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.