Multiple choice

Let $-\dfrac{\pi}{6}< \theta <-\dfrac{\pi}{12}$. 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 + 2 x\tan\theta - 1 = 0 $. If $\alpha_1 > \beta _2 $ , then $\alpha_1 + \beta_2$ equals

  1. $2(\sec\theta - \tan\theta)$
  2. $2\sec\theta$
  3. $-2\tan\theta$
  4. $0$
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C Correct answer
Explanation

For the first equation, the roots are sec(theta) +/- tan(theta). Given the range of theta, alpha1 = sec(theta) + tan(theta). For the second equation, the roots are -tan(theta) +/- sec(theta), so beta2 = -tan(theta) - sec(theta). Summing these gives alpha1 + beta2 = -2*tan(theta).

AI explanation

Solving the first equation for alpha_1 gives x = sec(theta) + tan(theta), since sec(theta) is negative and tan(theta) is positive, making this root the larger one. Solving the second equation for beta_2 gives x = tan(theta) - sec(theta), which is the positive root since it equals the reciprocal of alpha_1. Adding these two roots yields (sec(theta) + tan(theta)) + (tan(theta) - sec(theta)), which simplifies to -2tan(theta).

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