UPSC NDA I 2021 Mathematics Solved Paper

Concept A function $f$ from $X$ to $Y$ is onto (or surjective), if and only if for every element $y \in Y$ there is an element $x \in X$ with $f(x)=y$. In words: "Each element in the co-domain of has a pre-image" Mathematical Description: $f: X \rightarrow Y$ is onto$\Leftrightarrow \forall y \exists x, f(x)=y$ One-to-one Correspondence A function $f$ is a one-to-one correspondence (or bijection), if and only if it is both one-to-one and onto In words: "No element in the co-domain of has two (or more) pre images" (one-to-one)and "Each element in the co-domain of f has a pre-image" (onto). Calculation: 1. A function $f: Z \rightarrow Z$, defined by $f(x)=x+1$, is one-one as well as onto. $f(x)=x+1$ calculate $f(x_{1}):$ $f(x_{1})=x_{1}+1$ calculate $f(x_{2}):$ $f(x_{2})=x_{2}+1$ Now, $f(x_{1})=f(x_{2})$ $\Rightarrow x_{1}+1=x_{2}+1$ $\Rightarrow x_{1}=x_{2}$ So, $f$ is one-one function. Consider $f(x)=y$ $y=x+1$ $x=y-1$ $f(y-1)=y-1+1=y$ $f$ is onto. 2. A function $f: N \rightarrow N$, defined by $f(x)=x+1$, is one-one but not onto. $f(x)=x+1$ calculate $f(x_{1})$ : $f(x_{1})=x_{1}+1$ calculate $f(x_{2})$ : $f(x_{2})=x_{2}+1$ Now, $f(x_{1})=f(x_{2})$ $\Rightarrow x_{1}+1=x_{2}+1$ $\Rightarrow x_{1}=x_{2}$ So, $f$ is one-one function. Clearly, $f(x)=x+1 \geq 2$ for all $x \in N$ So, $f(x)$ does not assume values 1 . f is not an onto function. So, Both 1 and 2 are correct.