Theory Of Equations Ques 12
- 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
(2016 Adv.)
(a) $2(\sec \theta-\tan \theta)$
(b) $2 \sec \theta$
(c) $-2 \tan \theta$
(d) 0
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Answer:
Correct Answer: 12.(c)
Solution:
- Here, $x^{2}-2 x \sec \theta+1=0$ has roots $\alpha _1$ and $\beta _1$.
$ \begin{aligned} \therefore \quad \alpha _1, \beta _1 & =\frac{2 \sec \theta \pm \sqrt{4 \sec ^{2} \theta-4}}{2 \times 1} \\ & =\frac{2 \sec \theta \pm 2|\tan \theta|}{2} \end{aligned} $
Since, $\quad \theta \in (-\frac{\pi}{6},-\frac{\pi}{12})$,
i.e. $\quad \theta \in I V$ quadrant $=\frac{2 \sec \theta \mp 2 \tan \theta}{2}$
$\therefore \quad \alpha _1=\sec \theta-\tan \theta \operatorname{and} \beta _1=\sec \theta+\tan \theta\left[\operatorname{as} \alpha _1>\beta _1\right]$ and $x^{2}+2 x \tan \theta-1=0$ has roots $\alpha _2$ and $\beta _2$.
$ \begin{array}{lll} \text { i.e. } & \alpha _2, \beta _2 & =\frac{-2 \tan \theta \pm \sqrt{4 \tan ^{2} \theta+4}}{2} \\ \therefore & \alpha _2 & =-\tan \theta+\sec \theta \\ \text { and } & \beta _2 & =-\tan \theta-\sec \theta \quad\left[\operatorname{as} \alpha _2>\beta _2\right] \\ \text { Thus, } & \alpha _1+\beta _2 & =-2 \tan \theta \end{array} $
BETA
