# Number of values of the parameter alpha in [0, 2pi] for which the quadratic function,  (sin alpha) x^2 + 2 cos alpha x + 1/2 (cos alpha +sin alpha) is the square of a linear function is? (A) 2 (B) 3 (C) 4 (D) 1

Mar 5, 2017

See below.

#### Explanation:

If we know that the expression must be the square of a linear form then

$\left(\sin \alpha\right) {x}^{2} + 2 \cos \alpha x + \frac{1}{2} \left(\cos \alpha + \sin \alpha\right) = {\left(a x + b\right)}^{2}$

then grouping coefficients we have

$\left({\alpha}^{2} - \sin \left(\alpha\right)\right) {x}^{2} + \left(2 a b - 2 \cos \alpha\right) x + {b}^{2} - \frac{1}{2} \left(\sin \alpha + \cos \alpha\right) = 0$

so the condition is

$\left\{\begin{matrix}{a}^{2} - \sin \left(\alpha\right) = 0 \\ a b - \cos \alpha = 0 \\ {b}^{2} - \frac{1}{2} \left(\sin \alpha + \cos \alpha\right) = 0\end{matrix}\right.$

This can be solved obtaining first the values for $a , b$ and substituting.

We know that ${a}^{2} + {b}^{2} = \sin \alpha + \frac{1}{\sin \alpha + \cos \alpha}$ and
${a}^{2} {b}^{2} = {\cos}^{2} \alpha$ Now solving

${z}^{2} - \left({a}^{2} + {b}^{2}\right) z + {a}^{2} {b}^{2} = 0$. Solving and substituting for ${a}^{2} = \sin \alpha$ we obtain

$a = b = \pm \frac{1}{\sqrt[4]{2}} , \alpha = \frac{\pi}{4}$
$a = \pm \frac{\sqrt{2}}{\sqrt[4]{5}} , b = \pm \frac{1}{\sqrt{2} \sqrt[4]{5}} , \alpha = \pi - {\tan}^{-} 1 \left(2\right)$