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

Neither injective nor surjective.

Given:

$f:R \rightarrow R$ defined by $f(x)=1+x^2$

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:

Injection condition:

Let x,y be any two elements in domain R such that

\begin{aligned} &f(x)=f(y), f(x)=1+x^{2}, f(y)=1+y^{2} \\ &1+x^{2}=1+y^{2} \\ &x^{2}=y^{2} \\ &x=\pm y \end{aligned}

Therefore, f is not an injection.

Surjection test:

Let y be any two elements in domain (R) such that $f(x)=y$ for some element x in R.

\begin{aligned} &f(x)=y \\ &1+x^{2}=y \\ &x^{2}=y-1 \\ &x=\sqrt[2]{(y-1)} \\ &\text { For, } y=O \in R \\ &x=\pm \sqrt{-1}=\pm i \text { is not in } R \text { . } \end{aligned}

So, f is not a surjection and f is not a bijection.