Class 11 Maths Chapter 1 – Sets Miscellaneous Exercise

Miscellaneous Exercise

Q1. Decide, among the following sets, which sets are subsets of one and another:

$A=\{x:x\in R$ and $x$ $\text{satisfy }x^2-8x+12=0\}$ , $B=\{2,4,6\}$ , $C=\{2,4,6,8,\ldots\}$ , $D=\{6\}$ .

Sol. Given,

A=\{x:x\in\mathbb R,\ x^2-8x+12=0\}

Solving $x^2-8x+12=0$ ,

\begin{aligned} & x^2-2x-6x+12 =0 \\ & x(x-2)-6(x-2) =0 \\ & (x-2)(x-6) =0 \\ & x =2\text{ or }x=6 \end{aligned}

∴ $A=\{2,6\}$

The other sets are

\begin{aligned} & B =\{2,4,6\} \\ & C =\{2,4,6,8,\ldots\} \\ & D =\{6\} \end{aligned}

Comparing the elements of these sets,

D\subset A\subset B\subset C

∴ The required subset relations are

\begin{gathered} A\subset B,\quad A\subset C,\quad B\subset C\\ D\subset A,\quad D\subset B,\quad D\subset C \end{gathered}

Q2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

(i) If $x\in A$ and $A\in B$ , then $x\in B$

Sol. The given statement is false.

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

Here, $1\in A$ and $A\in B$ , but $1\notin B$ (the elements of $B$ are $\{1\}$ and $2$ ).

(ii) If $A\subset B$ and $B\in C$ , then $A\in C$

Sol. The given statement is false.

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

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

(iii) If $A\subset B$ and $B\subset C$ , then $A\subset C$

Sol. The given statement is true.

Proof: Let $x$ be any element of $A$ .

Since $A\subset B$ ,

x\in A\Rightarrow x\in B

Since $B\subset C$ ,

x\in B\Rightarrow x\in C

So, every element of $A$ is an element of $C$ .

∴ $A\subset C$

Hence proved.

(iv) If $A\not\subset B$ and $B\not\subset C$ , then $A\not\subset C$

Sol. The given statement is false.

For example, let $A=\{1,2\}$ , $B=\{2,3\}$ and $C=\{1,2,5\}$ .

Here, $A\not\subset B$ (as $1\in A$ and $1\notin B$ ) and $B\not\subset C$ (as $3\in B$ and $3\notin C$ ), but $A\subset C$ .

(v) If $x\in A$ and $A\not\subset B$ , then $x\in B$

Sol. The given statement is false.

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

Here, $1\in A$ and $A\not\subset B$ , but $1\notin B$ .

(vi) If $A\subset B$ and $x\notin B$ , then $x\notin A$

Sol. The given statement is true.

Proof: Given, $A\subset B$ and $x\notin B$ .

If possible, let $x\in A$ . Then $x\in B$ , as $A\subset B$ .

This contradicts $x\notin B$ . So our supposition is wrong.

∴ $x\notin A$

Hence proved.

Q3. Let $A$ , $B$ , and $C$ be the sets such that $A\cup B=A\cup C$ and $A\cap B=A\cap C$ . Show that $B=C$ .

Sol. Given,

A\cup B=A\cup C\qquad ...(1)

A\cap B=A\cap C\qquad ...(2)

To prove: $B=C$

Taking intersection with $B$ on both sides of (1),

\begin{aligned} & (A\cup B)\cap B=(A\cup C)\cap B\\ & B=(A\cap B)\cup(C\cap B)\\ & B=(A\cap B)\cup(B\cap C)\qquad ...(3) \end{aligned}

Taking intersection with $C$ on both sides of (1),

\begin{aligned} & (A\cup B)\cap C=(A\cup C)\cap C\\ & (A\cap C)\cup(B\cap C)=C \end{aligned}

(A\cap B)\cup(B\cap C)=C\qquad ...(4)

[Using (2)]

From (3) and (4),

\therefore B=C

Hence proved.

Q4. Show that the following four conditions are equivalent:

(i) $A\subset B$

(ii) $A-B=\phi$

(iii) $A\cup B=B$

(iv) $A\cap B=A$

Sol. We show that (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (iv) $\Rightarrow$ (i). Then each condition implies every other, so the four conditions are equivalent.

(i) $\Rightarrow$ (ii): Let $A\subset B$ . Then every element of $A$ is in $B$ , so no element of $A$ lies outside $B$ .

∴ $A-B=\phi$

(ii) $\Rightarrow$ (iii): Let $A-B=\phi$ . Then there is no element which is in $A$ but not in $B$ , i.e., every element of $A$ is in $B$ , so $A\subset B$ .

Hence, every element of $A\cup B$ is in $B$ , i.e., $A\cup B\subset B$ . Also, $B\subset A\cup B$ .

∴ $A\cup B=B$

(iii) $\Rightarrow$ (iv): Let $A\cup B=B$ . Since $A\subset A\cup B$ , we get $A\subset B$ .

So, every element of $A$ is in both $A$ and $B$ , i.e., $A\subset A\cap B$ . Also, $A\cap B\subset A$ .

∴ $A\cap B=A$

(iv) $\Rightarrow$ (i): Let $A\cap B=A$ . Since $A\cap B\subset B$ , we get $A\subset B$ .

Hence, the four conditions are equivalent.

Q5. Show that if $A\subset B$ , then $C-B\subset C-A$ .

Sol. Given, $A\subset B$ .

