The given relation is 2y=(x+1)(x−1).
Given: f(x)=1
∴ We need to evaluate x at y=1.
2(1)=(x+1)(x−1)
⇒2=x⋅x+x−x−1
⇒2+1=x2
⇒x2=3
Here, x can have two values.
For x=√3,−√3
Here, x is independent variable, having two sets of values that results to a single value, i.e., f(x)=1.
∴ This relation is a function.