If $5,5 \mathrm{r}, 5 \mathrm{r}^{2}$ are the lengths of the sides of a triangle, then $\mathrm{r}$ cannot be equal to:
Correct Option: , 4
$\mathrm{r}=1$ is obviously true.
Let $0<\mathrm{r}<1$
$\Rightarrow \quad r+r^{2}>1$
$\Rightarrow r^{2}+r-1>0$
$\left(r-\frac{-1-\sqrt{5}}{2}\right)\left(r-\left(\frac{-1+\sqrt{5}}{2}\right)\right)$
$\Rightarrow r-\frac{-1-\sqrt{5}}{2}$ or $r>\frac{-1+\sqrt{5}}{2}$
$r \in\left(\frac{\sqrt{5}-1}{2}, 1\right)$
$\frac{\sqrt{5}-1}{2} When $r>1$ $\Rightarrow \frac{\sqrt{5}+1}{2}>\frac{1}{\mathrm{r}}>1$ $\Rightarrow \mathrm{r} \in\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$ Now check options
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