Show that f : [- 1, 1] → R, given by f (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)

Solution:

A function is a process or a relation that associates each element 'a' of a non-empty set A, to a single element 'b' of another non-empty set B.

The inverse of a function f(x) is a function g(x) such that if f maps an element ′a′ to an element ′b′, g maps ′b′ to ′a′.

According to the given problem,

f : [- 1, 1] → R, given by f (x) = x/(x + 2)

For one-one

f (x) = f (y)

x/(x + 2) = 2y/(1 - y)

⇒ xy + 2x = xy + 2 y

⇒ 2x = 2 y

⇒ x = y

Therefore,

f is a one-one function.

It is clear that f : [- 1, 1] → R is onto.

Therefore,

f : [- 1, 1] → R is one-one and onto, and therefore,

the inverse of the function

f : [- 1, 1] → R 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 same x∈ [- 1, 1]

⇒ y = x/(x + 2)

⇒ xy + 2 y = x

⇒ x (1 - y) = 2 y

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)] = 2 [x/(x + 2)]/[1 - x/(x + 2)] = 2x/(x + 2 - x)] = 2x/2 = x

(fog )(x) = f (g (y)) = f [2y/(1 - y)] = [2y/(1 - y)]/[2y/(1 - y) + 2] = 2y/(2y + 2 - 2y) = 2y/2 = y

⇒ gof = I_{[- 1, 1]} and fog = I_{Range f}

⇒ f ^{- 1} = g

⇒ f ^-1 (y) = 2y/(1 - y), y ≠ 1

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NCERT Solutions for Class 12 Maths - Chapter 1 Exercise 1.3 Question 6

Show that f : [- 1, 1] → R, given by f (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)

Summary:

f : [- 1, 1] → R, given by f (x) = x/(x + 2) is one-one. the inverse of the function f : [- 1, 1] → Range f is equal to f ^-1 (y) = 2y/(1 - y), y ≠ 1