The domain for which the functions defined by $f(x)=3 x^2-1$ and $g(x)=3+x$ are equal to
Let $f$ and $g$ be two real functions given by
$$\text { and } \quad \begin{aligned} f & =\{(0,1),(2,0),(3,-4),(4,2),(5,1)\} \\ & =\{(1,0),(2,2),(3,-1),(4,4),(5,3)\} \end{aligned}$$
then the domain of $f \cdot g$ is given by........... .
$$\begin{aligned} \text{We have,}\quad f & =\{(0,1),(2,0),(3,-4),(4,2),(5,1)\} \\ \text{and}\quad g & =\{(1,0),(2,2),(3,-1),(4,4),(5,3)\} \end{aligned}$$
$\begin{array}{ll}\therefore & \text { Domain of } f=\{0,2,3,4,5\}, \\ \text { and } & \text { Domain of } g=\{1,2,3,4,5\}\end{array}$
$$\therefore$$ Domain of $(f . g)=$ Domain of $f~\cap$ Domain of $g=\{2,3,4,5\}$
Let $f=\{(2,4),(5,6),(8,-1),(10,-3)\}$ and $g=\{(2,5),(7,1),(8,4),(10,13),(11,5)\}$ be two real functions. Then, match the following.
| Column I | Column II | ||
|---|---|---|---|
| (i) | $f-g$ | (a) | $$\left\{ {\left( {2,{4 \over 5}} \right),\left( {8,{{ - 1} \over 4}} \right),\left( {10,{{ - 3} \over {13}}} \right)} \right\}$$ |
| (ii) | $f+g$ | (b) | $$\{ (2,20),(8, - 4),(10, - 39)\} $$ |
| (iii) | $f.g$ | (c) | $$\{ (2, - 1),(8, - 5),(10, - 16)\} $$ |
| (iv) | $\frac{f}{g}$ | (d) | $$\{ (2,9),(8,3),(10, - 10)\} $$ |
The domain of $f-g, f+g, f \cdot g, \frac{f}{g}$ is domain of $f \cap$ domain of $g$. Then, find their images.
$$\begin{aligned} \text{We have,}\quad f & =\{(2,4),(5,6),(8,1),(10,-3)\} \\ \text{and}\quad g & =\{(2,5),(7,1),(8,4),(10,13),(11,5)\} \end{aligned}$$
So, $f-g, f+g, f . g, \frac{f}{g}$ are defined in the domain (domain of $f \cap$ domain of $g$ )
$$\text { i.e., }\{2,5,8,10\} \cap\{2,7,8,10,11\} \Rightarrow\{2,8,10\}$$
$$\begin{array}{l} \text{(i)}\quad (f-g)(2)=f(2)-g(2) =4-5=-1 \\ (f-g)(8) =f(8)-g(8)=-1-4=-5 \\ (f-g)(10) =f(10)-g(10)=-3-13=-16 \\ \therefore \quad f-g =\{(2,-1),(8,-5),(10,-16)\} \end{array}$$
$$\begin{array}{rlrl} \text{(ii)}\quad (f+g)(2)=f(2)+g(2) & =4+5=9 \\ (f+g)(8) & =f(8)+g(8)=-1+4=3 \\ (f+g)(10) & =f(10)+g(10)=-3+13=10 \\ & f+g =\{(2,9),(8,3),(10,10)\} \end{array}$$
$$\begin{aligned} \text{(iii)}\quad (f \cdot g)(2)=f(2) \cdot g(2) & =4 \times 5=20 \\ (f \cdot g)(8) & =f(8) \cdot g(8)=-1 \times 4=-4 \\ (f \cdot g)(10) & =f(10) \cdot g(10)=-3 \times 13=-39 \\ \therefore \quad f g & =\{(2,20),(8,-4),(10,-39)\} \end{aligned}$$
$$\begin{aligned} \text{(iv)}\quad & \left(\frac{f}{g}\right)(2)=\frac{f(2)}{g(2)}=\frac{4}{5} \\ & \left(\frac{f}{g}\right)(8)=\frac{f(8)}{g(8)}=\frac{-1}{4} \\ & \left(\frac{f}{g}\right)(10)=\frac{f(10)}{g(10)}=\frac{-3}{13} \\ & \therefore \quad \frac{f}{g}=\left\{\left(2, \frac{4}{5}\right),\left(8,-\frac{1}{4}\right),\left(10, \frac{-3}{13}\right)\right\} \end{aligned}$$
Hence, the correct matches are (i) $\rightarrow$ (c), (ii) $\rightarrow$ (d), (iii) $\rightarrow$ (b), (iv) $\rightarrow$ (a).
The ordered pair $(5,2)$ belongs to the relation $$R=\{(x, y): y=x-5, x, y \in Z\}$$
If $P=\{1,2\}$, then $P \times P \times P=\{(1,1,1),(2,2,2),(1,2,2),(2,1,1)\}$