If f : (0, ∞) → (0, ∞) and f (x) = \(\frac{x}{1+x}\) , then t

1.	If f : (0, ∞) → (0, ∞) and f (x) = \(\frac{x}{1+x}\) , then the function 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» Answer: (b) = one - one but not onto ∀ (x, y) ∈ (0, ∞) f (x) = f (y) ⇒ \(\frac{x}{1+x}\) = \(\frac{y}{1+y}\) ⇒ x + xy = y + xy ⇒ x = y ⇒ f is one-one Let z = f (x) = \(\frac{x}{1+x}\) ⇒ z + zx = x ⇒ z = x (1 – z) ⇒ x = \(\frac{z}{1-z}\) ⇒ x is not defined for z = 1 and 1 ∈ (0, ∞) ∴ 1 ∈ (0, ∞) has no pre-image in (0, ∞) ⇒ f is not onto

If f : (0, ∞) → (0, ∞) and f (x) = \(\frac{x}{1+x}\) , then the function 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»

Answer: (b) = one - one but not onto

∀ (x, y) ∈ (0, ∞)

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

⇒ x + xy = y + xy

⇒ x = y

⇒ f is one-one

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

⇒ z + zx = x

⇒ z = x (1 – z) ⇒ x = \(\frac{z}{1-z}\)

⇒ x is not defined for z = 1 and 1 ∈ (0, ∞)

∴ 1 ∈ (0, ∞) has no pre-image in (0, ∞)

⇒ f is not onto