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If f : R → R be defined by f(x) = ex and g : R → R be defined by g(x) = x2. The mapping gof : R → R be defined by (gof)(x) = g[f(x)] ∀ x ∈ R, Then
(A) gof is bijective but f is not injective
(B) gof is injective and g is injective
(C) gof is injective but g is not bijective
(D) gof is surjective and g is surjective
If f : R → R be defined by f(x) = ex and g : R → R be defined by g(x) = x2. The mapping gof : R → R be defined by (gof)(x) = g[f(x)] ∀ x ∈ R, Then
(A) gof is bijective but f is not injective
(B) gof is injective and g is injective
(C) gof is injective but g is not bijective
(D) gof is surjective and g is surjective
The correct option (C) gof is injective but g is not bijective
Explanation:
f(x) = ex : R → R g(x) = x2 : R → R
g(f(x)) = g(ex) = (ex)2 = e2x ∀ x ∈ R
clearly g(f(x)) is injective and g(x) is not injective