Question

Let $-\frac{\pi}{6}<\theta<-\frac{\pi}{12}$. Suppose $\alpha_{1}$ and $\beta_{1}$ are the roots of the equation $x^{2}-2 x \sec \theta+1=0$ and $\alpha_{2}$ and $\beta_{2}$ are the roots of the equation $x ^{2}+2 x \tan \theta-1=0$. If $\alpha_{1}>\beta_{1}$ and $\alpha_{2}>\beta_{2}$, then $\alpha_{1}+\beta_{2}$ equals

• A. $2(\sec \theta-\tan \theta)$
• B. $2 \sec \theta$
• C. $-2 \tan \theta$
• D. $0$

Answer 1

Answer: (C)

$x^{2}-2 x \sec \theta+1=0$
Now roots are $x =\sec \theta \pm \tan \theta$
So $\alpha_{1}=\sec \theta-\tan \theta$
& $\beta_{1}=\sec \theta+\tan \theta$
$\begin{cases}\text { as } \theta \in\left(-\frac{\pi}{6},-\frac{\pi}{12}\right) \\ \& \alpha_{1}>\beta_{1}\end{cases}$
$\& x^{2}+2 x \tan \theta-1=0$
Now roots are $x =\pm \sec \theta-\tan \theta$
So,$\alpha_{2}=+\sec \theta-\tan \theta $
$\beta_{2}=-\sec \theta-\tan \theta$
$\therefore \alpha_{1}+\beta_{2}=-2 \tan \theta$