Question

The function $f:A\to B$ defined by $f\left(x\right)=-x^2+6x-8$ is a bijection if

• A. $A=(-\infty ,3]\text{and}B=(-\infty ,1]$
• B. $A=[3,\infty )\text{and}B=(-\infty ,1]$
• C. $A=(-\infty ,-3]\text{and}B=(-\infty ,-1]$
• D. $A=[-3,\infty )\text{and}B=[1,\infty )$

Answer 1

Answer: (A), (B)

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

Given : $f\left(x\right)=-x^2+6x-8$
$\Rightarrow f\left(x\right)=-(x-2)(x-4)$


$f\left(x\right)$ is bijective,
If domain is $(-\infty ,3],$ then range is $(-\infty ,1]$
If domain is $[3,\infty ),$ then range is $(-\infty ,1]$
If domain is $(-\infty ,-3],$ then range is $(-\infty ,-35]$
If domain is $[-3,\infty ),$ then range is $(-\infty ,1]$

From the given options $f\left(x\right)$ is bijective when
$A=(-\infty ,3]\text{and}B=(-\infty ,1]A=[3,\infty )\text{and}B=(-\infty ,1]$