f : R → R, f(x) = x2 + x is:(A) One-one one(B) One-one into(C) Many

1.	f : R → R, f(x) = x2 + x is:(A) One-one one(B) One-one into(C) Many-one onto(D) Many one onto
Answer» Answer is (D) Given: f : R → R and f(x) = x² + x where R is a set of real numbers. Let x₁, x₂ ∈ R are such that f(x₁) = f(x₂) f(x₁) = f(x₂) ⇒ x₁² + x₁ = x₂² + x₂ ⇒ x₁² – x₂² + x₁ – x₂ = 0 ⇒ (x₁ – x₂)(x₁ + x₂) + 1 (x₁ – x₂) = 0 ⇒ (x₁ – x₂) (x₁ + x₂ + 1} = 0 ⇒ x₁ = x₂, x₁ = -(x₂ + 1) ∀ x₁, x₂ ∈ R Here, element of set A relates to two elements of set B. So, it is a many-one function. Again, let y ∈ R (co-domain) If possible then let pre-image of y is x in domain R. then f(x) = y ⇒ x² + x = y ⇒ x(x + 1) = y ⇒ x = y, x = y – 1 If y < 1, then there is no real value of x. So, pre-image of many elements of R does not exist in domain R, so, f is an into function. Thus, f is many-one, into function.

f : R → R, f(x) = x2 + x is:(A) One-one one(B) One-one into(C) Many-one onto(D) Many one onto

Answer»

Answer is (D)

Given: f : R → R and f(x) = x² + x

where R is a set of real numbers.

Let x₁, x₂ ∈ R are such that f(x₁) = f(x₂)

f(x₁) = f(x₂)

⇒ x₁² + x₁ = x₂² + x₂

⇒ x₁² – x₂² + x₁ – x₂ = 0

⇒ (x₁ – x₂)(x₁ + x₂) + 1 (x₁ – x₂) = 0

⇒ (x₁ – x₂) (x₁ + x₂ + 1} = 0

⇒ x₁ = x₂, x₁ = -(x₂ + 1) ∀ x₁, x₂ ∈ R

Here, element of set A relates to two elements of set B.

So, it is a many-one function.

Again, let y ∈ R (co-domain)

If possible then let pre-image of y is x in domain R.
then f(x) = y

⇒ x² + x = y

⇒ x(x + 1) = y

⇒ x = y, x = y – 1

If y < 1, then there is no real value of x.

So, pre-image of many elements of R does not exist in domain R,

so, f is an into function.

Thus, f is many-one, into function.