Question

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

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

Answer 1

The correct option is A $10$

$(k+1){\tan }^{2}x-\sqrt{2}\lambda \tan x=1-k{\tan }^2(\alpha +\beta )=50$
$\tan \alpha$ and $\tan \beta$ are the roots of the equation $(k+1)x^2-\sqrt{2}\lambda x=1-k$

Now, $\tan \alpha +\tan \beta =\frac{\sqrt{2}\lambda }{k+1},\tan \alpha \tan \beta =\frac{k-1}{k+1}\Rightarrow {\left(\frac{\frac{\sqrt{2}\lambda }{k+1}}{1-\frac{k-1}{k+1}}\right)}^2=50\Rightarrow \frac{2{\lambda }^2}{4}=50\Rightarrow {\lambda }^2=100\Rightarrow \lambda =\pm 10$