NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.1

Chapter 2 Ex.2.1 Question 1

If \begin{align}\left( {\frac{x}{3} + 1,\;y - \frac{2}{3}} \right) = \left( {\fracabc{3},\frac{1}{3}} \right), \end{align} find the values of $$x$$ and $$y.$$

Solution

It is given that \begin{align}\left( {\frac{x}{3} + 1,\;y - \frac{2}{3}} \right) = \left( {\frac{5}{3},\frac{1}{3}} \right) \end{align}.

Since the ordered pairs are equal, the corresponding elements will also be equal.

Therefore, \begin{align}\frac{x}{3} + 1 = \frac{5}{3}\end{align} and \begin{align}y - \frac{2}{3} = \frac{1}{3} \end{align}

\begin{align}\frac{x}{3} + 1 &= \frac{5}{3} \\\Rightarrow \frac{x}{3} &= \frac{5}{3} - 1\\ \Rightarrow \frac{x}{3} &= \frac{{5 - 3}}{3}\\ \Rightarrow \frac{x}{3} &= \frac{2}{3}\\ \Rightarrow x &= 2 \end {align}\qquad and \begin{align}\qquad y - \frac{2}{3} &= \frac{1}{3}\\ \Rightarrow y &= \frac{1}{3} + \frac{2}{3}\\ \Rightarrow y &= 1\end{align}

Hence, $$x = 2$$ and $$y = 1$$

Chapter 2 Ex.2.1 Question 2

If the set $$A$$ has $$3$$ elements and the set $$B = \{ 3,4,5\}$$, then find the number of elements in $$\left( {A \times B} \right)$$?

Solution

It is given that set $$A$$ has $$3$$ elements and the set $$B = \{ 3,4,5\}$$

Number of elements in set $$A,$$ $$n\left( A \right) = 3$$

Number of elements in set $$B,$$ $$n\left( B \right) = 3$$

Number of elements in $$\left( {A \times B} \right)\;=\;$$ (Number of elements in $$A$$) $$\times$$ (Number of elements in $$B$$)

\begin{align}n\left( {A \times B} \right) &= n\left( A \right) \times n\left( B \right)\\ &= 3 \times 3\\ &= 9\end{align}

Thus, the number of elements in $$\left( {A \times B} \right)$$ is $$9.$$

Chapter 2 Ex.2.1 Question 3

If  $$G = \{ 7,8\}$$ and $$H = \{ 5,4,2\}$$, find $$G \times H$$ and $$H \times G$$.

Solution

It is given that $$G = \{ 7,8\}$$ and $$H = \{ 5,4,2\}$$

We know that the Cartesian product $$P \times Q$$ of two non-empty sets $$P$$ and $$Q$$ is defined as

$P \times Q = \left\{ {\left( {p,q} \right):p \in P,q \in Q} \right\}$

Therefore,

$G \times H = \left\{ {\left( {7,5} \right),\left( {7,4} \right),\left( {7,2} \right),\left( {8,5} \right),\left( {8,4} \right),\left( {8,2} \right)} \right\}$

$H \times G = \left\{ {\left( {5,7} \right),\left( {5,8} \right),\left( {4,7} \right),\left( {4,8} \right),\left( {2,7} \right),\left( {2,8} \right)} \right\}$

Chapter 2 Ex.2.1 Question 4

State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.

(i) If $$P = \left\{ {m,n} \right\}$$ and $$Q = \left\{ {n,m} \right\}$$, then $$P \times Q = \left\{ {\left( {m,n} \right),\left( {n,m} \right)} \right\}$$

(ii) If $$A$$ and $$B$$ are non-empty sets, then $$\left( {A \times B} \right)$$ is a non-empty set of ordered pairs $$\left( {x,y} \right)$$ such that $$x \in A$$ and $$y \in B$$.

(iii) If $$A = \left\{ {1,2} \right\}$$, $$B = \left\{ {3,4} \right\}$$, then $$A \times \left( {B \cap \phi } \right) = \phi$$

Solution

(i) False, $$P \times Q = \left\{ {\left( {m,n} \right),\left( {m,m} \right),\left( {n,n} \right),\left( {n,m} \right)} \right\}$$

(ii) True

(iii) True

Chapter 2 Ex.2.1 Question 5

If $$A = \left\{ { - 1,1} \right\}$$, find $$A \times A \times A$$.

