Question
If $$f:\left[ {0,\infty } \right) \to \left[ {0,\infty } \right)$$ and $$\,f\left( x \right) = \frac{x}{{1 + x}}$$ then $$f$$ is
A.
one-one and onto
B.
one-one but not onto
C.
onto but not one-one
D.
neither one-one nor onto
Answer :
one-one but not onto
Solution :
Given that $$f:\left[ {0,\infty } \right) \to \left[ {0,\infty } \right)$$
Such that $$\,f\left( x \right) = \frac{x}{{1 + x}}$$
Then $$f'\left( x \right) = \frac{{1 + x - x}}{{{{\left( {1 + x} \right)}^2}}} = \frac{1}{{{{\left( {1 + x} \right)}^2}}} > 0\forall x$$
$$\therefore f$$ is an increasing function $$ \Rightarrow f$$ is one-one.
Also, $${D_f} = \left[ {0,\infty } \right)$$
And for range let $$\frac{x}{{1 + x}} = y \Rightarrow x = \frac{y}{{1 - y}}$$
$$x \geqslant 0 \Rightarrow 0 \leqslant y < 1$$
$$\therefore {R_f} = \left[ {0,1} \right) \ne \,{\text{Co - domain}}$$
$$\therefore f$$ is not onto.