Question

Show that the function f : R → R defined by `f(x) = x/(x^2 + 1), ∀  x ∈ R` is neither one-one nor onto.

Answer 1

Given: `f : R → R; f(x) = x/(1 + x^2)`

To show that f is neither one-one nor onto

(i) f is one-one: Let x₁, x₂ ∈ R (domain)

And f(x₁) = f(x₂)

⇒ `(x_1)/(1 + x_1^2) = (x_2)/(1 + x_2^2)`

⇒ `x_1(1 + x_2^2) = x_2(1 + x_1^2)`

⇒ `x_1 + x_1 * x_2^2 - x_2 - x_2x_1^2 = 0`

⇒ (x₁ – x₂)(1 – x₁·x₂) = 0

Taking x₁ = 4, x₂ = `1/4` ∈ R

f(x₁) = f(4) = `4/17`

`f(x_2) = f(1/4) = 4/17`

∴ f not is one-one.

(ii) f is onto: Let y ∈ R (co-domain)

f(x) = y

⇒ `x/(1 + x^2) = y`

⇒ y·(1 + x²) = x

⇒ yx² + y – x = 0

⇒ `x = (1 +- sqrt(1 - 4y^2))/(2y)`

Since, x ∈ R,

∴ 1 – 4y² ≥ 0

⇒ `- 1/2 ≤ y ≤ 1/2`

So Range `(f) ∈ [- 1/2, 1/2]`

Range (f) ≠ R (Co-domain)

∴ f is not onto.

Hence, f is neither one-one nor onto.