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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)

1 Answer

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(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)

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