Question

If $f\left(x\right)=9^x-3^x+1$ $\forall xϵ(0,\infty ).$ Then the range of $f\left(x\right)$ is

• A. $(0,7)$
• B. $(0,4)$
• C. $(0,3)$
• D. None of these

Answer 1

The correct option is D None of these

$9^x-3^x+1={\left(3^x\right)}^2-3^x+1$

let $3^x=t$

$t^2-t+1=t^2-t+\frac{1}{4}-\frac{1}{4}+1={(t-\frac{1}{2})}^2+\frac{3}{4}\ge \frac{3}{4}={(3^x-\frac{1}{2})}^2+\frac{3}{4}\ge \frac{3}{4}\forall x\in (-\infty ,1)$

Hence the range of $9^x-3^x+1is[\frac{3}{4},7)$