Question

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

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

Answer 1

Answer: (C)

real and unequal roots

$\frac{1}{x-sin\alpha }+\frac{1}{x-sin\beta }+\frac{1}{x-sin\gamma }=0$
$\Rightarrow 3x^2+2(sin\alpha +sin\beta +sin\gamma )x+sin\alpha sin\beta +sin\beta sin\gamma +sin\gamma sin\alpha =0$
$\Delta =b^2-4ac$
$=si{n}^2\alpha +si{n}^2\beta +si{n}^2\alpha -sin\alpha sin\beta -sin\beta sin\gamma -sin\gamma sin\alpha =0$
$=\frac{1}{2}\left[\right(sin\alpha -sin\beta {)}^2+(sin\beta -sin\gamma {)}^2+(sin\gamma -sin\alpha {)}^2]$
$\Delta >0$
$\therefore$ Roots are real and unequal .