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

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

Miscellaneous Example 1 

Let $\mathbf{R}$ be the set of real numbers. Define the real function
$
f: \mathbf{R} \rightarrow \mathbf{R} \text { by } f(x)=x+10
$
and sketch the graph of this function.
Solution Here $f(0)=10, f(1)=11, f(2)=12, \ldots$, $f(10)=20$, etc., and
$f(-1)=9, f(-2)=8, \ldots, f(-10)=0$ and so on.
Therefore, shape of the graph of the given function assumes the form as shown in Fig 2.16.

Miscellaneous Example 2

Let $\mathrm{R}$ be a relation from $\mathbf{Q}$ to $\mathbf{Q}$ defined by $\mathrm{R}=\{(a, b): a, b \in \mathbf{Q}$ and $a-b \in \mathbf{Z}\}$. Show that
(i) $(a, a) \in \mathrm{R}$ for all $a \in \mathbf{Q}$
(ii) $(a, b) \in \mathrm{R}$ implies that $(b, a) \in \mathrm{R}$
(iii) $(a, b) \in \mathrm{R}$ and $(b, c) \in \mathrm{R}$ implies that $(a, c) \in \mathrm{R}$

Solution
(i) Since, $a-a=0 \in \mathbf{Z}$, if follows that $(a, a) \in \mathrm{R}$.
(ii) $(a, b) \in \mathbf{R}$ implies that $a-b \in \mathbf{Z}$. So, $b-a \in \mathbf{Z}$. Therefore, $(b, a) \in \mathrm{R}$
(iii) $(a, b)$ and $(b, c) \in \mathrm{R}$ implies that $a-b \in \mathbf{Z} . b-c \in \mathbf{Z}$. So, $a-c=(a-b)+(b-c) \in \mathbf{Z}$. Therefore, $(a, c) \in \mathbf{R}$

Miscellaneous Example 3

Let $f=\{(1,1),(2,3),(0,-1),(-1,-3)\}$ be a linear function from $\mathbf{Z}$ into $\mathbf{Z}$. Find $f(x)$.
Solution

Since $f$ is a linear function, $f(x)=m x+c$. Also, since $(1,1),(0,-1) \in \mathrm{R}$, $f(1)=m+c=1$ and $f(0)=c=-1$. This gives $m=2$ and $f(x)=2 x-1$.

Miscellaneous Example 4

Find the domain of the function $f(x)=\frac{x^2+3 x+5}{x^2-5 x+4}$
Solution

Since $x^2-5 x+4=(x-4)(x-1)$, the function $f$ is defined for all real numbers except at $x=4$ and $x=1$. Hence the domain of $f$ is $\mathbf{R}-\{1,4\}$.
Miscellaneous Example 5

The function $f$ is defined by

$
f(x)=\left\{\begin{array}{lr}
1-x, & x<0 \\
1 & , x=0 \\
x+1, & x>0
\end{array}\right.
$

Draw the graph of $f(x)$.
Solution

Here, $f(x)=1-x, x<0$, this gives
$
\begin{aligned}
f(-4) & =1-(-4)=5 ; \\
f(-3) & =1-(-3)=4 ; \\
f(-2) & =1-(-2)=3 \\
f(-1) & =1-(-1)=2 ; \text { etc, } \\
\text { and } f(1) & =2, f(2)=3, f(3)=4
\end{aligned}
$
and $f(1)=2, f(2)=3, f(3)=4$
$f(4)=5$ and so on for $f(x)=x+1, x>0$.
Thus, the graph of $f$ is as shown in Fig 2.17