Examples (Revised) - Chapter 2 - Relations & Functions - Ncert Solutions class 11 - Maths

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Chapter 2: Relations & Functions - NCERT Solutions for Class 11 Maths

Example 1

If $(x+1, y-2)=(3,1)$, find the values of $x$ and $y$.
Solution

Since the ordered pairs are equal, the corresponding elements are equal.
Therefore $\quad x+1=3$ and $y-2=1$.
Solving we get $\quad x=2$ and $y=3$.
Example 2

If $\mathrm{P}=\{a, b, c\}$ and $\mathrm{Q}=\{r\}$, form the sets $\mathrm{P} \times \mathrm{Q}$ and $\mathrm{Q} \times \mathrm{P}$.
Are these two products equal?
Solution

By the definition of the cartesian product,
$
\mathrm{P} \times \mathrm{Q}=\{(a, r),(b, r),(c, r)\} \text { and } \mathrm{Q} \times \mathrm{P}=\{(r, a),(r, b),(r, c)\}
$

Since, by the definition of equality of ordered pairs, the pair $(a, r)$ is not equal to the pair $(r, a)$, we conclude that $\mathrm{P} \times \mathrm{Q} \neq \mathrm{Q} \times \mathrm{P}$.
However, the number of elements in each set will be the same.
Example 3

Let $A=\{1,2,3\}, B=\{3,4\}$ and $C=\{4,5,6\}$. Find
(i) $\mathrm{A} \times(\mathrm{B} \cap \mathrm{C})$
(ii) $(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})$
(iii) $\mathrm{A} \times(\mathrm{B} \cup \mathrm{C})$
(iv) $(\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})$

Solution

(i) By the definition of the intersection of two sets, $(\mathrm{B} \cap \mathrm{C})=\{4\}$.
Therefore, $\mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=\{(1,4),(2,4),(3,4)\}$.
(ii) Now $(\mathrm{A} \times \mathrm{B})=\{(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)\}$ and $(\mathrm{A} \times \mathrm{C})=\{(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\}$

Therefore, $(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})=\{(1,4),(2,4),(3,4)\}$.
(iii) Since, $\quad(\mathrm{B} \cup \mathrm{C})=\{3,4,5,6\}$, we have $\mathrm{A} \times(\mathrm{B} \cup \mathrm{C})=\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3)$, $(3,4),(3,5),(3,6)\}$.
(iv) Using the sets $\mathrm{A} \times \mathrm{B}$ and $\mathrm{A} \times \mathrm{C}$ from part (ii) above, we obtain $(\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})=\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6)$, $(3,3),(3,4),(3,5),(3,6)\}$.

Example 4

If $\mathrm{P}=\{1,2\}$, form the set $\mathrm{P} \times \mathrm{P} \times \mathrm{P}$.
Solution

We have, $\mathrm{P} \times \mathrm{P} \times \mathrm{P}=\{(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1)$, $(2,2,2)\}$.

Example 5

If $\mathbf{R}$ is the set of all real numbers, what do the cartesian products $\mathbf{R} \times \mathbf{R}$ and $\mathbf{R} \times \mathbf{R} \times \mathbf{R}$ represent?
Solution

The Cartesian product $\mathbf{R} \times \mathbf{R}$ represents the set $\mathbf{R} \times \mathbf{R}=\{(x, y): x, y \in \mathbf{R}\}$ which represents the coordinates of all the points in two dimensional space and the cartesian product $\mathbf{R} \times \mathbf{R} \times \mathbf{R}$ represents the set $\mathbf{R} \times \mathbf{R} \times \mathbf{R}=\{(x, y, z): x, y, z \in \mathbf{R}\}$ which represents the coordinates of all the points in three-dimensional space.

Example 6

If $\mathrm{A} \times \mathrm{B}=\{(p, q),(p, r),(m, q),(m, r)\}$, find $\mathrm{A}$ and $\mathrm{B}$.
Solution
$\mathrm{A}=$ set of first elements $=\{p, m\}$
$\mathrm{B}=$ set of second elements $=\{q, r\}$.

