Let f:[0,1]→[–1,1] and g:[–1,1]→[0,2] be two functions such that g is injective and g∘f:[0,1]→[0,2] is surjective. Then
A
f must be injective but need not be surjective
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B
f must be surjective but need not be injective
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C
f must be bijective
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D
f must be a constant function
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Solution
The correct option is Bf must be surjective but need not be injective g∘f is surjective. ⇒ For each g(f(x))∈[0,2],∃f(x)∈Domain ofg ⇒f(x)∈[−1,1]=Rf ⇒f is surjective.
Given, g is injective. So, given x1,x2∈Dom(g∘f) g(f(x1))=g(f(x2))⇒f(x1)=f(x2) We don't have enough evidence to conclude x1=x2 further. So, we can't conclude about injectivity of f from the given conditions.