Let $$f,g:[-1, 2] \rightarrow R$$ be continuous functions which are twice differentiable on the interval (-1, 2). Let the valuesof f and g at the points —1, 0 and 2 be as given in the following table:

In each of the intervals (-1,0) and (0, 2) the function $$(f - 3g)"$$ never vanishes. Then the correct statement(s) is(are)

A

$$f'(x) - 3g'(x) = 0$$ has exactly three solutions in $$(-1, 0) \cup (0, 2)$$

B

$$f'(x) - 3g'(x) = 0$$ has exactly one solution in (-1, 0)

C

$$f'(x) - 3g'(x) = 0$$ has exactly one solution in (0, 2)

D

$$f'(x) - 3g'(x) = 0$$ has exactly two solutions in (-1, 0) and exactly two solutions in (0, 2)