Example 7

Let $\mathrm{A}=\{1,2,3,4,5,6\}$. Define a relation $\mathrm{R}$ from $\mathrm{A}$ to $\mathrm{A}$ by
$
\mathrm{R}=\{(x, y): y=x+1\}
$
(i) Depict this relation using an arrow diagram.
(ii) Write down the domain, codomain and range of $\mathrm{R}$.

Solution
(i) By the definition of the relation,
$
R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\} \text {. }
$

The corresponding arrow diagram is shown in Fig 2.5.
(ii) We can see that the
$
\begin{aligned}
& \text { domain }=\{1,2,3,4,5,\} \\
& \quad \text { Similarly, the range }=\{2,3,4,5,6\} \\
& \text { and the codomain }=\{1,2,3,4,5,6\} .
\end{aligned}
$

Example 8

The Fig 2.6 shows a relation
Fig 2.5 between the sets $\mathrm{P}$ and $\mathrm{Q}$. Write this relation (i) in set-builder form, (ii) in roster form. What is its domain and range?

Solution

It is obvious that the relation $\mathrm{R}$ is " $x$ is the square of $y$ ".
(i) In set-builder form, $\mathrm{R}=\{(x, y)$ : $x$ is the square of $y, x \in \mathrm{P}, y \in \mathbf{Q}\}$
(ii) In roster form, $\mathrm{R}=\{(9,3)$, $(9,-3),(4,2),(4,-2),(25,5),(25,-5)\}$

The domain of this relation is $\{4,9,25\}$.
The range of this relation is $\{-2,2,-3,3,-5,5\}$.
Note that the element 1 is not related to any element in set $P$. The set $\mathrm{Q}$ is the codomain of this relation.

The total number of relations that can be defined from a set $\mathrm{A}$ to a set $\mathrm{B}$ is the number of possible subsets of $\mathrm{A} \times \mathrm{B}$. If $n(\mathrm{~A})=p$ and $n(\mathrm{~B})=q$, then $n(\mathrm{~A} \times \mathrm{B})=p q$ and the total number of relations is $2^{p q}$.
Example 9

Let $\mathrm{A}=\{1,2\}$ and $\mathrm{B}=\{3,4\}$. Find the number of relations from $\mathrm{A}$ to $\mathrm{B}$.
Solution

We have,
$
\mathrm{A} \times \mathrm{B}=\{(1,3),(1,4),(2,3),(2,4)\} .
$

Since $n(\mathrm{~A} \times \mathrm{B})=4$, the number of subsets of $\mathrm{A} \times \mathrm{B}$ is $2^4$. Therefore, the number of relations from $\mathrm{A}$ into $\mathrm{B}$ will be $2^4$.
Remark

A relation $\mathrm{R}$ from $\mathrm{A}$ to $\mathrm{A}$ is also stated as a relation on $\mathrm{A}$.

Example 10

Let $\mathbf{N}$ be the set of natural numbers and the relation $\mathrm{R}$ be defined on $\mathrm{N}$ such that $\mathrm{R}=\{(x, y): y=2 x, x, y \in \mathbf{N}\}$.
What is the domain, codomain and range of $\mathrm{R}$ ? Is this relation a function?

Solution

The domain of $\mathrm{R}$ is the set of natural numbers $\mathbf{N}$. The codomain is also $\mathbf{N}$. The range is the set of even natural numbers.

Since every natural number $n$ has one and only one image, this relation is a function.

Example 11

Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not?
(i) $\mathrm{R}=\{(2,1),(3,1),(4,2)\}$, (ii) $\mathrm{R}=\{(2,2),(2,4),(3,3),(4,4)\}$
(iii) $\mathrm{R}=\{(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)\}$

Solution

(i) Since $2,3,4$ are the elements of domain of $R$ having their unique images, this relation $R$ is a function.
(ii) Since the same first element 2 corresponds to two different images 2 and 4 , this relation is not a function.
(iii) Since every element has one and only one image, this relation is a function.

Example 12

Let $\mathbf{N}$ be the set of natural numbers. Define a real valued function $f: \mathbf{N} \rightarrow \mathbf{N}$ by $f(x)=2 x+1$. Using this definition, complete the table given below.

