Question
A function $f: R \rightarrow R$ defined as $f(x)=x^2-4 x+5$ is:
✓
Answer
Given, $f(x)=x^2-4 x+5$
Here $f(0)=f(4)=5$
Hence, $f(x)$ is not one$-$one.
To check whether the function is onto or not, we have to find range of function.
$\text { Let } y=x^2-4 x+5 $
$\Rightarrow x^2-4 x+5-y=0$
$\therefore D=(4)^2-4(1)(5-y) \geq 0 \forall x \in R$
$\Rightarrow 16-20+4 y \geq 0 $
$\Rightarrow 4 y-4 \geq 0$
$\Rightarrow 4(y-1) \geq 0 $
$\Rightarrow y \geq 1$
Hence, range $=(1, \infty)$
Here, Co$-$domain $\neq$ Range
So, $f(x)$ is not onto.
Here $f(0)=f(4)=5$
Hence, $f(x)$ is not one$-$one.
To check whether the function is onto or not, we have to find range of function.
$\text { Let } y=x^2-4 x+5 $
$\Rightarrow x^2-4 x+5-y=0$
$\therefore D=(4)^2-4(1)(5-y) \geq 0 \forall x \in R$
$\Rightarrow 16-20+4 y \geq 0 $
$\Rightarrow 4 y-4 \geq 0$
$\Rightarrow 4(y-1) \geq 0 $
$\Rightarrow y \geq 1$
Hence, range $=(1, \infty)$
Here, Co$-$domain $\neq$ Range
So, $f(x)$ is not onto.
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