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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. 
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  • 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:
$x_1=\pm x_2$
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,
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



answered Mar 18, 2013 by meena.p
edited Mar 20, 2013 by thagee.vedartham

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