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 \(x \notin B\)

Which is a contradiction

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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