Question

lf f : [0, ∞) `rightarrow` [0, ∞) and f(x) = `x/(1 + x)`, then f is ______.

Options

• one-one and onto

• one-one but not onto

• onto but not one-one

• neither one-one nor onto

Answer 1

lf f : [0, ∞) `rightarrow` [0, ∞) and f(x) = `x/(1 + x)`, then f is one-one but not onto.

Explanation:

Here f : [0, ∞) `rightarrow` [0, ∞) i.e., domain is [0, ∞) and codomain is (0, ∞).

For one-one, f(x) = `x/(1 + x)`

`\implies` f'(x) = `1/(1 + x)^2 > 0, ∀  x ∈ [0, ∞)`

∴ f(x) is increasing in its domain.

Thus, f(x) is one-one in its domain.

For onto, we find the range

f(x) = `x/(1 + x)` i.e., y = `x/(1 + x)`

`\implies` y + yx = x

`\implies` x = `y/(1 - y)`

`\implies` `y/(1 - y) ≥ 0  "as"  x ≥ 0`

∴ 0 ≤ y ≠ 1

i.e., Range ≠ Codomain

∴ f(x) is one-one but not onto.