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If f x = 2 sin ⁡ x - 1 - 2 cot ⁡ x , then the value of f ' π 3 is equal to

Options

  1. A0
  2. B- 5 3
  3. C5 3
  4. D8 3

Correct answer

C. 5 3

Step-by-step solution

In the neighbourhood of x = π 3 , we know, 2 sin ⁡ x - 1 > 0 ⇒ f x = 2 sin ⁡ x - 1 - 2 cot ⁡ x Now, for the neighbourhood of x = π 3 2 sin ⁡ x - 1 - 2 cot ⁡ x = 2 3 2 - 1 - 2 3 = 3 - 2 3 - 1 = 1 3 - 1 < 0 ∴ f x = 2 cot ⁡ x - 2 sin ⁡ x - 1 Now, f ' x = 2 - c o s e c 2 x - 2 cos ⁡ x f ' π 3 = 2 - 2 3 2 - 2 1 2 = 2 4 3 - 1 = 5 3

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