Let $-\frac{\pi}{6} \beta_{1}$ and $\alpha_{2}>\beta_{2}$, then $\alpha_{1}+\beta_{2}$ equals

(A) $2(\sec \theta-\tan \theta)$

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Quadratic Equations and Inequalities

Let

- \frac{π}{6} < θ < - \frac{π}{12}

. Suppose

α_{1}

and

β_{1}

are the roots of the equation

x^{2} - 2 x sec θ + 1 = 0

and

α_{2}

and

β_{2}

are the roots of the equation

x^{2} + 2 x tan θ - 1 = 0

. If

α_{1} > β_{1}

and

α_{2} > β_{2}

, then

α_{1} + β_{2}

equals

Answer: (C)

For the first equation

x^{2} - 2 x sec θ + 1 = 0

, the roots are

α_{1}, β_{1} = sec θ \pm tan θ

. Since

α_{1} > β_{1}

and

- \frac{π}{6} < θ < - \frac{π}{12}

(so

sec θ > 0

and

tan θ < 0

, with

∣ tan θ ∣ < sec θ

), we have

α_{1} = sec θ - tan θ

and

β_{1} = sec θ + tan θ

.

For the second equation

x^{2} + 2 x tan θ - 1 = 0

, the roots are

α_{2}, β_{2} = - tan θ \pm sec θ

. Since

α_{2} > β_{2}

, we have

α_{2} = - tan θ + sec θ

and

β_{2} = - tan θ - sec θ

.

Therefore,

α_{1} + β_{2} = (sec θ - tan θ) + (- tan θ - sec θ) = - 2 tan θ

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Single correct option