NCERT Solutions
Class 12 Maths
Relations and Functions
Misc. Ex. Q4
Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) = x/(1+|x|), x ∈ R is one – one and onto function.
It is given that f: R → {x ∈ R: −1 < x < 1} is defined as f(x) = x/(1+|x|), x ∈ R.
For one – one:
Suppose f(x) = f(y), where x, y ∈ R.
⇒ x/(1 + |x|) = y/(1 + |y|)
It can be observed that if x is positive and y is negative,
Then, we have
x/(1 + x) = y/(1 – y)
⇒ 2xy = x − y
Since, x is positive and y is negative
x > y ⇒ x − y > 0
But, 2xy is negative.
Then 2xy ≠ x − y
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out.
So, x and y have to be either positive or negative.
When x and y are both positive, we have
f(x) = f(y)
⇒ x/(1 + x) = y/(1 + y)
⇒ x + xy = y + xy
⇒ x = y
When x and y are both negative, we have
f(x) = f(y)
⇒ x/(1 – x) = y/(1 – y)
⇒ x − xy = y − xy
⇒ x = y
So, f is one – one.
For onto:
Now, let y ∈ R such that −1 < y < 1
If y 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 y 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
So, f is onto.
Hence, f is one – one and onto.