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

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

Q32 / 75 · JEE Main 2020 (07 Jan Shift 1) · Mathematics

Let $α$ and $β$ be two real roots of the equation $(k + 1) {t a n}^{2} x - \sqrt{2} \cdot λ \tan x = (1 - k),$ where $k (\neq - 1)$ and $λ$ are real numbers. If ${t a n}^{2} (α + β) = 50,$ then a value of $λ$ is

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

Solution

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

\tan α + \tan β = \frac{\sqrt{2} λ}{k + 1}

t a n α \tan β = \frac{k - 1}{k + 1}

\tan (α + β) = \frac{\frac{\sqrt{2} λ}{k + 1}}{1 - \frac{k - 1}{k + 1}} = \frac{\sqrt{2} λ}{2} = \frac{λ}{\sqrt{2}}

\tan^{2} (α + β) = \frac{λ^{2}}{2} = 50

λ = \pm 10

Hence, the correct answer is option: 'B'.

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Let α and β be two real roots of the equation k+1tan2x-2⋅λtan⁡x=1-k, where k≠-1 and λ are real numbers. If tan2α+β=50, then a value of λ is

Solution

Relevant Questions

Let $α$ and $β$ be two real roots of the equation $(k + 1) {t a n}^{2} x - \sqrt{2} \cdot λ \tan x = (1 - k),$ where $k (\neq - 1)$ and $λ$ are real numbers. If ${t a n}^{2} (α + β) = 50,$ then a value of $λ$ is