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

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Chapter 1 - Sets | NCERT Solutions for Class 11 Maths

Miscellaneous Exercise Question 1.

Decide among the following sets, which sets are subsets of each another:
$A=\left\{x: x \in R\right.$ and $x$ satisfies $\left.x^2-8 x+12=0\right\}, B=\{2,4,6\}, C=\{2,4,6,8 \ldots \ldots\},. D=\{6\}$
Ans.wer

Given: $\mathrm{A}=\left\{x: x \in \mathrm{R}\right.$ and $x$ satisfies $\left.x^2-8 x+12=0\right\}$
$=\{x: x \in \mathrm{R}$ and $x$ satisfies $(x-6)(x-2)=0\}=\{2,6\}$
$B=\{2,4,6\}, C=\{2,4,6,8 \ldots \ldots\},. D=\{6\}$
$\therefore \mathrm{A} \subset \mathrm{B}, \mathrm{A} \subset \mathrm{C}, \mathrm{B} \subset \mathrm{C}, \mathrm{D} \subset \mathrm{A}, \mathrm{D} \subset \mathrm{B}$ and $\mathrm{D} \subset \mathrm{C}$
Miscellaneous Exercise Question 2.

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 $x \in \mathrm{A}$ and $\mathrm{A} \in \mathrm{B}$ then $x \in \mathrm{B}$
(ii) If $\mathrm{A} \subset \mathrm{B}$ and $\mathrm{B} \in \mathrm{C}$ then $\mathrm{A} \in \mathrm{C}$
(iii) If $\mathrm{A} \subset \mathrm{B}$ and $\mathrm{B} \subset \mathrm{C}$ then $\mathrm{A} \subset \mathrm{C}$
(iv) If $\mathrm{A} \not \subset \mathrm{B}$ and $\mathrm{B} \not \subset \mathrm{C}$ then $\mathrm{A} \not \subset \mathrm{C}$
(v) If $x \in \mathrm{A}$ and $\mathrm{A} \not \subset \mathrm{B}$ then $x \in \mathrm{B}$
(vi) If $\mathrm{A} \subset \mathrm{B}$ and $x \notin \mathrm{C}$ then $x \notin \mathrm{A}$

Answer.

(i) The statement is false.

Let $\mathrm{A}=\{1\}$ and $\mathrm{B}=\{\{1\}, 2\}$

Then $1 \in \mathrm{A}$ and $\mathrm{A} \in \mathrm{B}$ but $1 \notin \mathrm{B}$

(ii) The statement is false.

Let $\mathrm{A}=\{1\}$ and $\mathrm{B}=\{1,2\}, C\{\{1,2\}, 3\}$
Then $\mathrm{A} \subset \mathrm{B}$ and $\mathrm{B} \in \mathrm{C}$ but $\mathrm{A} \notin \mathrm{C}$
(iii) The statement is true.

Let $x \in \mathrm{A} \Rightarrow x \in \mathrm{B}(\because \mathrm{A} \subset \mathrm{B})$
$\Rightarrow x \in \mathrm{C}(\because \mathrm{B} \subset \mathrm{C})$
$\therefore x \in \mathrm{A} \Rightarrow x \in \mathrm{C} \therefore \mathrm{A} \subset \mathrm{C}$
(iv) The statement is false.

Let $\mathrm{A}=\{1,2\}, \mathrm{B}=\{2,3\}, \mathrm{C}\{1,2,5\}$
Then $\mathrm{A} \not \subset \mathrm{B}$ and $\mathrm{B} \not \subset \mathrm{C}$ but $\mathrm{A} \subset \mathrm{C}$
(v) The statement is false.

Let $\mathrm{A}=\{1,2\}$ and $\mathrm{B}=\{2,3,4,5\}$
Then $1 \in \mathrm{A}$ and $\mathrm{A} \not \subset \mathrm{B}$ but $1 \not \mathrm{B}$
(vi) The statement is true.

Let $x \in \mathrm{A} \Rightarrow x \in \mathrm{B}(\because \mathrm{A} \subset \mathrm{B})$

Now, $x \notin \mathrm{B} \Rightarrow x \notin \mathrm{A}$
Miscellaneous Exercise Question 3.

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

Answer.

