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Question: Let f, g: [-1, 2] → R continuous functions which are twice differentiable on the interval (-1, 2). L...

Let f, g: [-1, 2] → R 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 : [JEE 2015]

In each of the interval (-1, 0) and (0, 2) the function (f-3g)" never vanishea. 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 solutions in (-1, 0)

C

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

D

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

Answer

B, C

Explanation

Solution

Let h(x) = f(x) - 3g(x).

First, let's calculate the values of h(x) at the given points x = -1, 0, 2 using the provided table:

  • At x = -1: h(-1) = f(-1) - 3g(-1) = 3 - 3(0) = 3
  • At x = 0: h(0) = f(0) - 3g(0) = 6 - 3(1) = 3
  • At x = 2: h(2) = f(2) - 3g(2) = 0 - 3(-1) = 3

Since f and g are continuous functions on [-1, 2] and twice differentiable on (-1, 2), h(x) is also continuous on [-1, 2] and twice differentiable on (-1, 2).

Applying Rolle's Theorem for the interval (-1, 0): We observe that h(-1) = 3 and h(0) = 3. Since h(x) is continuous on [-1, 0] and differentiable on (-1, 0), and h(-1) = h(0), by Rolle's Theorem, there exists at least one value c1 ∈ (-1, 0) such that h'(c1) = 0. This means f'(c1) - 3g'(c1) = 0.

We are given that the function (f - 3g)'' never vanishes in the interval (-1, 0). This means h''(x) ≠ 0 for all x ∈ (-1, 0). If the second derivative h''(x) never vanishes in an interval, it implies that the first derivative h'(x) is strictly monotonic (either strictly increasing or strictly decreasing) in that interval. A strictly monotonic function can have at most one root in an interval. Since we have already established that h'(x) = 0 has at least one solution in (-1, 0), and h'(x) is strictly monotonic in this interval, it must have exactly one solution in (-1, 0). Therefore, f'(x) - 3g'(x) = 0 has exactly one solution in (-1, 0). This makes statement (B) correct.

Applying Rolle's Theorem for the interval (0, 2): We observe that h(0) = 3 and h(2) = 3. Since h(x) is continuous on [0, 2] and differentiable on (0, 2), and h(0) = h(2), by Rolle's Theorem, there exists at least one value c2 ∈ (0, 2) such that h'(c2) = 0. This means f'(c2) - 3g'(c2) = 0.

We are given that the function (f - 3g)'' never vanishes in the interval (0, 2). This means h''(x) ≠ 0 for all x ∈ (0, 2). Similar to the previous interval, if h''(x) never vanishes in (0, 2), then h'(x) is strictly monotonic in (0, 2). Since h'(x) = 0 has at least one solution in (0, 2), and h'(x) is strictly monotonic in this interval, it must have exactly one solution in (0, 2). Therefore, f'(x) - 3g'(x) = 0 has exactly one solution in (0, 2). This makes statement (C) correct.

Based on the analysis, both statements (B) and (C) are correct.

Explanation of the solution:

  1. Define h(x) = f(x) - 3g(x).
  2. Calculate h(-1) = 3, h(0) = 3, and h(2) = 3.
  3. Apply Rolle's Theorem to h(x) on [-1, 0]: Since h(-1) = h(0), there exists c1 ∈ (-1, 0) such that h'(c1) = 0.
  4. Apply Rolle's Theorem to h(x) on [0, 2]: Since h(0) = h(2), there exists c2 ∈ (0, 2) such that h'(c2) = 0.
  5. Given h''(x) = (f - 3g)'' ≠ 0 in (-1, 0) and (0, 2). This implies h'(x) is strictly monotonic in each of these intervals.
  6. A strictly monotonic function can have at most one root. Combining with step 3, h'(x) = 0 has exactly one solution in (-1, 0).
  7. Combining with step 4, h'(x) = 0 has exactly one solution in (0, 2).
  8. Thus, f'(x) - 3g'(x) = 0 has exactly one solution in (-1, 0) and exactly one solution in (0, 2).