JEE Main 2021 · 25 July · evening · Single correct · +4 / −1
Consider function f : A → B and g : B → C (A, B, C ⊆ R) such that (gof)⁻¹ exists, then :
- (a) f and g both are one-one
- (b) f and g both are onto
- (c) f is one-one and g is onto
- (d) f is onto and g is one-one
Answer and solution
Answer (c) f is one-one and g is onto
Solution
Maths sir · 5 steps
If gof is invertible, it is one-one and onto. Trace each property back to the function that carries it.
- Inverse means bijective. (g ° f)⁻¹ exists only when g ° f : A → C is both one-one and onto.
- One-one passes to f. Suppose f(x₁) = f(x₂). Apply g to get g(f(x₁)) = g(f(x₂)). Since g ° f is one-one, x₁ = x₂, so f is one-one.
- Onto passes to g. Take any c ∈ C. Since g ° f is onto, c = g(f(a)) for some a ∈ A. Then b = f(a) is an element of B with g(b) = c, so g is onto.
- Nothing more is forced. Take A = {1}, B = {1, 2}, C = {1}, with f(1) = 1 and g(1) = g(2) = 1. Here g ° f is a bijection, yet f is not onto and g is not one-one.
- So. The only guaranteed conclusion is that f is one-one and g is onto, which is option (c).
The idea to keep. In g ° f, one-one is inherited by the first function applied (f), and onto is inherited by the last function applied (g). The other two properties can fail.
Answer from the NTA final answer key. Solution by PrepWiser.