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JEE Main2017MathematicsContinuity and DifferentiabilityActual

The value of k which the function f x = 4 5 tan ⁡ 4 x tan ⁡ 5 x , 0 < x < π 2 k + 2 5 , x = π 2 is continuous at x = π 2 , is

Options

  1. A2 5
  2. B- 2 5
  3. C17 20
  4. D3 5

Correct answer

D. 3 5

Step-by-step solution

Given function is continuous at x = π 2 ⇒ f π 2 = f π + 2 = f π - 2   ⇒ k + 2 5 = 4 5 tan ⁡ 4 x tan ⁡ 5 x ⇒ k + 2 5 = 4 5 tan 2 π cot 5 π 2 ⇒ k + 2 5 = 4 5 0     ⇒ k + 2 5 = 1       ⇒ k = 3 5

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