Question

Let $-\frac{\pi }{6}<\theta <-\frac{\pi }{12}$, Suppose ${\alpha }_1$ and ${\beta }_1$ are the roots of the equation $x^2-2x\sec \theta +1=0$ and ${\alpha }_2$ and ${\beta }_2$ are the roots of the equation $x^2+2x\tan \theta -1=0$. If ${\alpha }_1>{\beta }_1$ and ${\alpha }_2>{\beta }_2$, then ${\alpha }_1+{\beta }_2$ equals to

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

Answer 1

Answer: (C)

$-2\tan \theta$

$x^2-2x\sec \theta +1=0$ has roots ${\alpha }_1,{\beta }_1$, out of which ${\alpha }_1>{\beta }_1$

So, $x=\frac{2\sec \theta \pm 2\tan \theta }{2}=\sec \theta \pm \tan \theta$

In the given range of $\theta ,\sec \theta$ is positive, while $\tan \theta$ is negative and so, ${\alpha }_1=\sec \theta -\tan \theta$

Similarly, for $x^2+2x\tan \theta -1=0,x=\frac{-2\tan \theta \pm 2\sec \theta }{2}=-\tan \theta \pm \sec \theta$

Since ${\alpha }_2>{\beta }_2,{\beta }_2=-\tan \theta -\sec \theta$

$\Rightarrow {\alpha }_1+{\beta }_2=-2\tan \theta$