Let $\mathrm{A} \subseteq \mathbb{R}, \mathrm{B} \subseteq \mathbb{R}$ and $\mathrm{f}: \mathrm{A}…

Let $\mathrm{A} \subseteq \mathbb{R}, \mathrm{B} \subseteq \mathbb{R}$ and $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ be defined by $\mathrm{f}(\mathrm{x})=\mathrm{x}^2$ $-3 \mathrm{x}+2$. If $\mathrm{f}$ is a bijection, then
  1. $A=(-\infty, 0], B=\left(-\infty, \frac{-1}{4}\right]$
  2. $\mathrm{A}=\left(-\infty, \frac{3}{2}\right], \mathrm{B}=\left[\frac{-1}{4}, \infty\right)$
  3. $\mathrm{A}=\left[\frac{3}{2}, \infty\right), \mathrm{B}=\left(-\infty, \frac{-1}{4}\right]$
  4. $A=(-\infty, \infty), B=\left[\frac{-1}{4}, \infty\right)$

Solution

The function $f$ defined $f: A \rightarrow B$ by $f(x)=x^2-3 x+2$ Now $\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-3$ For stationary point, $\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \mathrm{x}=\frac{3}{2}$ $f\left(\frac{3}{2}\right)=\frac{9}{4}-\frac{9}{2}+2=\frac{9-18+8}{4}=\frac{-1}{4}$ so for one-one $x \in\left[\frac{3}{2}, \infty\right)$ and $y \in\left[-\frac{1}{4}, \infty\right)$ or $x \in\left(-\infty, \frac{3}{2}\right]$ and $y \in\left[-\frac{1}{4}, \infty\right)$

Asked in: AP EAMCET 2023 (16 May Shift 2)

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