Let α and β be two real roots of the equation (k+1)tan2x-√2⋅λ tan x = (1-k), where k(≠ -1) and λ are real numbers. If tan2 (α + β )= 50, then a value of λ is :

Question

Let α and β be two real roots of the equation (k+1)tan2x-√2⋅λ tan x = (1-k), where k(≠ -1) and λ are real numbers. If tan2 (α + β )= 50, then a value of λ is : (A) 5√2 (B) 10√2 (C) 10 (D) 5

Tardigrade Admin · Accepted Answer

tanα + tanβ = (λ√2/k+1) tanα. tanβ = (k-1/k+1) tan (α +β ) =((λ √2/k+1)/1- (k-1)k+1) = (λ√2/2) = (λ /√2) ⇒ (λ 2 /2) = 50 ⇒ = 10 -10

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

$52$

$102$

$10$

$5$

$t an α + t an β = \frac{λ 2}{k + 1}$
$t an α . t an β = \frac{k - 1}{k + 1}$
$t an (α + β) = \frac{\frac{λ 2}{k + 1}}{1 - \frac{k - 1}{k + 1}} = \frac{λ 2}{2} = \frac{λ}{2}$
$\Rightarrow \frac{λ ^{2}}{2} = 50 \Rightarrow= 10& - 10$

Q. Let α and β be two real roots of the equation (k+1)tan2x−2​⋅λtanx=(1−k), where k(=−1) and λ are real numbers. If tan2(α+β)=50, then a value of λ is :

52​

102​

10

5

tanα+tanβ=k+1λ2​​ tanα.tanβ=k+1k−1​ tan(α+β)=1−k+1k−1​k+1λ2​​​=2λ2​​=2​λ​ ⇒2λ2​=50⇒=10&−10

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

$52$

$102$

$10$

$5$

$t an α + t an β = \frac{λ 2}{k + 1}$
$t an α . t an β = \frac{k - 1}{k + 1}$
$t an (α + β) = \frac{\frac{λ 2}{k + 1}}{1 - \frac{k - 1}{k + 1}} = \frac{λ 2}{2} = \frac{λ}{2}$
$\Rightarrow \frac{λ ^{2}}{2} = 50 \Rightarrow= 10& - 10$