Solution

The completed table is given by 

Example 13

Define the function $f: \mathbf{R} \rightarrow \mathbf{R}$ by $y=f(x)=x^2, x \in \mathbf{R}$. Complete the Table given below by using this definition. What is the domain and range of this function? Draw the graph of $f$.

Solution

The completed Table is given below: 

Domain of $f=\{x: x \in \mathbf{R}\}$. Range of $f=\left\{x^2: x \in \mathbf{R}\right\}$. The graph of $f$ is given by Fig 2.10

Example 14

Draw the graph of the function $\boldsymbol{f}: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=x^3, x \in \mathbf{R}$.
Solution

We have $f(0)=0, f(1)=1, f(-1)=-1, f(2)=8, f(-2)=-8, f(3)=27 ; f(-3)=-27$, etc. Therefore, $f=\left\{\left(x, x^3\right): x \in \mathbf{R}\right\}$. The graph of $f$ is given in Fig 2.11.

(iv) Rational functions are functions of the type $\frac{f(x)}{g(x)}$, where $f(x)$ and $g(x)$ are polynomial functions of $x$ defined in a domain, where $g(x) \neq 0$.

Example 15

Define the real valued function $f: \mathbf{R}-\{0\} \rightarrow \mathbf{R}$ defined by $f(x)=\frac{1}{x}$, $x \in \mathbf{R}-\{0\}$. Complete the Table given below using this definition. What is the domain and range of this function?

Solution

The completed Table is given by 

The domain is all real numbers except 0 and its range is also all real numbers except 0 . The graph of $f$ is given in Fig 2.12.

(v) The Modulus function The function $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=|x|$ for each $x \in \mathbf{R}$ is called modulus function. For each non-negative value of $x, f(x)$ is equal to $x$. But for negative values of $x$, the value of $f(x)$ is the negative of the value of $x$, i.e.,
$
f(x)=\left\{\begin{array}{l}
x, x \geq 0 \\
-x, x<0
\end{array}\right.
$

The graph of the modulus function is given in Fig 2.13.
(vi) Signum function The function $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by
$
f(x)=\left\{\begin{array}{l}
1, \text { if } x>0 \\
0, \text { if } x=0 \\
-1, \text { if } x<0
\end{array}\right.
$

is called the signum function. The domain of the signum function is $\mathbf{R}$ and the range is the set $\{-1,0,1\}$. The graph of the signum function is given by the Fig 2.14.

(vii) Greatest integer function The function $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=[x], x \in \mathbf{R}$ assumes the value of the greatest integer, less than or equal to $x$. Such a function is called the greatest integer function.

From the definition of $[x]$, we can see that
$
\begin{aligned}
& {[x]=-1 \text { for }-1 \leq x<0} \\
& {[x]=0 \text { for } 0 \leq x<1} \\
& {[x]=1 \text { for } 1 \leq x<2} \\
& {[x]=2 \text { for } 2 \leq x<3 \text { and }}
\end{aligned}
$
so on.
The graph of the function is shown in Fig 2.15.

Example 16

Let $f(x)=x^2$ and $g(x)=2 x+1$ be two real functions. Find
$
(f+g)(x),(f-g)(x),(f g)(x),\left(\frac{f}{g}\right)(x) .
$

Solution

We have,
$
\begin{aligned}
& (f+g)(x)=x^2+2 x+1,(f-g)(x)=x^2-2 x-1, \\
& (f g)(x)=x^2(2 x+1)=2 x^3+x^2,\left(\frac{f}{g}\right)(x)=\frac{x^2}{2 x+1}, x \neq-\frac{1}{2}
\end{aligned}
$

Example 17

Let $f(x)=\sqrt{x}$ and $g(x)=x$ be two functions defined over the set of nonnegative real numbers. Find $(f+g)(x),(f-g)(x),(f g)(x)$ and $\left(\frac{f}{g}\right)(x)$.
Solution

We have
$
\begin{aligned}
& (f+g)(x)=\sqrt{x}+x,(f-g)(x)=\sqrt{x}-x, \\
& (f g) x=\sqrt{x}(x)=x^{\frac{3}{2}} \text { and }\left(\frac{f}{g}\right)(x)=\frac{\sqrt{x}}{x}=x^{-\frac{1}{2}}, x \neq 0
\end{aligned}
$