To show: $C-B\subset C-A$

Let $x\in C-B$ .

\begin{aligned} &\Rightarrow x\in C\text{ and }x\notin B\\ &\Rightarrow x\in C\text{ and }x\notin A\quad[A\subset B]\\ &\Rightarrow x\in C-A \end{aligned}

∴ $C-B\subset C-A$

Q6. Show that for any sets $A$ and $B$ ,

$A=(A\cap B)\cup(A-B)$ and $A\cup(B-A)=(A\cup B)$

Sol. Let

A=\{1,2,3\},\quad B=\{2,3,4,5\}

(i) $A=(A\cap B)\cup(A-B)$

\begin{aligned} \text{RHS}&=(A\cap B)\cup(A-B)\\ &=\{2,3\}\cup\{1\}\\ &=\{1,2,3\}\\ &=A=\text{LHS} \end{aligned}

∴ LHS = RHS

(ii) $A\cup(B-A)=A\cup B$

\begin{aligned} \text{LHS}&=A\cup(B-A)\\ &=\{1,2,3\}\cup\{4,5\}\\ &=\{1,2,3,4,5\} \end{aligned}

\begin{aligned} \text{RHS}&=A\cup B\\ &=\{1,2,3\}\cup\{2,3,4,5\}\\ &=\{1,2,3,4,5\} \end{aligned}

∴ LHS = RHS

Alternative Sol. (Method 2)

We know that $A-B=A\cap B'$ , as both are the set of elements which are in $A$ and not in $B$ .

First result:

\begin{aligned} & \text{RHS} =(A\cap B)\cup(A-B) \\ & =(A\cap B)\cup(A\cap B') \end{aligned}

=A\cap(B\cup B')

[Using distributive law]

=A\cap U

[Using complement law]

\begin{aligned} & =A \\ & =\text{LHS} \end{aligned}

Second result:

\begin{aligned} & \text{LHS} =A\cup(B-A) \\ & =A\cup(B\cap A') \end{aligned}

=(A\cup B)\cap(A\cup A')

[Using distributive law]

=(A\cup B)\cap U

[Using complement law]

\begin{aligned} & =A\cup B \\ & =\text{RHS} \end{aligned}

Hence proved.

Q7. Using properties of sets, show that

(i) $A\cup(A\cap B)=A$

(ii) $A\cap(A\cup B)=A$ .

(i) Show that $A\cup(A\cap B)=A$

Sol. Using distributive law,

\begin{aligned} A\cup(A\cap B) &=(A\cup A)\cap(A\cup B)\\ &=A\cap(A\cup B)\\ &=A\qquad[A\subseteq A\cup B] \end{aligned}

∴ $A\cup(A\cap B)=A$

(ii) Show that $A\cap(A\cup B)=A$

Sol. Using distributive law,

\begin{aligned} A\cap(A\cup B) &=(A\cap A)\cup(A\cap B)\\ &=A\cup(A\cap B)\\ &=A\qquad\text{[From part (i)]} \end{aligned}

∴ $A\cap(A\cup B)=A$

Q8. Show that $A\cap B=A\cap C$ need not imply $B=C$ .

Sol. It is enough to give one example in which $A\cap B=A\cap C$ but $B\ne C$ .

Let $A=\{0,1\}$ , $B=\{0,2,3\}$ and $C=\{0,4,5\}$ .

$A\cap B=\{0,1\}\cap\{0,2,3\}=\{0\}$

$A\cap C=\{0,1\}\cap\{0,4,5\}=\{0\}$

So, $A\cap B=A\cap C$ , but $B\ne C$ (as $2\in B$ and $2\notin C$ ).

∴ $A\cap B=A\cap C$ need not imply $B=C$ .

Q9. Let $A$ and $B$ be sets. If $A\cap X=B\cap X=\phi$ and $A\cup X=B\cup X$ for some set $X$ , show that $A=B$ .

(Hints $A=A\cap(A\cup X)$ , $B=B\cap(B\cup X)$ and use Distributive law )

Sol. Given,

A\cap X=B\cap X=\phi\qquad ...(1)

A\cup X=B\cup X\qquad ...(2)

To prove: $A=B$

Taking intersection with $A$ on both sides of (2),

A\cap(A\cup X)=A\cap(B\cup X)

Using distributive law,

(A\cap A)\cup(A\cap X)=(A\cap B)\cup(A\cap X)

Using (1),

\begin{aligned} A\cup\phi&=(A\cap B)\cup\phi\\ A&=A\cap B\qquad ...(3) \end{aligned}

Taking intersection with $B$ on both sides of (2),

B\cap(A\cup X)=B\cap(B\cup X)

Using distributive law,

(B\cap A)\cup(B\cap X)=(B\cap B)\cup(B\cap X)

Using (1),

\begin{aligned} (B\cap A)\cup\phi&=B\cup\phi\\ B\cap A&=B\qquad ...(4) \end{aligned}

From (3) and (4),

\therefore A=B

Hence proved.

Q10. Find sets $A$ , $B$ and $C$ such that $A\cap B$ , $B\cap C$ and $A\cap C$ are non-empty sets and $A\cap B\cap C=\phi$ .

Sol. Let

$A=\{0,1\},\ B=\{1,2\},\ C=\{2,0\}$

Then,

\begin{aligned} A\cap B&=\{1\}\\ B\cap C&=\{2\}\\ A\cap C&=\{0\} \end{aligned}

∴ All three intersections are non-empty.

But,

A\cap B\cap C=\phi