Verify whether relations between the elements of the sets given below are
functions or not.
(1) $\mathrm{A}=\{1,2,3,4\}, \mathrm{B}=\{3,5,7,9\}$, and relation is $f(x)=2 x+1, x \in \mathrm{A}$
(2) $\mathrm{P}=\left\{-\frac{1}{2}, 0,1\right\}, \mathrm{S}=\{10\}$, and rule is $k(x)=10, x \in \mathrm{P}$
(3) $\mathrm{A}=\{2,5,6\}, \mathrm{B}=\left\{1, \frac{3}{2}, \frac{9}{5}, \frac{11}{7}, \frac{13}{6}\right\}$, and rule is $y=\frac{2 x-1}{x+1}, x \in \mathrm{A}$
(4) $\mathrm{B}=\{-1,0,1,3\}, \mathrm{C}=\{-5,-3,-1,1,3\}$, and rule is $h(x)=2 x+3, x \in \mathrm{B}$
✓
Answer
(1) We shall determine the image for all $x \in \mathbf{A}$.
The relation is $f(x)=2 x+1$
For $x=1, f(1)=2(1)+1=3$
For $x=2, f(2)=2(2)+1=5$
For $x=3, f(3)=2(3)+1=7$
For $x=4, f(4)=2(4)+1=9$
Thus, for each element of set $\mathrm{A}$, there exists one and only one element in set $\mathrm{B}$ i.e. each element of set $\mathrm{A}$ is related with one and only one element of set $\mathrm{B}$ by the relation $f(x)=2 x+1$. Hence, $f(x)=2 x+1$ is a function.
(2) We shall determine the image for all $x \in P$ for $k(x)=10$
For $x=-\frac{1}{2}, k\left(-\frac{1}{2}\right)=10$
For $x=0, k(0)=10$
For $x=1, k(1)=10$
So, for every $x \in P$, there exists an element " 10 " in set $\mathrm{S}$.
Hence the relation $k(x)=10$ is a function.
(3) We shall determine the image for all $x \in \mathrm{A}$, for $y=\frac{2 x-1}{x+1}$
For $x=2, y=\frac{2(2)-1}{2+1}=\frac{4-1}{3}=\frac{3}{3}=1$
For $x=5, y=\frac{2(5)-1}{5+1}=\frac{10-1}{6}=\frac{9}{6}=\frac{3}{2}$
For $x=6, y=\frac{2(6)-1}{6+1}=\frac{12-1}{7}=\frac{11}{7}$
So for every $x \in \mathrm{A}$, there exists a unique coresponding element in set $\mathrm{B}$.
Hence, the relation $y=\frac{2 x-1}{x+1}$ is a function.
(4) We shall determine the image for all $x \in$ B. The relation is $h(x)=2 x+3$
For $x=-1, h(-1)=2(-1)+3=-2+3=1$
For $x=0, h(2)=2(0)+3=0+3=3$
For $x=1, h(1)=2(1)+3=2+3=5$, which not an element of set C
So, there is no corresponding element for $x=1$ in set $\mathrm{C}$.
Hence, the relation $h(x)=2 x+3$ is not a function.
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