Question

Which of the following functions is NOT one-one?

विकल्प

• f : R `rightarrow` R defined by f(x) = 6x – 1.

• f : R `rightarrow` R defined by f(x) = x² + 7.

• f : R `rightarrow` R defined by f(x) = x³.

• f : R – {7} `rightarrow` R defined by f(x) = `(2x + 1)/(x - 7)`

Answer 1

f : R `rightarrow` R defined by f(x) = x² + 7.

Explanation:

(a) We have f(x) = 6x – 1, x ∈ R.

Let f(x₁) = f(x₂), x₁, x₂ ∈ R

`\implies` 6x₁ – 1 = 6x₂ – 1

`\implies` 6x₁ = 6x₂

`\implies` x₁ = x₂.

∴ ‘f’ is one-one.

(b) We have f(x) = x² + 7, x ∈ R

f(–2) = (–2)² + 7 = 11,

f(2) = (2)² + 7 = 11

∴ The images of distinct elements –2 and 2 of R are equal.

∴ ‘f’ cannot be one-one.

(c) f(x) = x³, x ∈ R.

Let f(x₁) = f(x₂), x₁, x₂ ∈ R

`\implies x_1^3 = x_2^3`

`\implies x_1^3 - x_2^3` = 0

`\implies (x_1 - x_2) (x_1^2 + x_1x_2 + x_2^2)` = 0

`\implies` x₁ – x₂ = 0

`\implies` x₁ = x₂,

because the other factor cannot be zero

∴ ‘f’ is one-one

(d) We have f(x) = `(2x + 1)/(x - 7), x ∈ R - {7}`.

Let f(x₁) = f(x₂), x₁, x₂ ∈ R – {7}.

`\implies (2x_1 + 1)/(x_1 - 7) = (2x_2 + 1)/(x_2 - 7)`

`\implies` – 15x₁ = – 15x₂

`implies` x₁ = x₂

∴ f is one-one.