Let f:[0,1] rightarrow mathbb{R} be an inject | vedclass

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(D) Given that $f:[0,1] \rightarrow \mathbb{R}$ is an injective continuous function with $-1 < f(0) < f(1) < 1$. Since $f$ is continuous and injective

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(D) Given that $f:[0,1] \rightarrow \mathbb{R}$ is an injective continuous function with $-1 Since $f$ is continuous and injective on $[0,1]$,it must be strictly monotonic. Given $f(0) The range of $f$ is $[f(0), f(1)]$,which is a subset of $(-1, 1)$. We are looking for functions $g:[-1,1] \rightarrow [0,1]$ such that $(g \circ f)(x) = x$ for all $x \in [0,1]$. This implies that for any $y \in [f(0), f(1)]$,$g(y) = f^{-1}(y)$. However,for $y \in [-1, 1] \setminus [f(0), f(1)]$,the function $g(y)$ can take any value in $[0, 1]$ because there is no restriction on $g$ for these values of $y$. Since the set $[-1, 1] \setminus [f(0), f(1)]$ is non-empty and contains infinitely many points,we can define $g(y)$ in infinitely many ways for $y$ in this set. Therefore,there are infinitely many such functions $g$.

Let $f:[0,1] \rightarrow \mathbb{R}$ be an injective continuous function that satisfies the condition $-1 < f(0) < f(1) < 1$. Then,the number of functions $g:[-1,1] \rightarrow [0,1]$ such that $(g \circ f)(x) = x$ for all $x \in [0,1]$ is

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