NCERT Solutions
Class 12 Maths
Relations and Functions
Ex. 1.3 Q8
Consider f: R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by f−1(y) = √(y – 4), where R+ is the set of all non-negative real numbers.
Given, f : R+ → [4, ∞) is given as f(x) = x2 + 4
For one – one:
Let f(x) = f(y)
⇒ x2 + 4 = y2 + 4
⇒ x2 = y2
⇒ x = y [as x = y ∈ R+]
So, f is a one – one function.
For onto:
For y ∈ [4, ∞), let y = x2 + 4
⇒ x2 = y − 4 ≥ 0 [as y ≥ 4]
⇒ x = √(y – 4) ≥ 0
Therefore, for any y ∈ [4, ∞), there exists x = √(y – 4) ∈ R+, such that
f(x) = f(√(y -4)) = (√(y -4))2 + 4 = y - 4 + 4 = y
So, f is onto.
Thus, f is one – one and onto and therefore, f−1 exists.
Let us define g: [4, ∞) → R+ by g(y) = √(y – 4)
Now,
(gof)(x) = g(f(x)) = g(x2 + 4) = √{x2 + 4 – 4} = √x2 = x
and
(fog)(y) = f(g(y)) = f(√(y - 4)) = (√(y – 4))2 + 4 = y - 4 + 4 = y
So, gof = fog = IR
Hence, f is invertible and the inverse of f is given by
f-1(y) = g(y) = √(y – 4)