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If $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=x^2-2 x-3$ then $f$ is
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Verified Answer
The correct answer is:
neither one-one nor onto
$f(x)=x^2-2 x-3$
for one one
Let $\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)$
$\begin{aligned} & x_1^2=2 x_1-3=x_2^2-2 x_2-3 \\ & \Rightarrow x_1^2-x_2^2-2\left(x_1-x_2\right)=0 \\ & \Rightarrow\left(x_1-x_2\right)\left[x_1+x_2-2\right]=0 \\ & \Rightarrow x_1=x_2 \text { or } x_1+x_2=2\end{aligned}$
So, $f(x)$ is not one-one
for onto
Let $y=x^2-2 x-3=0$
$\begin{aligned} & x^2-2 x-3-y=0 \\ & x=\frac{2 \pm \sqrt{4-4(3+y)}}{2}=1 \pm \sqrt{4+y} \\ & \therefore 4+y \geq 0=y \geq-4 \\ & \therefore \text { range }=-4 \uparrow \text { codomain }(R)\end{aligned}$
So, $f(x)$ is not onto
for one one
Let $\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)$
$\begin{aligned} & x_1^2=2 x_1-3=x_2^2-2 x_2-3 \\ & \Rightarrow x_1^2-x_2^2-2\left(x_1-x_2\right)=0 \\ & \Rightarrow\left(x_1-x_2\right)\left[x_1+x_2-2\right]=0 \\ & \Rightarrow x_1=x_2 \text { or } x_1+x_2=2\end{aligned}$
So, $f(x)$ is not one-one
for onto
Let $y=x^2-2 x-3=0$
$\begin{aligned} & x^2-2 x-3-y=0 \\ & x=\frac{2 \pm \sqrt{4-4(3+y)}}{2}=1 \pm \sqrt{4+y} \\ & \therefore 4+y \geq 0=y \geq-4 \\ & \therefore \text { range }=-4 \uparrow \text { codomain }(R)\end{aligned}$
So, $f(x)$ is not onto
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