NCERT Solutions
Class 12 Maths
Relations and Functions
Ex. 1.3 Q6
Show that f : [−1, 1] → R, given byf(x) = x/(x+2) is one – one. Find the inverse of the function f : [−1, 1] → Range f
(Hint: For y ∈ Range f, y = f(x) = x/(x+2), for some x in [−1, 1], i.e., x = 2y/(1−y)
Given, f : [−1, 1] → R is given as f(x) = x/(x+2)
For one – one:
Let f(x) = f(y)
⇒ x/(x + 2) = y/(y + 2)
⇒ xy +2x = xy +2y
⇒ 2x = 2y
⇒ x = y
So, f is a one – one function.
It is clear that f : [−1, 1] → Range f is onto.
So, f : [−1, 1] → Range f is one – one and onto and therefore, the inverse of the function
f : [−1, 1] → Range f exists.
Let g : Range f → [−1, 1] be the inverse of f.
Let y be an arbitrary element of range f.
Since f : [−1, 1] → Range f is onto, we have
y = f(x) for some x ∈ [−1, 1]
⇒ y = x/(x + 2)
⇒ xy + 2y = x
⇒ x(1-y) = 2y
⇒ x = 2y/(1 – y), y ≠ 1
Now, let us define g : Range f → [−1, 1] as
g(y) = 2y/(1 – y), y ≠ 1
Now, (gof)(x) = g(f(x))
= g(x/(x + 2))
= 2x/(x + 2)/{1 - x/(x + 2)}
= 2x/(x + 2 - x)
= 2x/2
= x
and (fog)(x) = f(g(x))
= f(2y/(1- y))
= 2y/(1- y)/{ 2y/(1- y) + 2}
= 2y/(2y + 2 – 2y)
= 2y/2
= y
So, gof = x = I[-1, 1] and fog = y = IRange f
So, f-1 = g
=> f-1(y) = 2y/(1 - y), y ≠ 1