Let α and β be two real roots of the equation (k + 1) ^ 2 x - 2 ⋅ λ tan x = (1 - k), where k(≠ - 1) and λ are real numbers. If ^ 2 (α + β) = 50, then a value of λ is

Correct answer: B. 10 (k + 1) ^ 2 x - 2 λ tan x + (k - 1) = 0 tan α + tan β = 2 λ k + 1 tanα tan β = (k - 1)/(k + 1) (α + β) = 2 λ k + 1 1 - (k - 1)/(k + 1) = 2 λ 2 = λ 2 ^ 2 (α + β) = λ ^ 2 2 = 50 λ = ± 10

JEE Main 20207 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 7 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2020 · 7 January · Morning Shift · Mathematics · Quadratic Equation

Let α and β be two real roots of the equation (k + 1) ^ 2 x - 2 ⋅ λ...

Let

\alpha

and

\beta

be two real roots of the equation \left(k + 1\right)\tan^{2}x - \sqrt{2} ⋅ \lambda tan x =

\left(1 - k\right),

where

k\left(\ne - 1\right)

and

\lambda

are real numbers. If

\tan^{2}\left(\alpha + \beta\right) = 50,

then a value of

\lambda

Options

A $10\sqrt{2}$
B $10$
C $5$
D $5\sqrt{2}$

Correct answer

B. $10$

Step-by-step solution

\left(k + 1\right)\tan^{2} x - \sqrt{2}\lambda tan x + \left(k - 1\right) = 0

tan \alpha + tan \beta = \dfrac{\sqrt{2}\lambda}{k + 1}

tan\alpha tan \beta = \dfrac{k - 1}{k + 1}

\tan \left(\alpha + \beta\right) = \dfrac{\dfrac{\sqrt{2}\lambda}{k + 1}}{1 - \dfrac{k - 1}{k + 1}} = \dfrac{\sqrt{2}\lambda}{2} = \dfrac{\lambda}{\sqrt{2}}

\tan^{2} \left(\alpha + \beta\right) = \dfrac{{\lambda}^{2}}{2} = 50

\lambda = \pm 10

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