FunctionHard
Question
If ′f′ and ′g′ are bijective functions and gof is defined, then, gof must be:
Options
A.injective
B.surjective
C.bijective
D.into only
Solution
f(g(x1)) = f(g(x2))
⇒ g(x1) = g(x2)
as f is one - one function
⇒ x1 = x2
as g is one - one function
hence f(g(x1)) = f(g(x2))
⇒ x1 = x2
⇒ f(g(x)) is one - one function
⇒ g(x1) = g(x2)
as f is one - one function
⇒ x1 = x2
as g is one - one function
hence f(g(x1)) = f(g(x2))
⇒ x1 = x2
⇒ f(g(x)) is one - one function
Create a free account to view solution
View Solution FreeMore Function Questions
For every pair of continuous functions f, g : [0, 1] →R such that max {f(x) : x ∈ [0, 1]} = max {g(x) : x ...If a2 + b2 + c2 = 1, then range of ab + bc + ca is-...Let f(x) = a x2 + b x + c, where a, b, c are rational and f : Z → Z, where Z is the set of integers. Then a + b is...If f(x) = sin ln , then...If f(x) = cos(ln x), then f(x) f(y) - has the value...