowbreak=\{x:x$ is the square of a natural $\text{number}\}$ , or $A=\{x:x=n^2,\text{ where }n\in N\}$ .

Eg 4. Write the set $\{\frac12,\frac23,\frac34,\frac45,\frac56,\frac67\}$ in the set-builder form.

Sol. In each element the numerator is one less than the denominator, and the numerators are $1,2,3,4,5,6$ .

So, each element is of the form $\frac{n}{n+1}$ , where $n$ is a natural number and $1\le n\le6$ .

∴

\{\frac12,\frac23,\frac34,\frac45,\frac56,\frac67\}

$=\{x:x=\frac{n}{n+1},$ where $n$ is a natural number $\text{and }1\le n\le6\}$

Eg 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) $\{P,R,I,N,C,A,L\}$

(ii) $\{0\}$

(iii) $\{1,2,3,6,9,18\}$

(iv) $\{3,-3\}$

(a) $\{x:x$ is a positive integer and is a divisor of $\text{18}\}$

(b) $\{x:x\text{ is an integer and }x^2-9=0\}$

(c) $\{x:x\text{ is an integer and }x+1=1\}$

(d) $\{x:x$ is a letter of the word $\text{PRINCIPAL}\}$

Sol. Writing each set on the right in roster form:

Set (a): the positive divisors of $18$ are $1,2,3,6,9,18$ , so it is $\{1,2,3,6,9,18\}$ .

Set (b): $x^2-9=0\Rightarrow x=3$ or $x=-3$ , so it is $\{3,-3\}$ .

Set (c): $x+1=1\Rightarrow x=0$ , so it is $\{0\}$ .

Set (d): PRINCIPAL has $9$ letters, but P and I are repeated; the distinct letters are P, R, I, N, C, A, L, so it is $\{P,R,I,N,C,A,L\}$ .

(i) $\{P,R,I,N,C,A,L\}$ is the set in (d).

∴ (i) matches (d).

(ii) $\{0\}$ is the set in (c).

∴ (ii) matches (c).

(iii) $\{1,2,3,6,9,18\}$ is the set in (a).

∴ (iii) matches (a).

(iv) $\{3,-3\}$ is the set in (b).

∴ (iv) matches (b).

Eg 6. State which of the following sets are finite or infinite:

(i) $\{x:x\in N\text{ and }(x-1)(x-2)=0\}$

Sol. $(x-1)(x-2)=0\Rightarrow x=1$ or $x=2$ , and both are natural numbers.

So, the given set $=\{1,2\}$ .

∴ It is finite.

(ii) $\{x:x\in N\text{ and }x^2=4\}$

Sol. $x^2=4\Rightarrow x=2$ or $x=-2$ , and only $2\in N$ .

So, the given set $=\{2\}$ .

∴ It is finite.

(iii) $\{x:x\in N\text{ and }2x-1=0\}$

Sol. $2x-1=0\Rightarrow x=\frac12$ , which is not a natural number.

So, the given set $=\phi$ .

∴ It is finite.

(iv) $\{x:x\in N\text{ and }x\text{ is prime}\}$

Sol. The given set is the set of all prime numbers, and there are infinitely many prime numbers.

∴ It is infinite.

(v) $\{x:x\in N\text{ and }x\text{ is odd}\}$

Sol. The given set is the set of all odd natural numbers $\{1,3,5,\ldots\}$ , which never ends.

∴ It is infinite.

Eg 7. Find the pairs of equal sets, if any, give reasons:

$A=\{0\}$ , $B=\{x:x>15\text{ and }x<5\}$ ,

$C=\{x:x-5=0\}$ , $D=\{x:x^2=25\}$ ,

$E=\{x:x$ is an integral positive root of the $\text{equation }x^2-2x-15=0\}$ .

Sol. Given, $A=\{0\}$ .

No number can be greater than $15$ and less than $5$ simultaneously.

∴ $B=\phi$

For $C$ , $x-5=0\Rightarrow x=5$ .

∴ $C=\{5\}$

For $D$ , $x^2=25\Rightarrow x=\pm5$ .

∴ $D=\{-5,5\}$

For $E$ , solving $x^2-2x-15=0$ ,

\begin{aligned} & x^2-5x+3x-15 =0 \\ & x(x-5)+3(x-5) =0 \\ & (x-5)(x+3) =0 \\ & x =5\text{ or }x=-3 \end{aligned}

Since only the positive integral root is $5$ ,

E=\{5\}

Clearly, $C=E=\{5\}$ , while the other sets have different elements.

∴ The only pair of equal sets is $\boldsymbol{C}$ and $\boldsymbol{E}$ .

Eg 8. Which of the following pairs of sets are equal? Justify your answer.

(i) $X$ , the set of letters in “ALLOY” and $B$ , the set of letters in “LOYAL”.