Solution

It is known that for any non-empty set $$A,$$ $$A \times A \times A$$ is defined as

$A \times A \times A = \left\{ {\left( {a,b,c} \right):a,b,c \in A} \right\}$

It is given that $$A = \left\{ { - 1,1} \right\}$$

Therefore,

$$A \times A \times A = \left\{ {\left( { - 1, - 1, - 1} \right),\left( { - 1, - 1,1} \right),\left( { - 1,1, - 1} \right),\left( { - 1,1,1} \right),\left( {1, - 1, - 1} \right),\left( {1, - 1,1} \right),\left( {1,1, - 1} \right),\left( {1,1,1} \right)} \right\}$$

Chapter 2 Ex.2.1 Question 6

If $$A \times B = \left\{ {\left( {a,x} \right),\left( {a,y} \right),\left( {b,x} \right),\left( {b,y} \right)} \right\}$$. Find $$A$$ and $$B$$.

Solution

It is given that $$A \times B = \left\{ {\left( {a,x} \right),\left( {a,y} \right),\left( {b,x} \right),\left( {b,y} \right)} \right\}$$

We know that the Cartesian product of two non-empty sets $$P$$ and $$Q$$ is defined as

$P \times Q = \left\{ {\left( {p,q} \right):p \in P,q \in Q} \right\}$

Therefore, $$A$$ is the set of all first elements and $$B$$ is the set of all second elements.

Thus, $$A = \left\{ {a,b} \right\}$$ and $$B = \left\{ {x,y} \right\}$$

Chapter 2 Ex.2.1 Question 7

Let $$A = \left\{ {1,2} \right\}$$, $$B = \left\{ {1,2,3,4} \right\}$$, $$C = \left\{ {5,6} \right\}$$ and $$D = \left\{ {5,6,7,8} \right\}$$. Verify that

(i) $$A \times \left( {B \cap C} \right) = \left( {A \times B} \right) \cap \left( {A \times C} \right)$$.

(ii) $$A \times C$$ is a subset of $$B \times D$$.

Solution

(i) To verify: $$A \times \left( {B \cap C} \right) = \left( {A \times B} \right) \cap \left( {A \times C} \right)$$

We have $$B \cap C = \left\{ {1,2,3,4} \right\} \cap \left\{ {5,6} \right\} = \phi$$

\begin{align}LHS &= A \times \left( {B \cap C} \right)\\ &= A \times \phi \\& = \phi\end{align}

Now,

\begin{align}A \times B &= \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}\\\\A \times C &= \left\{ {\left( {1,5} \right),\left( {1,6} \right),\left( {2,5} \right),\left( {2,6} \right)} \right\}\end{align}

Therefore,

\begin{align}RHS &= \left( {A \times B} \right) \cap \left( {A \times C} \right)\\ &= \phi\end{align}

Therefore, $$\text{L.H.S.} = \text{R.H.S}$$

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

(ii) To verify: $$A \times C$$ is a subset of $$B \times D$$

We have

\begin{align}A \times C &= \left\{ {\left( {1,5} \right),\left( {1,6} \right),\left( {2,5} \right),\left( {2,6} \right)} \right\}\\\\B \times D &= \left\{ \begin{array}{l}\left( {1,5} \right),\left( {1,6} \right),\left( {1,7} \right),\left( {1,8} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {2,7} \right),\left( {2,8} \right),\\\left( {3,5} \right),\left( {3,6} \right),\left( {3,7} \right),\left( {3,8} \right),\left( {4,5} \right),\left( {4,6} \right),\left( {4,7} \right),\left( {4,8} \right)\end{array} \right\} \end{align}

We can observe that all the elements of set $$A \times C$$ are the elements of set $$B \times D$$. Therefore, $$A \times C$$ is a subset of $$B \times D$$.

Chapter 2 Ex.2.1 Question 8

Let $$A = \left\{ {1,2} \right\}$$ and $$B = \left\{ {3,4} \right\}$$. Write $$A \times B$$. How many subsets will $$A \times B$$ have? List them.