Since, $\mathrm{A}=\mathrm{A} \cap(\mathrm{A} \cup \mathrm{B})$ and $\mathrm{A}=\mathrm{A} \cup(\mathrm{A} \cap \mathrm{B})$
Now, it is given that $\mathrm{A} \cup \mathrm{B}=\mathrm{A} \cup \mathrm{C}$ and $\mathrm{A} \cap \mathrm{B}=\mathrm{A} \cap \mathrm{C}$
$
\therefore B=B \cup(B \cap A)
$

$
\begin{aligned}
& =B \cup(A \cap B) \\
& =B \cup(A \cap C) \\
& =(B \cup A) \cap(B \cup C) \\
& =(A \cup B) \cap(B \cup C) \\
& =(A \cup C) \cap(B \cup C) \\
& =(C \cup A) \cap(C \cup B) \\
& =C \cup(A \cap B) \\
& =C \cup(A \cap C) \\
& =C \cup(C \cap A)=C \\
& \therefore B=C
\end{aligned}
$
Miscellaneous Exercise Question 4.

Show that the following four conditions are equivalent:
(i) $\mathrm{A} \subset \mathrm{B}$
(ii) $\mathrm{A}-\mathrm{B}=\phi$
(iii) $\mathrm{A} \cup \mathrm{B}=\mathrm{B}$
(iv) $\mathrm{A} \cap \mathrm{B}=\mathrm{A}$

Answer.

(i) $\Rightarrow$ (ii) $\mathrm{A}-\mathrm{B}=\{x: x \in \mathrm{A}$ and $x \notin \mathrm{B}\}$
Since $\mathrm{A} \subset \mathrm{B}$, Therefore $\mathrm{A}-\mathrm{B}=\phi$
(ii) $\Rightarrow$ (iii) $\mathrm{A}-\mathrm{B}=\phi$
$
\Rightarrow \mathrm{A} \subset \mathrm{B} \Rightarrow \mathrm{A} \cup \mathrm{B}=\mathrm{B}
$
(iii) $\Rightarrow$ (iv) $\mathrm{A} \cup \mathrm{B}=\mathrm{B}$
$
\Rightarrow \mathrm{A} \subset \mathrm{B} \Rightarrow \mathrm{A} \cap \mathrm{B}=\mathrm{A}
$

(iv) $\Rightarrow$ (i) $\mathrm{A} \cap \mathrm{B}=\mathrm{A} \Rightarrow \mathrm{A} \subset \mathrm{B}$

Therefore, (i) $\Leftrightarrow$ (ii) $\Leftrightarrow$ (iii) $\Leftrightarrow$ (iv)
Miscellaneous Exercise Question 5.

Show that if $\mathrm{A} \subset \mathrm{B}$, then $\mathrm{C}-\mathrm{B} \subset \mathrm{C}-\mathrm{A}$.

Answer.

Let $x \in \mathrm{C}-\mathrm{B}$
$
\begin{aligned}
& \Rightarrow x \in \mathrm{C} \text { and } x \notin \mathrm{B} \\
& \Rightarrow x \in \mathrm{C} \text { and } x \notin \mathrm{A}[\because \mathrm{A} \subset \mathrm{B}] \\
& \therefore x \in \mathrm{C}-\mathrm{A} \\
& \Rightarrow \mathrm{C}-\mathrm{B} \subset \mathrm{C}-\mathrm{A}
\end{aligned}
$
Miscellaneous Exercise Question 6.

Assume that $P(A)=P(B)$, show that $A=B$

Answer.

Let $x \in \mathrm{A}$
$
\begin{aligned}
& \Rightarrow\{x\} \in \mathrm{P}(\mathrm{A}) \\
& \Rightarrow\{x\} \in \mathrm{P}(\mathrm{B}) \\
& \Rightarrow x \in \mathrm{B} \\
& \therefore \mathrm{A} \subset \mathrm{B} \ldots \ldots . . \text { (i) }
\end{aligned}
$

Let $x \in \mathrm{B}$
$
\begin{aligned}
& \Rightarrow\{x\} \in \mathrm{P}(\mathrm{B}) \\
& \Rightarrow\{x\} \in \mathrm{P}(\mathrm{A}) \\
& \Rightarrow x \in \mathrm{A} \\
& \therefore \mathrm{B} \subset \mathrm{A} \ldots \ldots . . \text { (ii) }
\end{aligned}
$

From eq. (i) and (ii),

$\text { we have } \mathrm{A}=\mathrm{B}$

Miscellaneous Exercise Question 6.

Show that for any sets $A$ and $B, A=(A \cap B) \cup(A-B)$ and $A \cup(B-A)=(A \cup B)$

Answer.

