Multiple choice

lf the sum of the squares of the roots of the equation $x^2-(\sin\alpha-2)x-(1+\sin\alpha)=0$ is least, then $\alpha$ is equal to

  1. $\dfrac{\pi}{4}$
  2. $\dfrac{\pi}{3}$
  3. $\dfrac{\pi}{2}$
  4. $\dfrac{\pi}{6}$
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C Correct answer
Explanation

Let s = sin(alpha). The sum of the squares of the roots is (s - 2)^2 - 2[-(1 + s)] = s^2 - 2s + 6. This equals (s - 1)^2 + 5, which is least when sin(alpha) = 1, giving alpha = pi/2.

AI explanation

Let the roots of the quadratic equation be alpha and beta; their sum equals (sin(a) - 2) and their product equals -(1 + sin(a)). The sum of the squares of the roots is given by the identity (alpha + beta)^2 - 2(alpha)(beta). Substituting the sum and product gives (sin(a) - 2)^2 - 2(-(1 + sin(a))) equals sin^2(a) - 4sin(a) + 4 + 2 + 2sin(a), which simplifies to sin^2(a) - 2sin(a) + 6. To find the minimum of this expression, we complete the square to get (sin(a) - 1)^2 + 5, which reaches its least value when sin(a) equals 1. This occurs when a equals pi/2. The result is pi/2.

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