Solution

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

Therefore, $$A \times B = \left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}$$

Hence, $$n\left( {A \times B} \right) = 4$$

We know that if $$C$$ is a set with $$n\left( C \right) = m$$, then $$n\left[ {P\left( C \right)} \right] = {2^m}$$.

Therefore, the set $$A \times B$$ has $${2^4} = 16$$ subsets. These are

$\left\{ \begin{array}{l}\phi ,\left\{ {\left( {1,3} \right)} \right\},\left\{ {\left( {1,4} \right)} \right\},\left\{ {\left( {2,3} \right)} \right\},\left\{ {\left( {2,4} \right)} \right\},\\\left\{ {\left( {1,3} \right),\left( {1,4} \right)} \right\},\left\{ {\left( {1,3} \right),\left( {2,3} \right)} \right\},\left\{ {\left( {1,3} \right),\left( {2,4} \right)} \right\},\\\left\{ {\left( {1,4} \right),\left( {2,3} \right)} \right\},\left\{ {\left( {1,4} \right),\left( {2,4} \right)} \right\},\left\{ {\left( {2,3} \right),\left( {2,4} \right)} \right\},\\\left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right)} \right\},\left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,4} \right)} \right\},\\\left\{ {\left( {1,3} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\},\left\{ {\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\},\left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}\end{array} \right\}$

Chapter 2 Ex.2.1 Question 9

Let $$A$$ and $$B$$ be two sets such that $$n\left( A \right) = 3$$ and $$n\left( B \right) = 2$$. If $$\left( {x,1} \right),\left( {y,2} \right),\left( {z,1} \right)$$ are in $$A \times B$$, find $$A$$ and $$B,$$ where $$x,\; y$$ and $$z$$ are distinct elements.

Solution

It is given that $$n\left( A \right) = 3$$ and $$n\left( B \right) = 2$$; and $$\left( {x,1} \right),\left( {y,2} \right),\left( {z,1} \right)$$ are in $$A \times B$$.

We know that

$$\rm{A} =$$ Set of first elements of the ordered pair elements of $$A \times B.$$

$$\rm{B} =$$ Set of second elements of the ordered pair elements of $$A \times B.$$

Therefore, $$x,\; y,$$ and $$z$$ are the elements of  $$A;$$ $$1$$ and $$2$$ are the elements of $$B.$$

Since $$n\left( A \right) = 3$$ and $$n\left( B \right) = 2$$,

It is clear that $$A = \left\{ {x,y,z} \right\}$$ and $$B = \left\{ {1,2} \right\}.$$

Chapter 2 Ex.2.1 Question 10

The Cartesian product $$A \times A$$ has $$9$$ elements among which are found $$\left( { - 1,0} \right)$$ and $$\left( {0,1} \right)$$. Find the set $$A$$ and the remaining elements of $$A \times A$$.

Solution

We know that if $$n\left( A \right) = p$$ and $$n\left( B \right) = q$$, then $$n\left( {A \times B} \right) = pq.$$

Therefore, $$n\left( {A \times A} \right) = n\left( A \right) \times n\left( A \right)$$

It is given that $$n\left( {A \times A} \right) = 9$$

$n\left( A \right) \times n\left( A \right) = 9$

Hence,

$n\left( A \right) = 3$

The ordered pairs $$\left( { - 1,0} \right)$$ and $$\left( {0,1} \right)$$ are two of the nine elements of $$A \times A$$.

We know that $$A \times A = \left\{ {\left( {a,a} \right):a \in A} \right\}$$.

Therefore, $$–1,\; 0,$$ and $$1$$ are elements of $$A.$$

Since $$n\left( A \right) = 3,$$ it is clear that $$A = \left\{ { - 1,0,1} \right\}.$$

The remaining elements of set $$A \times A$$ are $$\left( { - 1, - 1} \right),\left( { - 1,1} \right),\left( {0, - 1} \right),\left( {0,0} \right),\, \left( {1, - 1} \right),\left( {1,0} \right),$$ and $$\left( {1,1} \right)$$

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