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Let f : R → B be a function defined by f x = t a n - 1 2 x 1 + x 2 , then f is both one-one and onto when B is in the interval

Options

  1. A0 , π 4
  2. B0 , π 3
  3. C- π 4 , π 4
  4. D- π 4 , π 4

Correct answer

C. - π 4 , π 4

Step-by-step solution

Let, y = 2 x 1 + x 2 ⇒ y x 2 - 2 x + y = 0 ∵ x ∈ R ⇒ D ≥ 0 → 4 - 4 y 2 ≥ 0 ⇒ y ∈ - 1,1 - 1 ≤ 2 x 1 + x 2 ≤ 1 ⇒ t a n - 1 - 1 ≤ t a n - 1 2 x 1 + x 2 ≤ t a n - 1 1 - π 4 ≤ f x ≤ π 4

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