The correct answer is: C. Mean of f(x) and g(x) are different; Variance of f(x) and g(x) are same.
The mean of a probability density function is given by:
$$\mu = \int_{-\infty}^{\infty} x f(x) dx$$
The variance of a probability density function is given by:
$$\sigma^2 = \int_{-\infty}^{\infty} (x - \mu)^2 f(x) dx$$
For $f(x)$ and $g(x)$ as given in the question, we have:
$$\mu_f = \int_{-a}^0 x \frac{x}{a} + 1 dx = \frac{a^2}{4} + 0 = \frac{a^2}{4}$$
$$\mu_g = \int_{-a}^0 x \frac{-x}{a} + 1 dx = -\frac{a^2}{4} + 0 = -\frac{a^2}{4}$$
Therefore, $\mu_f \neq \mu_g$.
The variance of $f(x)$ is given by:
$$\sigma^2_f = \int_{-a}^0 (x - \mu_f)^2 \frac{x}{a} + 1 dx = \frac{a^4}{16} + 0 = \frac{a^4}{16}$$
The variance of $g(x)$ is given by:
$$\sigma^2_g = \int_{-a}^0 (x - \mu_g)^2 \frac{-x}{a} + 1 dx = \frac{a^4}{16} + 0 = \frac{a^4}{16}$$
Therefore, $\sigma^2_f = \sigma^2_g$.
Therefore, the mean of $f(x)$ and $g(x)$ are different, but the variance of $f(x)$ and $g(x)$ are the same.