$f(-1) = e^{(-1)^2 -a + b} = 1 = e^0$ $\implies$ $1-a+b = 0$ . $f(-2) = e^{(-2)^2 -2a + b} = 1 = e^0$ $\implies$ $4-2a+b = 0$ . Resolvendo o sistema, temos: $a = 3$ e $b = 2$. Logo: $f(x) = e^{x^2 +3x + 2}$ e $g(x) = \large{\ln \left(\frac{x}{2} \right)}$ A composta $g \circ f $ $=$ $g(f(x))$ $=$ $\large{\ln \left(\frac{e^{x^2 +3x + 2}}{2} \right)}$ $=$ $x^2 +3x + 2 - \ln 2$ . Analisemos $g(f(x)) = 0$ : $x^2 +3x + 2 - \ln 2 = 0$ $\implies$ $x^2 +3x = \ln 2 - 2$ $\implies$ $x^2 +3x + \frac{9}{4} = \ln 2 - 2 + \frac{9}{4}$ Temos: $x^2 +3x + \frac{9}{4} = \ln 2 + \frac{1}{4} > 0$ $\implies$ $\large{(x+\frac{3}{2})^2}$ $=$ $\ln 2 + \frac{1}{4}$ Assim: $\large{x+\frac{3}{2}}$ $=$ $\pm \sqrt{\ln 2 + \frac{1}{4}}$ $\implies$ $x$ $=$ $-\frac{3}{2} \pm \sqrt{\ln 2 + \frac{1}{4}}$ $\in \space \mathbb{R}$ Ou seja, $g\circ f$ admite dois zeros reais distintos. Alternativa $\mathbb{(E)}$