Question

If $f\left(x\right)={\log }_{1/2}\left(\frac{x^2-6x+12}{4}\right)$ and $f\left(x\right)$ is non-negative, then $x$ can be

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A), (B), (C)

$4$

$f\left(x\right)={\log }_{1/2}\left(\frac{x^2-6x+12}{4}\right)\Rightarrow f\left(x\right)=-{\log }_2\left(\frac{x^2-6x+12}{4}\right)\Rightarrow f\left(x\right)=-\left[{\log }_2\right(x^2-6x+12)-{\log }_{2}4]\Rightarrow f\left(x\right)=2-{\log }_2(x^2-6x+12)\Rightarrow f\left(x\right)=2-{\log }_2\left[\right(x-3{)}^2+3]$
Domain of the function
$x\in R$

As it is given that the function is non negative, so
$f\left(x\right)\ge 0\Rightarrow 2-{\log }_2\left[\right(x-3{)}^2+3]\ge 0\Rightarrow 2\ge {\log }_2[(x-3{)}^2+3]\Rightarrow 4\ge (x-3{)}^2+3\Rightarrow 1\ge (x-3{)}^2\Rightarrow 1\ge (x-3)\ge -1\therefore x\in [2,4]$