Intersection of y = x and y = ke^x for Non-Positive k

We are finding roots of $g(x) = x - ke^x = 0$ for $k \le 0$.

1. Differentiating $g(x)$:
   $g'(x) = 1 - ke^x$
   Since $k \le 0$, $-ke^x \ge 0$ for all $x \in \mathbb{R}$, which implies $g'(x) \ge 1 > 0$.
   Thus, $g(x)$ is strictly increasing on $(-\infty, \infty)$.

2. Asymptotic limits:
   $\lim_{x \to -\infty} g(x) = -\infty - 0 = -\infty$
   $\lim_{x \to \infty} g(x) = \infty - (-\infty) = +\infty$

By the Intermediate Value Theorem and strict monotonicity, $g(x) = 0$ has exactly one real root. Hence, the curves intersect at one point.

Correct Answer: B