∴(gof)−1 exist ⇒ gof=g(f(x)) is bijective
⇒f must be one-one and g must be onto.
JEE Main 2021 — Mathematics Algebra
Consider function f:A→B and g:B→C(A,B,C⊆R) such that (gof)−1 exists, then:
Held on 25 Jul 2021 · Verified 6 Jul 2026.
f and g both are one-one
f and g both are onto
f is one-one and g is onto
f is onto and g is one-one
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