If f and g are functions such that fog is onto then
A
f is onto
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B
g is onto
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C
gof is onto
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D
Neither f nor g is onto
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Solution
The correct option is Af is onto Given that, fog is onto. ⇒ Range(fog(x))=Codomain f(g(x)) ⇒ Range of fog(x)= Codomain of f(x)...(1) Since Range of f(x)⊇ Range of fog(x) and Range of f(x)⊆ Codomain of f(x), Using (1) we get Range of f(x)=Codomain of f(x). Hence, f(x) is onto.