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Question
Show that the function f : R → {x ∈ R : −1 < x < 1} defined by f(x) = `x/(1 + |x|)`, x ∈ R is one-one and onto function.
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Solution
It is given that f : R → {x ∈ R : −1 < x < 1} is defined as f(x) = `x/(1+ |x|)`, x ∈ R.
Suppose f(x) = f(y), where x, y ∈ R.
⇒ `x/(1 +| x|) = y/(1 - |y|)`
⇒ 2xy = x − y
⇒ `x/(1 + x) = y/(1 - y) `
⇒ x + xy = y + xy
⇒ 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
∴ 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/(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 − yx
⇒ x = y
∴ f is one-one.
Now, 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.
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