Prove that function f : R → R, f (x) = x2 + x is a many-one into fun

1.	Prove that function f : R → R, f (x) = x2 + x is a many-one into function?
Answer» Many-one Let a₁, a₂be any two arbitrary elements of R, then f (a₁) = f (a₂) ⇒ a₁² + a₁ = a₂² + a₂ ⇒ a₁² – a₂² + a₁ + a₂ = 0 ⇒ (a₁ – a₂) (a₁ + a₂) + (a₁ – a₂) = 0 ⇒ (a₁ – a₂) (a₁ + a₂+ 1) = 0 ⇒ a₁ – a₂ = 0 or a₁ + a₂ + 1 = 0 ⇒ a₁ = a₂ or a₁ + a₂ = – 1 ∈ R ⇒ Both the inferences can be true. So, f (a₁) = f (a₂) does not necessarily imply a₁ = a₂ ⇒ f is many-one. Into Let y = x² + x, then for all y ∈ R, there does not exist all x ∈ R, as for y = – 1, – 2, ..., etc. There is no pre-image in R. Hence f is an into function. ⇒ f is many-one into function.

Prove that function f : R → R, f (x) = x2 + x is a many-one into function?

Answer»

Many-one

Let a₁, a₂be any two arbitrary elements of R,

then f (a₁) = f (a₂) ⇒ a₁² + a₁ = a₂² + a₂

⇒ a₁² – a₂² + a₁ + a₂ = 0

⇒ (a₁ – a₂) (a₁ + a₂) + (a₁ – a₂) = 0

⇒ (a₁ – a₂) (a₁ + a₂+ 1) = 0

⇒ a₁ – a₂ = 0 or a₁ + a₂ + 1 = 0

⇒ a₁ = a₂ or a₁ + a₂ = – 1 ∈ R

⇒ Both the inferences can be true.

So, f (a₁) = f (a₂) does not necessarily imply a₁ = a₂

⇒ f is many-one.

Into

Let y = x² + x, then for all y ∈ R, there does not exist all x ∈ R, as for y = – 1, – 2, ..., etc. There is no pre-image in R. Hence f is an into function.

⇒ f is many-one into function.

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