Let f(x) and g(x) be differentiable for 0 ≤ x ≤ 1 such that f (0) = 0, g(0) = 0, f(1) = 6. Let there exist a real number c in (0, 1) such that f'(c) = 2g'(c), then the value of g(1) must be

1 3  -2 -1

3

Applying Rolle's theorem to F(x) = f(x) – 2g(x), F(0) = 0, F(1) – 2g(1) ⇒ 0 = 6 − 2g(1) ⇒ g(1) = 3.