Let two functions f(x) and g(x) are defined on R arrow R such that f(x)= begincasesx2, x ∈ text irrational 2-x2, x ∈ text rational endcases and g(x)= begincases2-x2, x ∈ text irrational x2, x ∈ text rational endcases. Then the function f+g: R arrow R is

Question

Let two functions f(x) and g(x) are defined on R arrow R such that f(x)= begincasesx2, x ∈ text irrational 2-x2, x ∈ text rational endcases and g(x)= begincases2-x2, x ∈ text irrational x2, x ∈ text rational endcases. Then the function f+g: R arrow R is (A) injective as well as surjective. (B) injective but not surjective. (C) surjective but not injective. (D) neither surjective nor injective.

Tardigrade Admin · Accepted Answer

Clearly ( f + g )( x )= f ( x )+ g ( x )=2 ∀ x ∈ R = constant function Hence the function ( f + g )( x ) defined on R arrow R is neither surjective nor injective.]

Q. Let two functions $f (x)$ and $g (x)$ are defined on $R \to R$ such that $f (x) = {x^{2}, 2 - x^{2}, x \in irrational x \in rational$ and $g (x) = {2 - x^{2}, x^{2}, x \in irrational x \in rational$ . Then the function $f + g : R \to R$ is

injective as well as surjective.

injective but not surjective.

surjective but not injective.

neither surjective nor injective.

Clearly $(f + g) (x) = f (x) + g (x) = 2\forall x \in R =$ constant function
Hence the function $(f + g) (x)$ defined on $R \to R$ is neither surjective nor injective.]

Q. Let two functions f(x) and g(x) are defined on R→R such that f(x)={x2,2−x2,​x∈ irrational x∈ rational ​ and g(x)={2−x2,x2,​x∈ irrational x∈ rational ​. Then the function f+g:R→R is