Sol. $X=\{A,L,L,O,Y\}=\{A,L,O,Y\}$ and $B=\{L,O,Y,A,L\}=\{A,L,O,Y\}$ , as repeating an element does not change a set.

∴ $X=B$ , i.e., the sets are equal.

(ii) $A=\{n:n\in Z\text{ and }n^2\le4\}$ and $B=\{x:x\in R\text{ and }x^2-3x+2=0\}$ .

Sol. $n^2\le4\Rightarrow-2\le n\le2$ , so $A=\{-2,-1,0,1,2\}$ .

Solving $x^2-3x+2=0$ ,

\begin{aligned} & x^2-x-2x+2 =0 \\ & x(x-1)-2(x-1) =0 \\ & (x-1)(x-2) =0 \\ & x =1\text{ or }x=2 \end{aligned}

So, $B=\{1,2\}$ .

Here, $0\in A$ but $0\notin B$ .

∴ $A\ne B$ , i.e., the sets are not equal.

Eg 9. Consider the sets

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

Insert the symbol $\subset$ or $\not\subset$ between each of the following pair of sets:

(i) $\phi\ldots B$

(ii) $A\ldots B$

(iii) $A\ldots C$

(iv) $B\ldots C$

(i) $\phi$ is a subset of every set.

∴ $\phi\subset B$

(ii) $3\in A$ but $3\notin B$ .

∴ $A\not\subset B$

(iii) Both elements $1,3$ of $A$ belong to $C$ .

∴ $A\subset C$

(iv) Each element $1,5,9$ of $B$ belongs to $C$ .

∴ $B\subset C$

Eg 10. Let $A=\{a,e,i,o,u\}$ and $B=\{a,b,c,d\}$ . Is $A$ a subset of $B$ ? No. (Why?). Is $B$ a subset of $A$ ? No. (Why?)

Sol. Given, $A=\{a,e,i,o,u\}$ and $B=\{a,b,c,d\}$ .

Since $e\in A$ but $e\notin B$ ,

∴ $A\not\subset B$

Since $b\in B$ but $b\notin A$ ,

∴ $B\not\subset A$

Eg 11. Let $A$ , $B$ and $C$ be three sets. If $A\in B$ and $B\subset C$ , is it true that $A\subset C$ ?. If not, give an example.

Sol. No, it is not true.

For example, let $A=\{1\}$ , $B=\{\{1\},2\}$ and $C=\{\{1\},2,3\}$ .

Here, $A\in B$ , as $A=\{1\}$ is an element of $B$ ; and $B\subset C$ , as $\{1\}$ and $2$ are elements of $C$ .

But $1\in A$ and $1\notin C$ (the elements of $C$ are $\{1\}$ , $2$ and $3$ ).

∴ $A\not\subset C$

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

Sol.

A\cup B=\{2,4,6,8,10,12\}

Eg 13. Let $A=\{a,e,i,o,u\}$ and $B=\{a,i,u\}$ . Show that $A\cup B=A$

Sol. Every element of $B$ belongs to $A$ .

∴ $B\subset A$

A\cup B=\{a,e,i,o,u\}=A

Eg 14. Let $X=\{\text{Ram, Geeta, Akbar}\}$ be the set of students of Class XI, who are in school hockey team. Let $Y=\{\text{Geeta, David, Ashok}\}$ be the set of students from Class XI who are in the school football team. Find $X\cup Y$ and interpret the set.

Sol.

X\cup Y=\{\text{Ram, Geeta, Akbar, David, Ashok}\}

Thus, $X\cup Y$ represents the Class XI students who belong to the hockey team or the football team or both.

Eg 15. Consider the sets $A$ and $B$ of Example 12. Find $A\cap B$ .

Sol. From Example 12, $A=\{2,4,6,8\}$ and $B=\{6,8,10,12\}$ .

The elements common to both $A$ and $B$ are $6$ and $8$ .

∴ $A\cap B=\{6,8\}$

Eg 16. Consider the sets $X$ and $Y$ of Example 14. Find $X\cap Y$ .

Sol. From Example 14, $X=\{\text{Ram, Geeta, Akbar}\}$ and $Y=\{\text{Geeta, David, Ashok}\}$ .

Geeta is the only element common to both sets.

∴ $X\cap Y=\{\text{Geeta}\}$

Eg 17. Let $A=\{1,2,3,4,5,6,7,8,9,10\}$ and $B=\{2,3,5,7\}$ . Find $A\cap B$ and hence show that $A\cap B=B$ .

Sol. Every element of $B$ belongs to $A$ .

∴ $B\subset A$

A\cap B=\{2,3,5,7\}=B

Eg 18. Let $A=\{1,2,3,4,5,6\}$ , $B=\{2,4,6,8\}$ . Find $A-B$ and $B-A$ .

