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,

• A. 'f' is a function and 'g' is not a function
• B. 'g' is a function and 'f' is not a function
• C. Both 'f' and 'g' are functions
• D. neither 'f' nor 'g' are functions

Answer 1

The correct option is A

'f' is a function and 'g' is not a function

A relation 'R' from 'A' to 'B' is said to be function if every element of 'A' has an unique image in 'B'.
The relation 'f' 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 1st interval: $f\left(3\right)=3^2=9$
and in 2nd interval: $f\left(3\right)=3\times 3=9$
Value of f(3) is same in both the intervals, so f(x) is a function.

The relation 'g' 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 1st interval: $g\left(2\right)=2^2=4$
and in 2nd interval: $g\left(2\right)=3\times 2=6$

For x=2, g(2) has two values. So g(x) is not a function.
Hence 'f' is a function but 'g' is not a function.