Multiple choice

lf $ 0<\alpha<\beta<\gamma <\displaystyle \dfrac{\pi}{2}$ , then the equation $\displaystyle \dfrac{1}{x-\sin\alpha}+\dfrac{1}{x-\sin\beta}+\dfrac{1}{x-\sin\gamma}=0$ has

  1. imaginary roots

  2. real and equal roots

  3. real and unequal roots

  4. rational roots
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let f(x) = 1/(x-sin(alpha)) + 1/(x-sin(beta)) + 1/(x-sin(gamma)). Since sin(alpha) < sin(beta) < sin(gamma), the function has vertical asymptotes at these values. Between each asymptote, the function goes from -infinity to +infinity, ensuring three real roots.

AI explanation

Multiplying the given equation by the product of the denominators yields the quadratic equation $3x^2 - 2(\sin\alpha + \sin\beta + \sin\gamma)x + (\sin\alpha\sin\beta + \sin\beta\sin\gamma + \sin\gamma\sin\alpha) = 0$. The discriminant of this quadratic is $4S_1^2 - 12S_2$, where $S_1$ is the sum of the sines and $S_2$ is the sum of their pairwise products. By the Cauchy-Schwarz inequality, $(\sin\alpha + \sin\beta + \sin\gamma)^2 > 3(\sin\alpha\sin\beta + \sin\beta\sin\gamma + \sin\gamma\sin\alpha)$ because $0 < \alpha < \beta < \gamma < \pi/2$ makes the variables distinct, meaning the discriminant is strictly positive. Therefore, the equation has real and unequal roots.

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