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Let f x = 2 tan 3 ⁡ x - 6 tan 2 ⁡ x + 1 + s g n e x ,   ∀ x ∈ - π 4 , π 4 . Then the positive difference between the least value and the local maximum value of the function is (where sgn f ( x ) represents the signum function)

Options

  1. A7
  2. B8
  3. C9
  4. D10

Correct answer

B. 8

Step-by-step solution

Here, sgn e x = 1 Let tan ⁡ x = t ⇒ t ∈ - 1 , 1 ∵ x ∈ - π 4 , π 4 ∴ f t = 2 t 3 - 6 t 2 + 2 ⇒ f ' t = 2 3 t 2 - 6 t = 6 t t - 2 ∴ f - 1 = 2 - 1 - 3 + 1 = - 6 f 1 = 2 1 - 3 + 1 = - 2 f 0 = 2 ∴ Least value = - 6 Local maximum value = 2 ∴ the required positive difference = 8

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