Multiple choice

If $1$ lies between the roots of the equation $\displaystyle 3x^{2}-3\sin \alpha x-2\cos ^{2}\alpha = 0$, then $\alpha $ lies in the interval

  1. $\displaystyle \left ( 0,\frac{\pi }{2} \right )$
  2. $\displaystyle \left ( \frac{\pi }{12},\frac{\pi }{2} \right )$
  3. $\displaystyle \left ( \frac{\pi }{6},\frac{5\pi }{6} \right )$
  4. $\displaystyle \left ( \frac{\pi }{6},\frac{\pi }{2} \right )\cup \left ( \frac{\pi }{2},\frac{5\pi }{6} \right )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let f(x) = 3x^2 - 3sin(a)x - 2cos^2(a). If 1 lies between the roots, then f(1) < 0. So, 3 - 3sin(a) - 2cos^2(a) < 0. Substituting cos^2(a) = 1 - sin^2(a), we get 3 - 3sin(a) - 2 + 2sin^2(a) < 0, which is 2sin^2(a) - 3sin(a) + 1 < 0. Factoring gives (2sin(a) - 1)(sin(a) - 1) < 0. This holds when 1/2 < sin(a) < 1. This corresponds to a in (pi/6, pi/2) U (pi/2, 5pi/6).

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