Question

Let $\alpha$ and $\beta$ be two real roots of the equation $\left(k+1\right)tan^{2}x-\sqrt{2}\cdot\lambda tan x = \left(1-k\right)$, where $k\left(\ne -1\right)$ and $\lambda$ are real numbers. If $tan^{2} \left(\alpha + \beta \right)= 50$, then a value of $\lambda$ is :

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

Answer 1

Answer: (C)

$tan\alpha + tan\beta = \frac{\lambda\sqrt{2}}{k+1}$
$tan\alpha.\, tan\beta = \frac{k-1}{k+1}$
$tan \left(\alpha +\beta \right) =\frac{\frac{\lambda \sqrt{2}}{k+1}}{1- \frac{k-1}{k+1}} = \frac{\lambda\sqrt{2}}{2} = \frac{\lambda }{\sqrt{2}}$
$\Rightarrow \frac{\lambda ^{2} }{2} = 50 \Rightarrow = 10\& -10$