Show that function f : R → {x ∈ R : - 1 < x < 1} be defined by f (x) = x/(1 + |x|), x ∈ R is one-one and onto function

Solution:

f : R → {x ∈ R : - 1 < x < 1} be defined by f (x) = x/(1 + |x|), x ∈ R.

For one-one:

f (x) = f (y) where x, y ∈ R

x/(1 + |x|) = y/(1 + |y|)

If x is positive and y is negative,

⇒ 2xy = x - y

Since, x is positive and y is negative,

x > y

⇒ x - y > 0

2xy is negative.

2xy ≠ x - y

The case of x is positive and y is negative, can be ruled out.

⇒ x and y have to be either positive or negative.

If x and y are positive,

f (x) = f (y)

x/(1 + x) = y/(1 + y)

⇒ x - xy = y - xy

⇒ x = y

⇒ f is one-one.

For onto:

Let y ∈ R such that - 1 < y < 1.

If x is negative, then there exists.

x = y/(1 + y) ∈ R such that

f (x) = f (y/(1 + y)) = (y/(1 + y))/[1 + |y/(1 + y)|]

= (y/(1 + y))/[1 + (- y/(1 + y))] = y/(1 + y - y) = y

If x is positive, then there exists x = y/(1 + y) ∈ R such that

f (x) = f (y/(1 - y)) = (y/(1 - y))/[1 + |y/(1 - y)|]

= (y/(1 - y))/[1 + (- y/(1 - y))]

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

⇒ f is onto.

Hence, f is one-one and onto

---

NCERT Solutions for Class 12 Maths - Chapter 1 Exercise ME Question 4

Show that function f : R → {x ∈ R : - 1 < x < 1} be defined by f (x) = x/(1 + |x|), x ∈ R is one-one and onto function

Summary:

Hence we have concluded that function f : R → {x ∈ R : - 1 < x < 1} be defined by f (x) = x/(1 + |x|), x ∈ R is one-one and onto function