Question

Let f, g : [-1, 2] ➝ℝ be continuous functions which are twice differentiable on the interval (-1, 2). Let the values of f and g at the points -1, 0 and 2 be as given in the following table:
x= -1 x=0 x=2
f(x) 3 6 0
g(x) 0 1 -1

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) ∪ (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)

Answer 1

Answer:

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