If $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=x^2-2 x-3$ then $f$ is

If $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=x^2-2 x-3$ then $f$ is
  1. one-one but not onto
  2. onto but not one-one
  3. neither one-one nor onto
  4. a bijection

Solution

$f(x)=x^2-2 x-3$ 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

Asked in: AP EAMCET 2022 (08 Jul Shift 1)

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