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JEE Advanced Mathematics Trigonometric Equations 2022 JEE Advanced 2022 (Paper 1)

JEE Advanced Mathematics Question (2022) — Solution

Question

Let M denote the determinant of a square matrix M . Let g : 0 , π 2 → ℝ  be the function defined by g θ = f θ - 1 + f π 2 - θ - 1  where  f θ = 1 2 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 + sin π cos θ + π 4 tan θ - π 4 sin θ - π 4 - cos π 2 log e 4 π cot θ + π 4 log e π 4 tan π Let p x  be a quadratic polynomial whose roots are the maximum and minimum values of the function g θ , and p 2 = 2 - 2 . Then, which of the following is/are TRUE ?

Options

  1. A. P 3 + 2 4 < 0
  2. B. P 1 + 3 2 4 > 0
  3. C. P 5 2 - 1 4 > 0
  4. D. P 5 - 2 4 < 0

Answer

C. P 5 2 - 1 4 > 0

Step-by-step solution

Given, f θ = 1 2 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 + sin π cos θ + π 4 tan θ - π 4 sin θ - π 4 - cos π 2 log e 4 π cot θ + π 4 log e π 4 tanπ ⇒ f θ = 1 2 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 + 0 cos θ + π 4 tan θ - π 4 sin θ - π 4 0 log e 4 π - tan θ - π 4 - log e 4 π 0 Here we used  cos θ + π 4 = - sin θ - π 4 And  tan θ - π 4 = - cot θ + π 4 And  log e 4 π = - log e π 4 Also  sin π = - cos π 2 = tan π = 0 So,  ⇒ f θ = 1 2 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 + skew symmetric ⇒ f θ = 1 + sin 2 θ So,  g θ = sin θ + cos θ Now maximum and minimum values are  2  and  1  respectively. Now quadratic polynomial will be  P x = a x - 2 x - 1 , where  a ∈ R - 0 , But given  P 2 = 2 - 2  , so  a = 1 . ∴   P x = x - 2 x - 1 Now solving all options we get, P 3 + 2 4 = 3 - 3 2 4 · 2 - 1 4 < 0 P 1 + 3 2 4 = 1 - 2 4 · 3 2 - 3 4 < 0 P 5 2 - 1 4 = 2 - 1 4 · 5 2 - 5 4 > 0 P 5 - 2 4 = 5 - 5 2 4 1 - 2 4 > 0

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