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# In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.(ii) $f : R\; \to R\; defined \; by \; f(x)\; =\; 1+x^2$

Note: This is the 2nd  part of a  2 part question, which is split as 2 separate questions here.

Toolbox:
• A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given : $f: R \to R$ defined as $f(x)=1+x^2$

Let $x_1\,x_2 \;be\;two\;elements\;\in R$ such that

Step1: Injective or One-One function:

$f(x_1)=f(x_2)$

$=>1+x_1^2=1+x_2^2$

$x_1^2=x_2^2$

$x_1=\pm x_2$

This is not imply $x_1=x_2$ as$x_1 and x_2$ can take both +ve and -ve values

f is not one one

Step 2: Surjective or On-to function:

Consider two elements 1 and -1 in R,

$f(x)=1+x^2$

We see that $f(1) =1+1^2$ =2

also $f(-1) = 1+{-1}^2=2$

but 1is not equal to-1

so f is not onto

Solution: $f: R \to R \qquad f(x)=1+x^2$ is neither one-one nor onto

edited Mar 20, 2013