Let f:[0,1] rightarrow [-1,1] and g:[-1,1] ri | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $h(x) = g(f(x))$. We are given that $h: [0,1] \rightarrow [0,2]$ is surjective. Since $h$ is surjective,the range of $h$ is equal to its codom

Vedclass · Accepted Answer

$f$ must be surjective but need not be injective

Vedclass · Answer

(B) Let $h(x) = g(f(x))$. We are given that $h: [0,1] ightarrow [0,2]$ is surjective. Since $h$ is surjective,the range of $h$ is equal to its codomain,which is $[0,2]$. Since $h(x) = g(f(x))$,the range of $h$ is a subset of the range of $g$. Thus,the range of $g$ must contain $[0,2]$. However,the codomain of $g$ is $[0,2]$,so the range of $g$ must be exactly $[0,2]$. This implies that $g$ is surjective. Since $g$ is given as injective and is now shown to be surjective,$g$ is a bijection. For $h = g \circ f$ to be surjective,$f$ must be surjective. If $f$ were not surjective,there would exist some $y \in [-1,1]$ such that $y 
otin 	ext{Range}(f)$. Since $g$ is a bijection,$g(y)$ would not be in the range of $g \circ f$,contradicting the surjectivity of $h$. Therefore,$f$ must be surjective.

Let $f:[0,1] \rightarrow [-1,1]$ and $g:[-1,1] \rightarrow [0,2]$ be two functions such that $g$ is injective and $g \circ f: [0,1] \rightarrow [0,2]$ is surjective. Then,

Similar Questions

$f: N \rightarrow N, f(x) = x^3$ is . . . . . . .

The function $f: R \rightarrow R$ defined by $f(x)=\frac{x}{\sqrt{1+x^2}}$ is

The mapping $f: R \to R$ defined as $f(x) = \cos x, x \in R$ is:

Let $f : R \rightarrow R$ be defined as $f(x) = 3^{-|x|} - 3^x + \operatorname{sgn}(e^{-x}) + 2$ (where $\operatorname{sgn}(x)$ denotes the signum function of $x$). Then which one of the following is correct?

Let $f: R \rightarrow R$ be defined as $f(x)=x^{4}$. Choose the correct answer.

Vedclass Test Series

Exam Paper Generator

Online Exam Module