Sol. The elements of $A$ which are not in $B$ are $1,3,5$ .

A-B=\{1,3,5\}

The element of $B$ which is not in $A$ is $8$ .

B-A=\{8\}

Hence, $A-B\ne B-A$ .

Eg 19. Let $V=\{a,e,i,o,u\}$ and $B=\{a,i,k,u\}$ . Find $V-B$ and $B-V$

Sol. The elements $e,o$ belong to $V$ but not to $B$ .

V-B=\{a,e,i,o,u\}-\{a,i,k,u\}\allowbreak=\{e,o\}

The element $k$ belongs to $B$ but not to $V$ .

B-V=\{a,i,k,u\}-\{a,e,i,o,u\}\allowbreak=\{k\}

∴ $V-B=\{e,o\}$ and $B-V=\{k\}$ . Note that $V-B\ne B-V$ .

Eg 20. Let $U=\{1,2,3,4,5,6,7,8,9,10\}$ and $A=\{1,3,5,7,9\}$ . Find $A'$ .

Sol. $A'$ is the set of the elements of $U$ which are not in $A$ .

A'=U-A\allowbreak=\{1,2,3,4,5,6,7,8,9,10\}-\{1,3,5,7,9\}\allowbreak=\{2,4,6,8,10\}

Eg 21. Let $U$ be universal set of all the students of Class XI of a coeducational school and $A$ be the set of all girls in Class XI. Find $A'$ .

Sol. $A'$ is the set of those students of Class XI who are not girls. In a coeducational school, such students are boys.

∴ $A'$ is the set of all boys in Class XI.

Eg 22. Let $U=\{1,2,3,4,5,6\}$ , $A=\{2,3\}$ and $B=\{3,4,5\}$ .

Find $A'$ , $B'$ , $A'\cap B'$ , $A\cup B$ and hence show that $(A\cup B)'=A'\cap B'$ .

Sol. Given, $U=\{1,2,3,4,5,6\}$ , $A=\{2,3\}$ and $B=\{3,4,5\}$ .

\begin{aligned} A' &= U-A &&= \{1,2,3,4,5,6\}-\{2,3\} \\ &&&= \{1,4,5,6\} \end{aligned}

\begin{aligned} B' &= U-B &&= \{1,2,3,4,5,6\}-\{3,4,5\} \\ &&&= \{1,2,6\} \end{aligned}

\begin{aligned} A'\cap B' &= \{1,4,5,6\}\cap\{1,2,6\} \\ &= \{1,6\}\qquad...(1) \end{aligned}

\begin{aligned} A\cup B &= \{2,3\}\cup\{3,4,5\} \\ &= \{2,3,4,5\} \end{aligned}

\begin{aligned} (A\cup B)' &= U-(A\cup B) \\ &= \{1,2,3,4,5,6\}-\{2,3,4,5\} \\ &= \{1,6\}\qquad...(2) \end{aligned}

From (1) and (2), $(A\cup B)'=A'\cap B'$ .

Hence proved.

Eg 23. Show that the set of letters needed to spell “CATARACT” and the set of letters needed to spell “TRACT” are equal.

Sol. Let $X$ be the set of letters in “CATARACT” and $Y$ be the set of letters in “TRACT”. A repeated letter is written only once.

$X=\{C,A,T,R\}$

$Y=\{T,R,A,C,T\}=\{T,R,A,C\}$

Every element of $X$ is in $Y$ and every element of $Y$ is in $X$ .

∴ $X=Y$

Hence proved.

Eg 24. List all the subsets of the set $\{-1,0,1\}$ .

Sol. Let $A=\{-1,0,1\}$ .

Since $A$ has $3$ elements, the number of subsets is $2^3=8$ .

∴ The subsets of $A$ are

$\varnothing$ , $\{-1\}$ , $\{0\}$ , $\{1\}$ , $\{-1,0\}$ , $\{-1,1\}$ , $\{0,1\}$ , $\{-1,0,1\}$

Eg 25. Show that $A\cup B=A\cap B$ implies $A=B$

Sol. Given, $A\cup B=A\cap B$ .

Let $a$ be any element of $A$ .

a\in A\Rightarrow a\in A\cup B

Since $A\cup B=A\cap B$ ,

\begin{array}{l} \displaystyle \Rightarrow a\in A\cap B \\[0.4em] \displaystyle \Rightarrow a\in B \end{array}

So, $A\subset B\qquad...(1)$

Let $b$ be any element of $B$ .

b\in B\Rightarrow b\in A\cup B

Since $A\cup B=A\cap B$ ,

\begin{array}{l} \displaystyle \Rightarrow b\in A\cap B \\[0.4em] \displaystyle \Rightarrow b\in A \end{array}

So, $B\subset A\qquad...(2)$

From (1) and (2), $A=B$ .

Hence proved.

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