
Problem 1. Find all value of the real parameter $a$ such that the equation

$$
9^{t}-4 a 3^{t}+4-a^{2}=0
$$

has an unique root in the interval $(0,1)$.


Solution: After the substitutions $x=3^{t}$ and $f(x)=x^{2}-4 a x+$ $4-a^{2}$ the problem is equivalent to: find all values of $a$ such that the equation $f(x)=0$ has an unique root in the interval $(1,3)$. It is easily seen that if $f(1)=0$ or $f(3)=0$, then $a=1, a=-5$ or $a=-13$ are not solutions of the problem. Therefore the equation $f(x)=0$ has an unique root in the interval $(1,3)$ when $f(1) \cdot f(3)<0$ or $D_{f}=0,2 a \in(1,3)$. It follows now that $f(1) . f(3)<0 \Longleftrightarrow(a-$ $1)^{2}(a+5)(a+13)<0 \Longleftrightarrow a \in(-13,-5)$ or $5 a^{2}-4=0,2 a \in(1,3)$, and so $a=\frac{2 \sqrt{5}}{5}$.
