Miscellaneous Exercise Sets - NCERT Class 11 Maths

Chapter 1 Ex.1.ME Question 1

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

\begin{align}A &= \left\{ {x:x \in {\bf{R}}{\text{ and}}\;x\;{\text{satisfy }}{x^2}-8x + 12 = 0} \right\},\\B &= \left\{ {2,4,6} \right\},\\C &= \left\{ {2,4,6,8 \ldots } \right\}\\D &= \left\{ 6 \right\}.\end{align}

Solution

$$A = \left\{ {x:x \in {\bf{R}}{\text{ and}}\;x\;{\text{satisfy }}{x^2}-8x + 12 = 0} \right\}$$

$$2$$ and $$6$$ are the only solutions of  $${x^2}-8x + 12 = 0$$

Hence,

\begin{align}A &= \left\{ {2,6} \right\}, \\B &= \left\{ {2,4,6} \right\},\\C &= \left\{ {2,4,6,8 \ldots } \right\}, \\D &= \left\{ 6 \right\}\end{align}

Therefore,  $$D \subset A \subset B \subset C$$

Hence,  $$A \subset B,\;\;A \subset C,\;\;B \subset C,\;\;D \subset A,\;\;D \subset B,\;\;D \subset C$$

Chapter 1 Ex.1.ME 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 $$x \in A$$ and $$A \in B$$, then $$x \in B$$

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

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

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

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

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

Solution

(i) False.

Let $$A = \left\{ {2,3} \right\}$$ and $$B = \left\{ {1,\left\{ {2,3} \right\},4} \right\}$$

Now, $$2 \in A$$ and $$A \in B$$

But, $$2 \notin B$$

(ii) False.

Let $$A = \left\{ 2 \right\},\;\;B = \left\{ {1,2} \right\}$$ and $$C = \left\{ {0,\left\{ {1,2} \right\},3} \right\}$$

Now, $$2 \in \left\{ {1,2} \right\}$$ and $$\left\{ {1,2} \right\} \in \left\{ {0,\left\{ {1,2} \right\},3} \right\}$$

Hence, $$A \subset B$$ and $$B \in C$$

But, $$2 \notin \left\{ {0,\left\{ {1,2} \right\},3} \right\}$$

(iii) True.

It is given that; $$A \subset B$$ and $$B \subset C$$

Let $$x \in A$$

Now,

\begin{align}& x\in B \qquad \left[ \because A\subset B \right] \\& x\in C\ \ \ \ \qquad \left[ \because B\subset C \right] \\\end{align}

Hence, $$A \subset C$$ proved.

(iv) False.

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

Here, $$\left\{ {1,2} \right\} \not\subset \left\{ {3,4} \right\}$$ and $$\left\{ {3,4} \right\} \not\subset \left\{ {0,1,2,5} \right\}$$

However, $$\left\{ {1,2} \right\} \subset \left\{ {0,1,2,5} \right\}$$

Hence, $$A \not\subset B$$ and $$B \not\subset C$$

But, $$A \subset C$$

(v) False

Let $$A = \left\{ {1,2,3} \right\}$$ and $$B = \left\{ {3,4,5} \right\}$$

Here, $$3 \in \left\{ {1,2,3} \right\}$$ and $$\left\{ {1,2,3} \right\} \not\subset \left\{ {3,4,5} \right\}$$

However, $$3 \notin B$$

Hence, $$x \in A$$ and $$A \not\subset B$$,

But, $$x \notin B$$

(vi) True

It is given that; $$A \subset B$$ and$$x \notin B$$, then$$x \notin A$$

Let $$x \in A$$, if possible

Now,

$$x\in B\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because A\subset B \right]$$

But it is given that $$x \notin B$$

So, $$x \notin A$$

Hence, if $$A \subset B$$ and $$x \notin B$$, then $$x \notin A$$ proved.

Chapter 1 Ex.1.ME Question 3

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

Solution

It is given that $$A,\, B$$ and $$C$$ be the sets such that $$A \cup B = A \cup C$$ and $$A \cap B = A \cap C$$.

Let $$x \in B$$

Therefore, $$x \in A \cup B$$

Since, $$A \cup B = A \cup C$$; $$x \in A \cup C$$

Hence, $$x \in A$$ or $$x \in C$$

If $$x \in A$$

Then $$x \in A \cap B$$, since $$x \in B$$

Now, $$A \cap B = A \cap C$$

Therefore, $$x \in A \cap C$$

That means, $$x \in A$$ and $$x \in C$$

So, $$B \subset C$$

Similarly, we can show that $$C \subset B$$

Hence, $$B = C$$ proved.

