The number of roots of $e^x + 0.5x^2 - 2 = 0$ in the range $[-5, 5]$ is

Let $f(x) = e^x + 0.5x^2 - 2$.

1. First and Second Derivatives:

   $$f^{\prime}(x) = e^x + x$$

   $$f^{\prime\prime}(x) = e^x + 1$$

   Since $e^x > 0$ for all real $x$, $f^{\prime\prime}(x) = e^x + 1 > 0$ for all $x \in \mathbb{R}$. Because $f(x)$ is strictly convex everywhere, it can have at most 2 real roots.

2. Evaluating $f(x)$ at specific points:

   • $f(-5) = e^{-5} + 0.5(-5)^2 - 2 \approx 0.0067 + 12.5 - 2 = 10.5067 > 0$

   • $f(0) = e^0 + 0 - 2 = 1 - 2 = -1 < 0$

   • $f(2) = e^2 + 0.5(2)^2 - 2 = e^2 + 2 - 2 = e^2 \approx 7.389 > 0$

3. Applying Intermediate Value Theorem:

   • Since $f(-5) > 0$ and $f(0) < 0$, there exists at least one root in $(-5, 0)$.

   • Since $f(0) < 0$ and $f(2) > 0$, there exists at least one root in $(0, 2) \subset [-5, 5]$.

Since $f(x)$ can have at most 2 real roots, there are exactly 2 roots in the interval $[-5, 5]$.

Hence, option (C) is correct.