Answer :

(i) $\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\}$

Since $2,5,8,11,14$, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain $=\{2,5,8,11,14,17\}$ and range $=\{1\}$

(ii) $\{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\}$

Since $2,4,6,8,10,12$, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain $=\{2,4,6,8,10,12,14\}$ and range $=\{1,2,3,4,5,6,7\}$

(iii) $\{(1,3),(1,5),(2,5)\}$

Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.

Answer :

(i) $f(x)=-|x|, x \in R$

We know that $|x|=\begin{cases} x, x \geq 0 \\ -x, x<0 \end{cases} .$

$\therefore f(x)=-|x|=\begin{cases} -x, x \geq 0 \\ x, x<0 \end{cases} .$

Since $f(x)$ is defined for $x \in \mathbf{R}$, the domain of fis $\mathbf{R}$.

It can be observed that the range of $f(x)=-|x|$ is all real numbers except positive real numbers.

$\therefore$ The range of fis $(- \infty, 0]$.

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

Since $\sqrt{9-x^{2}}$ is defined for all real numbers that are greater than or equal to 3 and less than or equal to 3 , the domain of $f(x)$ is $\{x$ : -3 $ x \leq 3\}$ or [- 3,3$]$.

For any value of $x$ such that $-3 \leq x \leq 3$, the value of $f(x)$ will lie between 0 and 3 .

$\therefore$ The range of $f(x)$ is $\{x: 0 \leq x \leq 3\}$ or $[0,3]$.

Answer :

The given function is $f(x)=2 x-5$.

Therefore,

(i) $f(0)=2 \times 0-5=0-5=-5$

(ii) $f(7)=2 \times 7-5=14-5=9$

(iii) $f(-3)=2 \times(-3)-5=-6-5=-11$

Answer :

The given function is

$ t(C)=\dfrac{9 C}{5}+32 $

Therefore,

(i)

$ \begin{aligned} & t(0)=\dfrac{9 \times 0}{5}+32=0+32=32 \\ & t(28)=\dfrac{9 \times 28}{5}+32=\dfrac{252+160}{5}=\dfrac{412}{5} \end{aligned} $

(ii)

(iii)

$ t(-10)=\dfrac{9 \times(-10)}{5}+32=9 \times(-2)+32=-18+32=14 $

(iv) It is given that $t(C)=212$

$\therefore 212=\dfrac{9 C}{5}+32$

$\Rightarrow \dfrac{9 C}{5}=212-32$

$\Rightarrow \dfrac{9 C}{5}=180$

$\Rightarrow 9 C=180 \times 5$

$\Rightarrow C=\dfrac{180 \times 5}{9}=100$

Thus, the value of $t$, when $t(C)=212$, is 100 .

Answer :

(i) $f(x)=2 - 3 x, x \in \mathbf{R}, x>0$

The values of $f(x)$ for various values of real numbers $x>0$ can be written in the tabular form as

Thus, it can be clearly observed that the range of fis the set of all real numbers less than 2.

i.e., range of $f=(- \infty, 2)$

Alter:

Let $x>0$

$\Rightarrow 3 x>0$

$\Rightarrow 2 - 3 x<2$

$\Rightarrow f(x)<2$

$\therefore$ Range of $f=(- \infty, 2)$

(ii) $f(x)=x^{2}+2, x$, is a real number

The values of $f(x)$ for various values of real numbers $x$ can be written in the tabular form as

Thus, it can be clearly observed that the range of fis the set of all real numbers greater than 2 .

i.e., range of $f=[2, \infty)$

Alter:

Let $x$ be any real number.

Accordingly,

$x^{2} \geq 0$

$\Rightarrow x^{2}+2 \geq 0+2$

$\Rightarrow x^{2}+2 \geq 2$

$\Rightarrow f(x) \geq 2$

$\therefore$ Range of $f=[2, \infty)$

(iii) $f(x)=x, x$ is a real number

It is clear that the range of fis the set of all real numbers.

$\therefore$ Range of $f=\mathbf{R}$