Exercise 2.3 (Revised) - Chapter 2 - Relations & Functions - Ncert Solutions class 11 - Maths

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

Ex 2.3 Question 1.

Which of the following are functions? Give reasons. If it is a function determine its domain and range.
(i) $\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\}$
(ii) $\{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\}$
(iii) $\{(1,3),(1,5),(2,5)\}$

Answer.

(i) Given Relation is $\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\}$
All values of $x$ are distinct. Each value of $x$ has a unique value of $y$.
Therefore, the relation is a function.
$\therefore$ Domain of function $=\{2,5,8,11,14,17\}$
Range of function $=\{1\}$
(ii) Given: Relation is $\{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\}$

All values of $x$ are distinct. Each value of $x$ has a unique value of $y$.
Therefore, the relation is a function.
$\therefore$ Domain of function $=\{2,4,6,8,10,12,14\}$
Range of function $=\{1,2,3,4,5,6,7\}$
(iii) Given: Relation is $\{(1,3),(1,5),(2,5)\}$

This relation is not a function because there is an element 1 which is associated to two elements 3 and 5.
Ex 2.3 Question 2.

Find the domain and range of the following real functions:

(i) $f(x)=-|x|$
(ii) $f(x)=\sqrt{9-x^2}$

Answer.

(i) Given: $f(x)=-|x|$. The function is defined for all real values of $x$
$\therefore$ Domain of the function $=\mathrm{R}$
Now, when $x<0$, then $|x|=-x$
$
\therefore f(x)=-(-x), x<0
$

When $x=0,|x|=0$
$
\therefore f(x)=-|0|=0
$

When $x>0,|x|=x$
$
\therefore f(x)=-x<0
$

Therefore, $f(x) \leq 0$ for all real values of $x$.
$\therefore$ Range of function $=(-\infty, 0]$

(ii) Given: $f(x)=\sqrt{9-x^2}$.

The function is not defined when $9-x^2<0$.
$
\begin{aligned}
& \therefore \text { Domain of function }=\left\{x: 9-x^2 \geq 0\right\}=\left\{x: x^2-9 \leq 0\right\} \\
& =\{x:(x+3)(x-3) \leq 0\}=[-3,3]
\end{aligned}
$
$\therefore$ Range of function $=[0,3]$
Ex 2.3 Question 3.

A function $f$ is defined by $f(x)=2 x-5$. Write down the values of:

(i) $f(0)$
(ii) $f(7)$
(iii) $f(-3)$

Answer.

Given: $f(x)=2 x-5$
(i) Putting $x=0$,
$
f(0)=2 \times 0-5=-5
$
(ii) Putting $x=7=$
$
f(7)=2 \times 7-5=14-5=9
$
(iii) Putting $x=-3$.
$
f(-3)=2 \times(-3)-5=6-5=-11
$
Ex 2.3 Question 4.

The function $t$ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(C)=\frac{9 C}{5}+32$. Find:
(i) $t(0)$
(ii) $t(28)$
(iii) $t(-10)$
(iv) The value of $\mathrm{C}$ when $t(\mathrm{C})=212$

Answer.

Given: $t(\mathrm{C})=\frac{9 \mathrm{C}}{5}+32$
(i) Putting $\mathrm{C}=0, t(0)=\frac{9 \times 0}{5}+32=32$
(ii) Putting $\mathrm{C}=28, t(28)=\frac{9 \times 28}{5}+32=\frac{252+160}{5}=\frac{412}{5}$
(iii) Putting $\mathrm{C}=-10, t(-10)=\frac{9 \times(-10)}{5}+32=-18+32=14$

(iv) Putting $t(\mathrm{C})=212,212=\frac{9 \mathrm{C}}{5}+32$
$
\begin{aligned}
& \Rightarrow \frac{9 \mathrm{C}}{5}=212-32 \\
& \Rightarrow \frac{9 \mathrm{C}}{5}=180 \\
& \Rightarrow \mathrm{C}=180 \times \frac{5}{9}=100
\end{aligned}
$
Ex 2.3 Question 5.

Find the range of each of the following functions:
(i) $f(x)=2-3 x, x \in \mathbb{R}, x>0$
(ii) $f(x)=x^2+2, x$ is a real number.
(iii) $f(x)=x, x$ is a real number.

Answer.

(i) Given: $f(x)=2-3 x, x \in \mathrm{R}$ and $x>0$
$
\begin{aligned}
& \therefore 3 x>0 \Rightarrow-3 x<0 \\
& \Rightarrow 2-3 x<2
\end{aligned}
$
$\therefore$ Range of function
$
=\{a \in \mathbb{R}: a<2\}=(-\infty, 2)
$
(ii) Given: $f(x)=x^2+2, x \in \mathrm{R}$
$
\begin{aligned}
& \therefore x^2 \geq 0 \text { for } x \in \mathrm{R} \\
& \Rightarrow x^2+2 \geq 2
\end{aligned}
$

$\therefore$ Range of function
$
=\{a \in \mathrm{R}: a \geq 2 \forall a \in \mathrm{R}\}=[2, \infty)
$
(iii) Given: $f(x)=x, x \in \mathrm{R}$
$\therefore$ Range of function $=\mathrm{R}$