Chapter 1 Ex.1.ME Question 4

Show that the following four conditions are equivalent:

(i) $$A \subset B$$

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

(iii) $$A \cup B = B$$

(iv) $$A \cap B = A$$

Solution

(i) $$A \subset B$$

If possible, suppose $$A-B \ne \phi$$

This means that there exists $$x \in A$$ and $$x \ne B$$, which is not possible as $$A \subset B$$.

Hence, if $$A \subset B$$, then, $$A-B = \phi$$

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

Let $$x \in A$$

Clearly, $$x \in B$$ because if $$x \notin B$$, then $$A-B \ne \phi$$

Hence, if $$A-B = \phi$$, then, $$A \subset B$$

Therefore,

\begin{align}\left( i \right) \Leftrightarrow \left( {ii} \right)\\A \subset B \Leftrightarrow A--B = \phi\end{align}

(iii) $$A \cup B = B$$

It is clear that, $$B \subset \left( {A \cup B} \right)$$ since, $$A \subset B$$

Let $$x \in \left( {A \cup B} \right)$$

Hence, $$x \in A$$ or $$x \in B$$

If $$x \in A$$

Then, $$x\in B \qquad \left[ \because A\subset B \right]$$

Therefore, $$\left( {A \cup B} \right) \subset B$$

If $$x \in B$$

Then, $$A \cup B = B$$

Hence, if $$A \subset B$$, then, $$A \cup B = B$$

Conversely, $$A \cup B = B$$

Let $$x \in A$$

Then,

\begin{align}& x\in \left( A\cup B \right) \quad \left[ \because A\subset \left( A\cup B \right) \right] \\& x\in B \quad \left[ \because A\cup B=B \right] \\\end{align}

Therefore, $$A \subset B$$

Hence, if $$A \cup B = B$$, then, $$A \subset B$$

Therefore,

\begin{align}\left( i \right) \Leftrightarrow \left( {iii} \right)\\A \subset B \Leftrightarrow A \cup B = B\end{align}

(iv) $$A \cap B = A$$

It is clear that, $$\left( {A \cap B} \right) \subset A$$ since, $$A \subset B$$

Let $$x \in A$$

Then,

$$x\in B \qquad \left[ \because A\subset B \right]$$

Therefore, $$x \in \left( {A \cap B} \right)$$ and then $$A \subset \left( {A \cap B} \right)$$

Hence, $$A = \left( {A \cap B} \right)$$

Conversely, $$A \cap B = A$$

Let $$x \in A$$

Then,

$$x \in \left( {A \cap B} \right)$$

So, $$x \in A$$ and $$x \in B$$

Therefore, $$A \subset B$$

Hence, if $$A \cap B = A$$, then, $$A \subset B$$

Therefore,

\begin{align}\left( i \right) \Leftrightarrow \left( {iv} \right)\\A \subset B \Leftrightarrow A \cap B = A\end{align}

Hence, it is proved that $$A \subset B \Leftrightarrow A-B = \phi \Leftrightarrow A \cup B = B \Leftrightarrow A \cap B = A$$.

Chapter 1 Ex.1.ME Question 5

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

Solution

Let $$A \subset B$$ and $$x \in \left( {C-B} \right)$$

Then,

$$x \in C$$ and $$x \notin B$$

Now,

$$x\notin A \qquad \left[ \because A\subset B \right]$$

Hence, $$x \in \left( {C-A} \right)$$

Therefore, $$C-B \subset C-A$$

Chapter 1 Ex.1.ME Question 6

Assume that $$P\left( A \right) = P\left( B \right)$$. Show that $$A = B$$.

Solution

Let $$P\left( A \right) = P\left( B \right)$$

Let $$x \in A$$

$$A \in P\left( A \right) = P\left( B \right)$$

Therefore, $$x \in C$$, for some $$C \in P\left( B \right)$$

Now, $$C \subset B$$

Hence, $$x \in B$$

So, $$A \subset B$$

Similarly, $$B \subset A$$

Therefore, $$A = B$$ proved.

Chapter 1 Ex.1.ME Question 7

Is it true that for any sets $$A$$ and $$B,$$ $$P\left( A \right)\cup P\left( B \right)=P\left( A\cup B \right)$$?

Justify your answer.$$A = B$$

Solution

False.

