If $R_3=\{(x,|x|) \mid x$ is a real number $\}$ is a relation, then find domain and range of $R_3$.
We have, $R_3=\{(x,|x|) \mid x$ is real number$\}$
clearly, domain of $R_3=R$
Since, image of any real number under $R_3$ is positive real number or zero.
$$\therefore \quad \text { Range of } R_3=R^{+} \cup\{0\} \text { or }(0, \infty)$$
Is the given relation a function? Give reason for your answer.
(i) $h=\{(4,6),(3,9),(-11,6),(3,11)\}$
(ii) $f=\{(x, x) \mid x$ is a real number $\}$
(iii) $g=\left\{\left(x, \frac{1}{x}\right) x\right.$ is a positive integer $\}$
(iv) $s=\left\{\left(x, x^2\right) \mid x\right.$ is a positive integer $\}$
(v) $t=\{(x, 3) \mid x$ is a real number $\}$
(i) We have, $h=\{(4,6),(3,9),(-11,6),(3,11)\}$.
Since, 3 has two images 9 and 11. So, it is not a function.
(ii) We have, $f=\{(x, x) \mid x$ is a real number.
We observe that, every element in the domain has unique image. So, it is a function.
(iii) We have, $g=\left\{\left.\left(x, \frac{1}{x}\right) \right\rvert\, x\right.$ is a positive integer$\}$
For every $x$, it is a positive integer and $\frac{1}{x}$ is unique and distinct. Therefore, every element in the domain has unique image. So, it is a function.
(iv) We have, $s=\left\{\left(x, x^2\right) \mid x\right.$ is a positive integer$\}$
Since, the square of any positive integer is unique. So, every element in the domain has unique image. Hence, it is a function.
(v) We have, $t=\{(x, 3) \mid x$ is a real number$\}$.
Since, every element in the domain has the image 3. So, it is a constant function.
If $f$ and $g$ are real functions defined by $f(x)=x^2+7$ and $g(x)=3 x+5$. Then, find each of the following.
(i) $f(3)+g(-5)$
(ii) $f\left(\frac{1}{2}\right) \times g(14)$
(iii) $f(-2)+g(-1)$
(iv) $f(t)-f(-2)$
(v) $\frac{f(t)-f(5)}{t-5}$, if $t \neq 5$
Given, $f$ and $g$ are real functions defined by $f(x)=x^2+7$ and $g(x)=3 x+5$.
(i) $f(3)=(3)^2+7=9+7=16$ and $g(-5)=3(-5)+5=-15+5=-10$
$$\therefore f(3)+g(-5)=16-10=6$$
(ii) $f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+7=\frac{1}{4}+7=\frac{29}{4}$
and $g(14)=3(14)+5=42+5=47$
$$\therefore \quad f\left(\frac{1}{2}\right) \times g(14)=\frac{29}{4} \times 47=\frac{1363}{4}$$
(iii) $$\begin{aligned} & f(-2)=(-2)^2+7=4+7=11 \text { and } g(-1)=3(-1)+5=-3+5=2 \\ & \therefore \quad f(-2)+g(-1)=11+2=13 \end{aligned}$$
$\begin{aligned} \text { (iv) } f(t)=t^2+7 \text { and } f(-2) & =(-2)^2+7=4+7=11 \\ \therefore \quad f(t)-f(-2) & =t^2+7-11=t^2-4\end{aligned}$
$\begin{aligned} & \text { (v) } f(t)=t^2+7 \text { and } f(5)=5^2+7=25+7=32 \\ & \therefore \quad \frac{f(t)-f(5)}{t-5} \text {, if } t \neq 5\end{aligned}$
$\begin{aligned} & =\frac{t^2+7-32}{t-5} \\ & =\frac{t^2-25}{t-5}=\frac{(t-5)(t+5)}{(t-5)} \\ & =t+5\quad [\because t \neq 5]\end{aligned}$
Let $f$ and $g$ be real functions defined by $f(x)=2 x+1$ and $g(x)=4 x-7$.
(i) For what real numbers $x, f(x)=g(x)$ ?
(ii) For what real numbers $x, f(x)< g(x)$ ?
We have, $$f(x)=2 x+1 \text { and } g(x)=4 x-7$$
(i) $\because$ $$f(x)=g(x)$$
$$\begin{aligned} & \Rightarrow \quad 2 x+1=4 x-7 \Rightarrow 2 x=8 \\ & \therefore \quad x=4 \end{aligned}$$
(ii) $\because\quad f(x)< g(x)$
$\begin{array}{lc}\Rightarrow & 2 x+1<4 x-7 \\ \Rightarrow & 2 x-4 x+1<4 x-7-4 x \\ \Rightarrow & -2 x+1<-7 \\ \Rightarrow & -2 x<-7-1 \\ \Rightarrow & -2 x<-8 \\ \Rightarrow & \frac{-2 x}{-2}>\frac{-8}{-2} \\ \therefore & x>4\end{array}$
If $f$ and $g$ are two real valued functions defined as $f(x)=2 x+1$ and $g(x)=x^2+1$, then find
(i) $f+g$
(ii) $f-g$
(iii) $f g$
(iv) $\frac{f}{g}$
We have, $f(x)=2 x+1$ and $g(x)=x^2+1$
$$\begin{aligned} \text{(i)}\quad (f+g)(x) & =f(x)+g(x) \\ & =2 x+1+x^2+1=x^2+2 x+2 \end{aligned}$$
$$\begin{aligned} \text{(i)}\quad (f-g)(x) & =f(x)-g(x)=(2 x+1)-\left(x^2+1\right) \\ & =2 x+1-x^2-1=2 x-x^2=x(2-x) \end{aligned}$$
$$\begin{aligned} \text{(iii)}\quad (f g)(x) & =f(x) \cdot g(x)=(2 x+1)\left(x^2+1\right) \\ & =2 x^3+2 x+x^2+1=2 x^3+x^2+2 x+1 \end{aligned}$$
(iv) $\frac{f}{g}(x)=\frac{f(x)}{g(x)}=\frac{2 x+1}{x^2+1}$