If f : I → I be defined by f(x) = [x + 1], where [.] denotes the greatest integer function, then f - 1(x) is equal to

x – 1 [x + 1] $\frac{1}{[x – 1]}$ $\frac{1}{x + 1}$

x – 1

since the domain of f is I, hence f(x) is one – one and onto and [x] = x ⇒ f(x) = x + 1 ⇒ f – 1(x) = x – 1 Hence (1) is the correct answer.