A car is called flex if it can be fueled with gasoline or ethanol. Consider that the prices of ethanol and gasoline are, respectively, R$ 1.59 and R$ 2.49 per liter.

(a) Suppose a flex car can travel $12.3 \mathrm{~km}$ per liter of gasoline, which we denote as $12.3 \mathrm{~km} / \mathrm{l}$. What must be the $\mathrm{km} / \mathrm{l}$ ratio of this car for ethanol to make the use of ethanol financially more advantageous than gasoline?

(b) If the performance of a flex car is $x \mathrm{~km} / \mathrm{l}$ with gasoline and $\left(\frac{x}{2}+1\right) \mathrm{km} / \mathrm{l}$ with ethanol, write the expression for the function $g(x)$ that gives the cost of this car traveling $100 \mathrm{~km}$ using gasoline and the expression for the function $a(x)$ that gives the cost of this car traveling $100 \mathrm{~km}$ using ethanol.

(c) For the cost to be the same, whether using ethanol or gasoline, what must be the $\mathrm{km} / \mathrm{l}$ ratio for gasoline and for ethanol?

(d) Under what condition is the use of ethanol more financially advantageous than gasoline? Provide a numerical example that satisfies the condition.
(a) With gasoline, the car travels $\frac{12.3}{2.49}=4.94 \mathrm{~km}$ per $\mathrm{R} \$ 1.00$. For ethanol to be more advantageous, the car needs to travel more than $4.94 \mathrm{~km}$ with $\mathrm{R} \$ 1.00$. Therefore, if the performance with ethanol is $y \mathrm{~km} / \mathrm{I}$, we need $\frac{y}{1.59}>4.94$, which implies $y>7.85$. In other words, the performance with ethanol must be greater than $7.85 \mathrm{~km} / \mathrm{I}$.

(b) Observe that $g(x)=2.49 \frac{100}{x}=\frac{249}{x}$ and $a(x)=1.59 \frac{100}{\frac{x}{2}+1}=\frac{318}{x+2}$.

(c) We need to have $a(x)=g(x)$, that is, $\frac{249}{x}=\frac{318}{x+2}$, which leads to $x=7.22 \mathrm{~km} / \mathrm{l}$, which must be the performance with gasoline. With ethanol, the car must travel

$$
\frac{7.22}{2}+1=3.61 \mathrm{~km} / \mathrm{l}
$$

(d) Assuming the car's performance is $x \mathrm{~km} / \mathrm{I}$ with gasoline and $y \mathrm{~km} / \mathrm{I}$ with ethanol, and considering a journey of $L \mathrm{~km}$, we need the cost with gasoline to be greater than the cost with ethanol:

$$
2.49 \frac{L}{x}>1.59 \frac{L}{y} \Rightarrow 2.49 y>1.59 x \Rightarrow y>0.64 x
$$

since $x$ and $y$ are positive values.

An example is a car that travels $10 \mathrm{~km} / \mathrm{I}$ with gasoline, which would need to travel more than $6.4 \mathrm{~km} / \mathrm{I}$ with ethanol for the use of ethanol to be more advantageous.

Note. The values determined in the solution were rounded to the second decimal place.