Let $$A = \left\{ {1,2} \right\}$$ and $$B = \left\{ {2,3} \right\}$$

Hence, $$A \cup B = \left\{ {1,2,3} \right\}$$

$$P\left( A \right) = \left\{ {\phi ,\left\{ 1 \right\},\left\{ 2 \right\},\left\{ {1,2} \right\}} \right\}$$

$$P\left( B \right) = \left\{ {\phi ,\left\{ 2 \right\},\left\{ 3 \right\},\left\{ {2,3} \right\}} \right\}$$

Now,

\begin{align}P\left( A \right) \cup P\left( B \right) &= \left\{ {\phi ,\left\{ 1 \right\},\left\{ 2 \right\},\left\{ {1,2} \right\}} \right\} \cup \left\{ {\phi ,\left\{ 2 \right\},\left\{ 3 \right\},\left\{ {2,3} \right\}} \right\}\\&= \left\{ {\phi ,\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 3 \right\},\left\{ {1,2} \right\},\left\{ {2,3} \right\}} \right\}\end{align}

$$P\left( {A \cup B} \right) = \left\{ {\phi ,\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 3 \right\},\left\{ {1,2} \right\},\left\{ {2,3} \right\},\left\{ {1,2,3} \right\}} \right\}$$

We can see that

$$P\left( A \right) \cup P\left( B \right) \ne P\left( {A \cup B} \right)$$

Therefore, it is not true that for any sets $$A$$ and $$B,$$ $$P\left( A \right) \cup P\left( B \right) = P\left( {A \cup B} \right)$$.

Chapter 1 Ex.1.ME Question 8

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

$$A = \left( {A \cap B} \right) \cup \left( {A-B} \right)$$ and $$A \cup \left( {B - A} \right) = \left( {A \cup B} \right)$$

Solution

To prove: $$A = \left( {A \cap B} \right) \cup \left( {A-B} \right)$$

Let $$x \in A$$

Case I

$$x \in A \cap B$$

Then, $$x \in \left( {A \cap B} \right) \subset \left( {A \cup B} \right) \cup \left( {A-B} \right)$$

Case II

$$x \notin \left( {A \cap B} \right)$$

Then, $$x \notin A$$ or$$x \notin B$$

$$x \notin \left( {A-B} \right) \subset \left( {A \cup B} \right) \cup \left( {A-B} \right)$$

$$A \subset \left( {A \cap B} \right) \cup \left( {A-B} \right) \qquad \ldots \left( 1 \right)$$

It is clear that

$$A \cap B \subset A$$ and $$\left( {A--B} \right) \subset A$$

$$\left( {A \cap B} \right) \cup \left( {A--B} \right) \subset A \qquad \ldots \left( 2 \right)$$

From (1) and (2), we obtain

$$A = \left( {A \cap B} \right) \cup \left( {A-B} \right)$$

To prove: $$A \cup \left( {B - A} \right) = \left( {A \cup B} \right)$$

Let $$x \in A \cup \left( {B-A} \right)$$

$$x \in A$$ or $$x \in \left( {B-A} \right)$$

$$x \in A$$ or ($$x \in B$$ and $$x \notin A$$)

($$x \in A$$ or $$x \in B$$) and ($$x \in A$$ or$$x \notin A$$)

$$x \in \left( {A \cup B} \right)$$

$$A \cup \left( {B-A} \right) \subset \left( {A \cup B} \right) \qquad \ldots \left( 3 \right)$$

Next, we show that.

Let $$y\in \left( A\cup B \right)$$$$\left( {A \cup B} \right) \subset A \cup \left( {B-A} \right)$$

$$y \in A$$ or $$y \in B$$

($$y \in A$$ or $$y \in B$$) and ($$y \in A$$ or$$y \notin A$$)

$$y \in A$$ or ($$y \in B$$ and $$y \notin A$$)

$$y \in A \cup \left( {B-A} \right)$$

$$A \cup B \subset A \cup \left( {B-A} \right) \qquad \ldots \left( 4 \right)$$

Hence, from (3) and (4), we obtain

$A \cup \left( {B - A} \right) = \left( {A \cup B} \right)$.

