Question

Let $\tan \alpha$ and $\tan \beta$ be 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 a value of $\lambda$ is:

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

Answer 1

The correct option is C

$10$

Finding the value of $\lambda$:

The given equation is $(k+1){\tan }^{2}x-\sqrt{2}\lambda \tan x+(k-1)=0$, and $\tan \alpha$ and $\tan \beta$ are two real roots then

$$\begin{gathered} \tan \alpha +\tan \beta =\frac{-\left(-\sqrt{2}\lambda \right)}{k+1}=\frac{\sqrt{2}\lambda }{k+1}......\left(i\right) \\ \tan \alpha .\tan \beta =\frac{k-1}{k+1}....\left(ii\right) \end{gathered}$$

$$\begin{gathered} \because \tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ =\frac{\frac{\sqrt{2}\lambda }{k+1}}{1-\left(\frac{k-1}{k+1}\right)}[from(i)\&(ii)] \\ =\frac{\sqrt{2}\lambda }{\left(k+1\right)-\left(k-1\right)} \\ =\frac{\sqrt{2}\lambda }{2} \\ \Rightarrow \tan \left(\alpha +\beta \right)=\frac{\lambda }{\sqrt{2}} \\ \Rightarrow {\tan }^2\left(\alpha +\beta \right)=\frac{{\lambda }^2}{2}\left[\text{squaring both sides}\right] \\ \Rightarrow 50=\frac{{\lambda }^2}{2}\left[\because {\tan }^2\left(\alpha +\beta \right)=50\text{is given}\right] \\ \Rightarrow {\lambda }^2=100 \\ \Rightarrow \lambda =\pm 10 \end{gathered}$$

Hence C is the correct .