  #### Please solve RD Sharma class 12 Chapter Functions exercise 2.1 question 5 sub question (xvii) maths textbook solution.

Neither injective nor surjective.

Given:

defined by

Hint:

Injective (one-one): function means every element in the domain has a distinct image in the  co-domain.

Surjective(Onto):  function means every element in the co-domain has at least one  pre image in the domain of function.

Solution:

Let x and y be any two elements in domain (R), such that f(x) = f(y).

xy2+ x = x2y + y

xy2 − x2y + x − y    = 0

−xy(−y + x) + 1(x − y) = 0

(x − y) (1 – xy) = 0

x = y or x =

So, f is not an injection.

Surjection test:

Let y be any element in co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

yx2 – x + y = 0

x((-1) ± √(1 - 4x2))/(2y) if y ≠ 0

= (1 ± √(1 - 4y2))/(2y), which may not be in R

For example, if y = 1, then (1 ± √(1-4))/(2y) = (1 ± i√3)/2, which is not in R Therefore, f is not surjection and f is not bijection.

#### lovekush 