Question

The maximum value of ${\left(\frac{1}{2}\right)}^{x^2-3x+2}$ is
$(x\in R)$

• A. $\frac{1}{4}$
• B. $\sqrt[4]{42}$
• C. $0$
• D. not defined

Answer 1

The correct option is B $\sqrt[4]{42}$

Let $f\left(x\right)=x^2-3x+2$,
$a=1(>0)$, $b=-3$, $c=2$
Minimum value of $f\left(x\right)$,
$\frac{4ac-b^2}{4a}=-\frac{1}{4}$

$\therefore$ Range of $f\left(x\right)$ is $[-\frac{1}{4},\infty )$
$\Rightarrow$ Range of ${\left(\frac{1}{2}\right)}^{f\left(x\right)}$ is $(0,{\left(\frac{1}{2}\right)}^{\frac{-1}{4}}]=(0,\sqrt[4]{42}]$