Question

The function $f:A\to B$ defined by $f\left(x\right)=-x^2+6x-8$ is a bijection if
(a) $A=(-\infty ,3]\mathrm{and}B=(-\infty ,1]$
(b) $A=[-3,\infty )\mathrm{and}B=(-\infty ,1]$
(c) $A=(-\infty ,3]\mathrm{and}B=[1,\infty )$
(d) $A=[3,\infty )\mathrm{and}B=[1,\infty )$

Answer 1

(a) $A=(-\infty ,3]\mathrm{and}B=(-\infty ,1]$


$$\begin{gathered} f\left(x\right)=-x^2+6x-8,\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{function} \\ \mathrm{And}\mathrm{the}\mathrm{domain}\mathrm{of}\mathrm{polynomial}\mathrm{function}\mathrm{is}\mathrm{real}\mathrm{number}. \\ \therefore x\in R \end{gathered}$$

$$\begin{gathered} f\left(x\right)=-x^2+6x-8 \\ =-\left(x^2-6x+8\right) \\ =-\left(x^2-6x+9-1\right) \\ =-{\left(x-3\right)}^2+1 \\ \mathrm{Maximum}\mathrm{value}\mathrm{of}-{\left(x-3\right)}^2\mathrm{woud}\mathrm{be}0 \\ \therefore \mathrm{Maximum}\mathrm{value}\mathrm{of}-{\left(x-3\right)}^2+1\mathrm{woud}\mathrm{be}1 \\ \therefore \mathit{}f\left(x\right)\mathit{}\in (-\infty ,1] \end{gathered}$$



$$\begin{gathered} \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{from}\mathrm{the}\mathrm{given}\mathrm{graph}\mathrm{that}\mathrm{function}\mathrm{is}\mathrm{symmetrical}\mathrm{about}x=3 \\ \&\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{bijective}. \\ \mathrm{So},x\mathrm{would}\mathrm{be}\mathrm{either}(-\infty ,3]\mathrm{or}[3,\infty ) \\ \mathrm{The}\mathrm{correct}\mathrm{option}\mathrm{which}\mathrm{satisfy}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{both}\mathrm{is}: \end{gathered}$$
$A=(-\infty ,3]\mathrm{and}B=(-\infty ,1]$