Since $(A \cap B) \cup(A-B)=(A \cap B) \cup\left(A \cap B^{\prime}\right)$
$\Rightarrow(\mathrm{A} \cap \mathrm{B}) \cup(\mathrm{A}-\mathrm{B})=\mathrm{A} \cap \mathrm{B} \cup \mathrm{B}^{\prime}=\mathrm{A} \cap \mathrm{U}=\mathrm{A}$
Therefore, $\mathrm{A}=(\mathrm{A} \sim \mathrm{B}) \cup(\mathrm{A}-\mathrm{B})$
Also $\mathrm{A} \cup(\mathrm{B}-\mathrm{A})=\mathrm{A} \cup\left(\mathrm{B} \cap \mathrm{A}^{\prime}\right)=(\mathrm{A} \cup \mathrm{B}) \cap\left(\mathrm{A} \cup \mathrm{A}^{\prime}\right)=(\mathrm{A} \cup \mathrm{B}) \cap \mathrm{U}=\mathrm{A} \cup \mathrm{B}$
Therefore, $\mathrm{A} \cup(\mathrm{B}-\mathrm{A})=\mathrm{A} \cup \mathrm{B}$
Miscellaneous Exercise Question 7.

Using properties of sets, show that:
(i) $\mathrm{A} \cup(\mathrm{A} \cap \mathrm{B})=\mathrm{A}$
(ii) $\mathrm{A} \cap(\mathrm{A} \cup \mathrm{B})=\mathrm{A}$

Answer.

(i) If $A \subset B$, then $A \cap B=B$

Also $\mathrm{A} \cap \mathrm{B} \subset \mathrm{A}$
$\therefore \mathrm{A} \cup(\mathrm{A} \cap \mathrm{B})=\mathrm{A}$
(ii) If $\mathrm{A} \subset \mathrm{B}$, then $\mathrm{A} \cap \mathrm{B}=\mathrm{A}$

Also $\mathrm{A} \subset \mathrm{A} \cup \mathrm{B}$
$\therefore \mathrm{A} \cap(\mathrm{A} \cup \mathrm{B})=\mathrm{A}$
Miscellaneous Exercise Question 8.

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

Answer.

Let $A=\{1,2,3,4\}, B=\{2,3,4,5,6\}$ and $C=\{2,3,4,9,10\}$
$\therefore A \cap B=\{2,3,4\}$
And $\mathrm{A} \cap \mathrm{C}=\{2,3,4\}$

Therefore, we have $\mathrm{A} \cap \mathrm{B}=\mathrm{A} \cap \mathrm{C}$

But $\mathrm{B} \neq \mathrm{C}$
Miscellaneous Exercise Question 9.

Let $A$ and $B$ sets. If $A \cap X=B \cap X=\phi$ and $A \cup X=B \cup X$ for some set $X$. Show that $\mathrm{A}=\mathrm{B}$.

Also $\mathrm{A} \cup \mathrm{X}=\mathrm{B} \cup \mathrm{X}$
$
\begin{aligned}
& \Rightarrow \mathrm{B} \cap(\mathrm{A} \cup \mathrm{X})=\mathrm{B} \cap(\mathrm{B} \cup \mathrm{X}) \\
& \Rightarrow(\mathrm{B} \cap \mathrm{A}) \cup(\mathrm{B} \cap \mathrm{X})=\mathrm{B} \\
& \Rightarrow(\mathrm{B} \cap \mathrm{A}) \cup \phi=\mathrm{B}
\end{aligned}
$
$
\Rightarrow \mathrm{B} \cap \mathrm{A}=\mathrm{B} \Rightarrow \mathrm{B} \subset \mathrm{A}
$
From eq. (i) and (ii), we have $\mathrm{A}=\mathrm{B}$

Miscellaneous Exercise Question 10.

Find sets $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ such that $\mathrm{A} \cap \mathrm{B}, \mathrm{B} \cap \mathrm{C}$ and $\mathrm{A} \cap \mathrm{C}$ are non-empty sets and $\mathrm{A} \cap$ $\mathbf{B} \cap \mathbf{C}=\phi$

Answer.

Let $\mathrm{A}=\{1,2\}, \mathrm{B}=\{1,4\}$ and $\mathrm{C}=\{2,4\}$
$\therefore \mathrm{A} \cap \mathrm{B}=\{1\} \neq \phi \mathrm{B} \cap \mathrm{C}=\{4\} \neq \phi$
And $\mathrm{A} \cap \mathrm{C}=\{2\} \neq \phi$
But $\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}=\phi$