Chapter 1 Ex.1.ME Question 9

Using properties of sets show that

(i) $$A \cup \left( {A \cap B} \right) = A$$

(ii) $$A \cap \left( {A \cup B} \right) = A$$

Solution

(i) $$A \cup \left( {A \cap B} \right) = A$$

We know that

$$A \subset A$$

$$A \cap B \subset A$$

$A \cup \left( {A \cap B} \right) \subset A \qquad \ldots \left( 1 \right)$

Also,

$A \subset A \cup \left( {A \cap B} \right) \qquad \ldots \left( 2 \right)$

From (1) and (2),

$$A \cup \left( {A \cap B} \right) = A$$

(ii) $$A \cap \left( {A \cup B} \right) = A$$

\begin{align}A \cap \left( {A \cup B} \right) &= \left( {A \cap A} \right) \cup \left( {A \cap B} \right)\\&= A \cup \left( {A \cap B} \right)\\&= A \qquad \left[ {{\rm{from }}\left( {\rm{i}} \right)} \right]\end{align}

Chapter 1 Ex.1.ME Question 10

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

Solution

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

Accordingly,

and $$A\cap C=\left\{ 0 \right\}$$$$A \cap B = \left\{ 0 \right\}$$

Here, $$A \cap B = A \cap C = \left\{ 0 \right\}$$

However, $$B\ne C \qquad \left[ \because 2\in B\text{ and }2\notin C \right]$$

Chapter 1 Ex.1.ME Question 11

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 $$A = A \cap \left( {A \cup X} \right),\;B = B \cap \left( {B \cup X} \right)$$ and use distributive law)

Solution

Let $$A$$ and $$B$$ be two sets such that $$A \cap X = B \cap X = \phi$$ and $$A \cup X = B \cup X$$ for some set $$X.$$

We know that

\begin{align}& A=A\cap \left( A\cup X \right) \\& =A\cap \left( B\cup X \right) \qquad \left[ \because A\cup X=B\cup X \right] \\& =\left( A\cap B \right)\cup \left( A\cap X \right) \\& =\left( A\cap B \right)\cup \phi \qquad \left[ \because \left( A\cap X \right)=\phi \right] \\& =\left( A\cap B \right) \qquad \qquad \qquad \ldots \left( 1 \right)\end{align}

Now,

\begin{align}& B=B\cap \left( B\cup X \right) \\& =B\cap \left( A\cup X \right) \qquad \left[ \because A\cup X=B\cup X \right] \\& =\left( B\cap A \right)\cup \left( B\cap X \right) \\& =\left( B\cap A \right)\cup \phi \qquad \left[ \because \left( B\cap X \right)=\phi \right] \\& =\left( A\cap B \right) \qquad \qquad \qquad \ldots \left( 2 \right)\end{align}

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

Hence proved.

Chapter 1 Ex.1.ME Question 12

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

Solution

Let $$A = \left\{ {1,2} \right\},\;B = \left\{ {2,3} \right\}$$ , and $$C = \left\{ {1,3} \right\}$$ .

We can see that

$$A \cap B = \left\{ 2 \right\},\;B \cap C = \left\{ 3 \right\}$$ and $$A \cap C = \left\{ 1 \right\}$$

Hence, $$A \cap B,\;B \cap C$$ and $$A \cap C$$ are non-empty.

However, $$A \cap B \cap C = \phi$$

Chapter 1 Ex.1.ME Question 13

In a survey of $$600$$ students in a school, $$150$$ students were found to be taking tea and $$225$$ taking coffee, $$100$$ were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Solution

Let $$U$$ be the set of all students who took part in the survey

$$T$$ be the set of students taking tea.

$$C$$ be the set of students taking coffee.

Students taking neither tea nor coffee, $$\left( {T' \cap C'} \right)$$

It is given that

$$n\left( U \right) = 600,\;\;\;n\left( T \right) = 150,\;\;\;n\left( C \right) = 225,\;\;\;n\left( {T \cap C} \right) = 100$$

We know that

\begin{align}n\left( {T' \cap C'} \right) &= n{\left( {T \cup C} \right)^\prime }\\& = n\left( U \right)-n\left( {T \cup C} \right)\\ &= n\left( U \right)-\left[ {n\left( T \right) + n\left( C \right)-n\left( {T \cap C} \right)} \right]\\ &= 600-\left[ {150 + 225-100} \right]\\ &= 600-275\\& = 325\end{align}

Hence, $$325$$ students were taking neither tea nor coffee.

Chapter 1 Ex.1.ME Question 14

In a group of students $$100$$ students know Hindi, $$50$$ know English and $$25$$ know both. Each of the students knows either Hindi or English. How many students are there in the group?

Solution

Let $$U$$ be the set of all students in the group.

$$E$$ be the set of all students who know English.

$$H$$ be the set of all students who know Hindi.

