Question

If \(0<\alpha<\beta<\gamma<\frac{\pi}{2}\) then the equation \(\frac{1}{x-\sin\alpha}+\frac{1}{x-\sin\beta}+\frac{1}{x-\sin\gamma}=0\) has:

• A. real and unequal roots
• B. imaginary roots
• C. real and equal roots
• D. rational roots

Hint:

For any equation structured as $\sum_{i=1}^n \frac{1}{x - a_i} = 0$ with sorted constants $a_1 < a_2 < \dots < a_n$, there will always be exactly one real root hidden inside every intermediate open interval $(a_i, a_{i+1})$. The roots alternate perfectly between the poles!

Answer 1

The Correct Option is A

Concept: The tracking of real roots across a rational function equation containing distinct singularities can be performed efficiently using the Intermediate Value Theorem (IVT). If a continuous function changes signs within an interval $(a, b)$, it must cross the zero axis at least once inside that interval. Step 1: Establish the relative order of the vertical asymptotes.
Let our given function expression be defined as: $$f(x) = \frac{1}{x-\sin \alpha}+\frac{1}{x-\sin \beta}+\frac{1}{x-\sin \gamma}$$ The three poles or vertical asymptotes occur where the individual denominators vanish: $x = \sin\alpha$, $x = \sin\beta$, and $x = \sin\gamma$. We are given that the angles lie entirely within the first quadrant ($0 < \alpha < \beta < \gamma < \frac{\pi}{2}$). Since $\sin x$ is strictly increasing in this quadrant, their values maintain a strict order: $$\sin\alpha < \sin\beta < \sin\gamma$$

Step 2: Analyze functional sign changes in the first interval $(\sin\alpha, \sin\beta)$.
Let us check the limits of $f(x)$ at the boundaries of this first interval:
• As $x \rightarrow \sin\alpha^+$, the term $\frac{1}{x-\sin\alpha} \rightarrow +\infty$, causing the whole function to blow up positively: $\lim_{x \rightarrow \sin\alpha^+} f(x) = +\infty$.
• As $x \rightarrow \sin\beta^-$, the term $\frac{1}{x-\sin\beta} \rightarrow -\infty$, causing the whole function to drop negatively: $\lim_{x \rightarrow \sin\beta^-} f(x) = -\infty$. Because $f(x)$ is perfectly continuous inside $(\sin\alpha, \sin\beta)$ and scales from positive infinity down to negative infinity, the Intermediate Value Theorem guarantees that at least one real root $x_1$ lies inside this open interval.

Step 3: Analyze functional sign changes in the second interval $(\sin\beta, \sin\gamma)$.
Let us apply the same limiting steps to the boundaries of our second adjacent interval:
• As $x \rightarrow \sin\beta^+$, the dominant term is now $\frac{1}{x-\sin\beta} \rightarrow +\infty \implies \lim_{x \rightarrow \sin\beta^+} f(x) = +\infty$.
• As $x \rightarrow \sin\gamma^-$, the terminal term is $\frac{1}{x-\sin\gamma} \rightarrow -\infty \implies \lim_{x \rightarrow \sin\gamma^-} f(x) = -\infty$. Once again, the continuous function flips signs entirely across the boundary limits, confirming that a second distinct real root $x_2$ must exist inside $(\sin\beta, \sin\gamma)$.

Step 4: Determine the maximum polynomial root degree.
If we simplify the rational expression into a standard polynomial equation by multiplying by the shared common denominator $\prod (x - \sin\theta) = 0$, the equation reduces down to a quadratic equation of degree 2. By fundamental algebra, a second-degree polynomial can possess a maximum of 2 roots. Since we have already isolated two real and separate roots ($x_1 < \sin\beta < x_2$), the roots are definitively real and unequal.