4.2- Zadanie 4.2: Kurs maturalny z matematyki rozszerzonej

Rozwiązanie

$x^{2} - (2 m - 1) x + m - \frac{1}{2} = 0$

$(1) m \in (- \infty, \frac{1}{2}) \cup (\frac{3}{2}, \infty)$

$(2) x_{1} \neq = 0 \land x_{2} \neq = 0 \to x_{1} \cdot x_{2} \neq = 0$
$m - \frac{1}{2} \neq = 0 \to m \neq = \frac{1}{2}$

$(3) \frac{1}{x _{1}^{2}} + \frac{1}{x _{2}^{2}} > 4$
$\frac{1}{x _{1}^{2}} + \frac{1}{x _{2}^{2}} = \frac{x _{2}^{2}}{x _{1}^{2} x _{2}^{2}} + \frac{x _{1}^{2}}{x _{1}^{2} x _{2}^{2}} =$
$= \frac{x _{1}^{2} + 2 x _{1} x _{2} - 2 x _{1} x _{2} + x _{2}^{2}}{( x _{1} x _{2} ) ^{2}} =$
$= \frac{( x _{1} + x _{2} ) ^{2} - 2 x _{1} x _{2}}{( x _{1} x _{2} ) ^{2}}$

$\frac{( x _{1} + x _{2} ) ^{2} - 2 x _{1} x _{2}}{( x _{1} x _{2} ) ^{2}} > 4$
$\frac{( 2 m - 1 ) ^{2} - 2 ( m - \frac{1}{2} )}{( m - \frac{1}{2} ) ^{2}} > 4 \cdot (m - \frac{1}{2})^{2}$

$4 m^{2} - 4 m + 1 - 2 m + 1 > 4 (m^{2} - m + \frac{1}{4})$
$4 m^{2} - 6 m + 2 > 4 m^{2} - 4 m + 1$
$- 2 m < - 1$
$m > \frac{1}{2}$
$m \in (\frac{1}{2}, \infty)$