Question

Let x₁, x₂, x₃ be the solutions of `tan^-1((2x + 1)/(x + 1)) + tan^-1((2x - 1)/(x - 1))` = 2tan^–1(x + 1) where x₁ < x₂ < x₃ then 2x₁ + x₂ + x₃² is equal to ______.

Options

• 0.00

• 1.00

• 2.00

• 3.00

Answer 1

Let x₁, x₂, x₃ be the solutions of `tan^-1((2x + 1)/(x + 1)) + tan^-1((2x - 1)/(x - 1))` = 2tan^–1(x + 1) where x₁ < x₂ < x₃ then 2x₁ + x₂ + x₃² is equal to 1.00.

Explanation:

Let α + β = 2γ  ...(i)

where tanα = `(2x + 1)/(x + 1)`, tanβ = `(2x - 1)/(x - 1)` and tanγ = x + 1

From tan(α + β) = tan2γ, we get

`(tanα + tanβ)/(1 - tanαtanβ) = (2tanγ)/(1 - tan^2γ)`

⇒ `((2x + 1)/(x + 1) + (2x - 1)/(x - 1))/(1 - (4x^2 - 1)/(x^2 - 1)) = (2(x + 1))/(1 - (x + 1)^2`

⇒ `(2x^2 - x - 1 + 2x^2 + x - 1)/(x^2 - 1 - 4x^2 + 1) = (2(x + 1))/(-x^2 - 2x)`

⇒ `(2(2x^2 - 1))/(-3x^2) = (2(x + 1))/(-(x^2 + 2x)`

x = 0

or 2x³ + 4x² – x – 2 = 3x² + 3x

⇒ 2x³ + x² – 4x – 2 = 0

⇒ x²(2x + 1) – 2(2x + 1) = 0

⇒ (x² – 2)(2x + 1) = 0

⇒ x = `sqrt(2), -sqrt(2)` or x = `-1/2`

x = `-sqrt(2)` is rejecred   ∵ It does not satisfy (i)

⇒ x₁ = `-1/2`, x₂ = 0 and x₃ = `sqrt(2)`

⇒ 2x₁ + x₂ + x₃² = 1