(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)
Let H (x) = f (x) – 3g (x)
H (-1) = H (0) = H (2) = 3.
Applying Rolle’s Theorem in the interval [-1, 0]
H''(x) = f'(x) – 3g'(x) = 0 for at least one c ∈ (-1, 0).
As H'(x) never vanishes in the interval
⇒ Exactly one c ∈ (-1, 0) for which H(x) = 0
Similarly, apply Rolle’s Theorem in the interval [0, 2].
⇒ H'(x) = 0 has exactly one solution in (0, 2)