Question

Show that the function f : R → { x ∈ R : – 1 – x < 1} defined by f(x) = $\frac{x}{1+|x|}$ x ∈ R is one-one and onto.

Answer 1

(i) Let x >0, f(x) = $\frac{x}{1+x}$
So, f(x1) = f(x2)
⇒ $\frac{x_1}{1+x_1}$ = $\frac{x_2}{1+x_2}$
or x1(1 + x2) = x2(1 + x1)
or x1 + x1x2 = x2 + x1x2
⇒ x1 = x2
When x< 0, f(x) = $\frac{x}{1-x}$
So, f(x1) = f(x2)
⇒ $\frac{x_1}{1-x_1}$ = $\frac{x_2}{1-x_2}$
or x1(1 – x2) = x2(1 – x1)
or x1 – x1x2 = x2 – x2x1
⇒ x1 = x2

(ii) As -1 < x < 1, f(x) = $\frac{x}{1+|x|}$ lies between and – $\frac{1}{2}$ and $\frac{1}{2}$ When x >0, y = $\frac{x}{1+x}$.
∴ (1 + x)y = x
or xy – x = – y
∴ x(y-1) = – y
or x = $\frac{-y}{y-1}$ = $\frac{1}{1-y}$.
When x < 0, f{x) = $\frac{x}{1-x}$ = y
∴ x = y – xy
x(1 + y) = y
∴ x = $\frac{y}{1+y}$
In both cases, each value of y in codomain of f has a unique value in its domain.
Hence, f is onto.
Thus, f is one-one and onto.