Let f: [0, 1] → R be an injective continuous function that satisifes the condition -1

Question

Let f: [0, 1] → R be an injective continuous function that satisifes the condition -1 < f (0) < f (1) < 1 Then, the number of functions g: [-1, 1] → [0, 1] such that (gof) (x) =x for all x ∈[0,1] is (A) 0 (B) 1 (C) more than 1, but finite (D) infinite

Tardigrade Admin · Accepted Answer

We have, -1< f (0)< f (1)< 1 g=[-1, 1] → [0,1] go f (x)=x, ∀ x ∈ [0, 1] Only condition that g (x) should satisfy for go f (x)=x, ∀ x ∈ [0, 1] is that g(x) should attain all values in [0,1] when range of f (x) a subset of (-1,1) is used as image for g(x). Thus, there can be infinite such functions g(x) with domain [-1,1] and range [0, 1]

Q. Let f:[0,1]→R be an injective continuous function that satisifes the condition −1<f(0)<f(1)<1 Then, the number of functions g:[−1,1]→[0,1] such that (gof)(x)=x for all x∈[0,1] is

0

1

more than 1, but finite

infinite

Q. Let $f : [0, 1] \to R$ be an injective continuous function that satisifes the condition $- 1 < f (0) < f (1) < 1$ Then, the number of functions $g : [- 1, 1] \to [0, 1]$ such that $(g o f) (x) = x$ for all $x \in [0, 1]$ is