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

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(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

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$f$ must be surjective but need not be injective

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(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.

List-$I$	List-$II$
$(A) f(-5)+f(0)+f(-1)$	$(I) 16$
$(B) f(f(5)+10f(-3))$	$(II) 40$
$(C) f(f(-4))$	$(III) -31$
$(D) f(f(f(1)))$	$(IV) -12$
	$(V) 19$

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,

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