Question

If the relation $f$ is defined by $f\left(x\right)=\{\begin{array}{cc}{x}^2,& 0\le x\le 3\\ 3x,& 3\le x\le 10\end{array}$
and the relation g is defined by $g\left(x\right)=\{\begin{array}{cc}{x}^2,& 0\le x\le 2\\ 3x,& 2\le x\le 10\end{array}$
then show that ${}^{\prime }{f}^{\prime }$ is a function but 'g' is not a function. [5 marks]

Answer 1

A relation ${}^{\prime }{R}^{\prime }$ from ${}^{\prime }{A}^{\prime }$ to ${}^{\prime }{B}^{\prime }$ is said to be function if every element of ${}^{\prime }{A}^{\prime }$ has an unique image in ${}^{\prime }{B}^{\prime }.$
The relation ${}^{\prime }{f}^{\prime }$ is given as $f\left(x\right)=\{\begin{array}{cc}{x}^2,& 0\le x\le 3\\ 3x,& 3\le x\le 10\end{array}$

The value $3$ belongs to both the intervals of $f\left(x\right)$.
In the $1$st interval: $f\left(3\right)=3^2=9$
and in $2$nd interval: $f\left(3\right)=3\times 3=9$
Value of $f\left(3\right)$ is same in both the intervals, so $f\left(x\right)$ is a function. [2 marks]

The relation ${}^{\prime }{g}^{\prime }$ is given as$g\left(x\right)=\{\begin{array}{cc}{x}^2,& 0\le x\le 2\\ 3x,& 2\le x\le 10\end{array}$

The value $2$ belongs to both the intervals of $g\left(x\right)$.
In the $1$st interval: $g\left(2\right)=2^2=4$
and in $2$nd interval: $g\left(2\right)=3\times 2=6$

For $x=2,g\left(2\right)$ has two values. So $g\left(x\right)$ is not a function. [2 marks]
Hence ${}^{\prime }{f}^{\prime }$ is a function but 'g' is not a function. [1 mark]