The function f : [0, ∞) → [0, ∞) defined by \(f(x) = \frac{2x}{1

1.	The function f : [0, ∞) → [0, ∞) defined by \(f(x) = \frac{2x}{1+2x}\) is(a) one-one and onto (b) one-one but not onto (c) not one-one but onto (d) neither one-one nor onto
Answer» Answer : (b) one-one but not onto For all (x1, x2) ∈ [0, ∞) f (x1) = f (x2) ⇒\(\frac{2x_1}{1+2x_1}\) = \(\frac{2x_2}{1+2x_2}\) ⇒ 2x₁ + 4x₁x₂ = 2x₂ + 4x₁x₂ ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂ ⇒ f is one-one. Let y = f(x) = \(\frac{2x}{1+2x}\) ⇒ y + 2xy = 2x ⇒ y = 2x – 2xy = 2x (1 – y) ⇒ x = \(\frac{y}{2(1-y)}\) x is not defined when (1 – y) = 0, i.e., y = 1 ∈ [0, ∞) ⇒ f is not onto ∴ f is one-one, not onto.

The function f : [0, ∞) → [0, ∞) defined by \(f(x) = \frac{2x}{1+2x}\) is(a) one-one and onto (b) one-one but not onto (c) not one-one but onto (d) neither one-one nor onto

Answer»

Answer : (b) one-one but not onto

For all (x1, x2) ∈ [0, ∞)

f (x1) = f (x2)

⇒\(\frac{2x_1}{1+2x_1}\) = \(\frac{2x_2}{1+2x_2}\)

⇒ 2x₁ + 4x₁x₂ = 2x₂ + 4x₁x₂

⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂

⇒ f is one-one.

Let y = f(x) = \(\frac{2x}{1+2x}\)

⇒ y + 2xy = 2x

⇒ y = 2x – 2xy = 2x (1 – y)

⇒ x = \(\frac{y}{2(1-y)}\)

x is not defined when (1 – y) = 0, i.e., y = 1 ∈ [0, ∞)

⇒ f is not onto

∴ f is one-one, not onto.