If the quadratic polynomial, $\displaystyle y=(\cot \alpha )x^{2}+2\left ( \sqrt{\sin \alpha } \right )x +\frac{1}{2}\tan \alpha, \ \ \alpha \, \varepsilon \, [0,2\pi ],$ can take negative values for all $\displaystyle x \, \varepsilon \, R,$ then the value of $\displaystyle \alpha $ must in the interval
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$\displaystyle \left ( \frac{5 \pi}{6},\pi \right )$
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$\displaystyle \left ( \frac{5 \pi}{6}, \pi \right )\cup \left ( \frac{11 \pi}{6},2 \pi\ \right )$
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$\displaystyle \left ( \frac{11\pi}{6},2 \pi \right )$
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$\displaystyle \left ( 0,\frac{\pi}{6} \right )\cup \left ( \frac{5 \pi}{6}, \pi \right )$
A
Correct answer
Explanation
For a quadratic to be negative for all x, the coefficient of x^2 (cot alpha) must be negative and the discriminant must be negative. cot alpha < 0 implies alpha is in (pi/2, pi) or (3pi/2, 2pi). Solving the discriminant condition leads to the interval (5pi/6, pi).