Let _ and _ be the distinct roots of 2x^2+( )x-1=0, (0,2 ).

Question

Let ${\alpha }_{	heta }$ and ${\beta }_{	heta }$ be the distinct roots of $2{x}^{2}+(\cos 	heta )x-1=0,	heta \in (0,2\pi )$. If m and M are the minimum and the maximum values of ${\alpha }_{4}^{	heta }+{\beta }_{4}^{	heta }$, then $16(M+m)$ equals :

Accepted Answer

To find the sum of the fourth powers of the roots $\alpha_	heta$ and $\beta_	heta$ of the quadratic equation $2x^2 + (\cos 	heta)x - 1 = 0$, we start analyzing the expression $\alpha_	heta^4 + \beta_	heta^4$.

The equation can be rewritten with its roots using:

$    \alpha + \beta = -\frac{\cos 	heta}{2}, \quad \alpha \beta = -\frac{1}{2}    $

We need to calculate $\alpha^2 + \beta^2$ and $\alpha^2 \beta^2$:

$    \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{\cos 	heta}{2}ight)^2 - 2\left(-\frac{1}{2}ight) = \frac{\cos^2 	heta}{4} + 1    $

$    \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(-\frac{1}{2}ight)^2 = \frac{1}{4}    $

Substitute these into:

$    \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2 \beta^2 = \left(\frac{\cos^2 	heta}{4} + 1ight)^2 - \frac{1}{2}    $

Maximize and minimize $\left(\frac{\cos^2 	heta}{4} + 1ight)^2$:

Zero of $\cos	heta$ leads to:

$      \left(\frac{0^2}{4} + 1ight)^2 = 1      $

Max value $\cos^2 	heta = 1$:

$      \left(\frac{1}{4} + 1ight)^2 = \left(\frac{5}{4}ight)^2 = \frac{25}{16}      $

Substitute back:

$      	ext{Max: } \frac{25}{16} - \frac{1}{2} = \frac{25}{16} - \frac{8}{16} = \frac{17}{16}      $

$      	ext{Min: } 1 - \frac{1}{2} = \frac{1}{2}      $

Finally, compute $16(M + m)$:

$    16\left(\frac{17}{16} + \frac{1}{2}ight) = 16\left(\frac{17}{16} + \frac{8}{16}ight) = 16 	imes \frac{25}{16} = 25    $

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