Best answer
(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.
Answer: \(f(x) = \dfrac{x}{1+|x|}\) is one-one and onto from R to \(\{y : -1 < y < 1\}\).
Sign and size of f(x): f(x) has the same sign as x, and \(|f(x)| = \dfrac{|x|}{1+|x|} < 1\). So f really maps R into \((-1, 1)\).
One-one: let \(f(x_1) = f(x_2)\). Then \(x_1, x_2\) have the same sign (or are both 0).
A positive and a negative x can never have equal images, because their images have opposite signs. So f is one-one.
Onto: take any y with \(-1 < y < 1\).
Every y in \((-1,1)\) has a pre-image, so f is onto. Hence f is one-one and onto.
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