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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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