It is given that

$$n\left( H \right) = 100,\;\;\;n\left( E \right) = 50,\;\;\;n\left( {H \cap E} \right) = 25$$

We know that

\begin{align}n\left( {H \cup E} \right) &= n\left( H \right) + n\left( E \right)-n\left( {H \cap E} \right)\\ &= 100 + 50-25\\ &= 150-25\\ &= 125\end{align}

Hence, there are $$125$$ students in the group.

Chapter 1 Ex.1.ME Question 15

In a survey of $$60$$ people, it was found that $$25$$ people read newspaper $$H$$, $$26$$ read newspaper $$T$$, $$26$$ read newspaper $$I$$, $$9$$ read both $$H$$ and $$I$$, $$11$$ read both $$H$$ and $$T$$, $$8$$ read both $$T$$ and $$I$$, $$3$$ read all three newspapers. Find:

(i) the number of people who read at least one of the newspapers.

(ii) the number of people who read exactly one newspaper.

Solution

Let $$H$$ be the set of people who read newspaper $$H.$$

$$T$$ be the set of people who read newspaper $$T.$$

$$I$$ be the set of people who read newspaper $$I.$$

It is given that

\begin{align}&n\left( H \right) = 25,\;\;\;n\left( T \right) = 26,\;\;\;n\left( I \right) = 26,\;\;\;n\left( {H \cap T} \right) = 11,\;\;\;n\left( {H \cap I} \right) = 9,\\&n\left( {T \cap I} \right) = 8,\;\;\;n\left( {H \cap T \cap I} \right) = 3 \end{align}

(i) The number of people who read at least one of the newspapers.

We know that

\begin{align}n\left( {H \cup T \cup I} \right) &= n\left( H \right) + n\left( T \right) + n\left( I \right)\\&\quad-n\left( {H \cap T} \right)-n\left( {H \cap I} \right)-n\left( {T \cap I} \right) + n\left( {H \cap T \cap I} \right)\\& = 25 + 26 + 26-11-9-8 + 3\\ &= 80 - 28\\& = 52\end{align}

Hence, $$52$$ people read at least one of the newspapers.

(ii) Let $$x$$ people read newspapers $$H$$ and $$T$$ only ypeople read newspapers $$T$$ and $$I$$ only and $$z$$ people read newspapers $$H$$ and $$I$$ only.

Now, draw the Venn diagram for the given problem

We can see that,

\begin{align}n\left( {H \cap T \cap I} \right) = a = 3\\n\left( {H \cap T} \right) = x + a = 11\\n\left( {H \cap I} \right) = z + a = 9\\n\left( {T \cap I} \right) = y + a = 8\end{align}

Now,

\begin{align}\left( {x + a} \right) + \left( {y + a} \right) + \left( {z + a} \right) &= 11 + 8 + 9\\x + 3 + y + 3 + z + 3 &= 28\\x + y + z &= 28 - 9\\x + y + z &= 19\end{align}

Number of people who read exactly two newspapers $$= x + y + z = 19$$

Number of people who read two or more newspapers $$= 19 + 3 = 22$$

Therefore,

Number of people who read $$3$$ exactly one newspaper $$= 52 - 22 = 30$$

Hence, $$30$$ people read exactly one newspaper.

Chapter 1 Ex.1.ME Question 16

In a survey it was found that $$21$$ people liked product $$A,$$ $$26$$ liked product $$B$$ and $$29$$ liked product $$C.$$ If $$14$$ people liked products $$A$$ and $$B,$$ $$12$$ people liked products $$C$$ and $$A,$$ $$14$$ people liked products $$B$$ and $$C$$ and $$8$$ liked all the three products. Find how many liked product $$C$$ only.

Solution

Let $$A$$ be the set of people who liked product $$A.$$

$$B$$ be the set of people who liked product $$B.$$

$$C$$ be the set of people who liked product $$C.$$

It is given that

\begin{align}&n\left( A \right) = 21,\;\;\;n\left( B \right) = 26,\;\;\;n\left( C \right) = 29,\;\;\;n\left( {A \cap B} \right) = 14,\\&n\left( {A \cap C} \right) = 12,\;\;\;n\left( {B \cap C} \right) = 14,\;\;\;n\left( {A \cap B \cap C} \right) = 8 \end{align}

Let’s draw the Venn diagram for the given problem

We can see that

Number of people who like product $$C$$ only

\begin{align}&= n\left( C \right) - \left( {4 + 8 + 6} \right)\\&= 29 - 18\\&= 11\end{align}

Hence, $$11$$ liked product $$C$$ only.

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