1. **State the problem:** Find all values of $m$ such that the inequalities hold for every real number $x$: 1. $m x (x - 2) < m^2 x^2 + 2$ 2. $2x(2x + 1) - 1 < m x (m x + 1)$ 2. **Analyze the first inequality:** Rewrite the inequality: $$m x (x - 2) < m^2 x^2 + 2$$ Expand the left side: $$m x^2 - 2 m x < m^2 x^2 + 2$$ Bring all terms to one side: $$m x^2 - 2 m x - m^2 x^2 < 2$$ Group $x^2$ terms: $$(m - m^2) x^2 - 2 m x < 2$$ Define $A = m - m^2 = m(1 - m)$, so: $$A x^2 - 2 m x < 2$$ Rewrite as: $$A x^2 - 2 m x - 2 < 0$$ For the inequality to hold for all $x$, the quadratic function $$f(x) = A x^2 - 2 m x - 2$$ must be strictly less than zero for all real $x$. 3. **Condition for $f(x) < 0$ for all $x$:** - If $A > 0$, the parabola opens upward, so $f(x)$ will be negative only between roots, not for all $x$. - If $A < 0$, the parabola opens downward, and $f(x)$ is less than zero outside the roots. - To have $f(x) < 0$ for all $x$, the quadratic must have no real roots and open downward, so: 1. $A < 0$ 2. Discriminant $D < 0$ 4. **Calculate discriminant $D$ of $f(x)$:** $$D = (-2 m)^2 - 4 imes A imes (-2) = 4 m^2 + 8 A = 4 m^2 + 8 m (1 - m) = 4 m^2 + 8 m - 8 m^2 = -4 m^2 + 8 m$$ 5. **Apply conditions:** - $A = m (1 - m) < 0$ implies $m (1 - m) < 0$. This inequality holds when $m$ is between 0 and 1: $$0 < m < 1$$ - $D < 0$: $$-4 m^2 + 8 m < 0 \\ 8 m - 4 m^2 < 0 \\ 4 m (2 - m) < 0$$ Since $4 > 0$, the inequality reduces to: $$m (2 - m) < 0$$ This holds when $m < 0$ or $m > 2$. 6. **Combine both conditions:** - From $A < 0$: $0 < m < 1$ - From $D < 0$: $m < 0$ or $m > 2$ There is no overlap, so no $m$ satisfies both simultaneously. Therefore, no $m$ makes the inequality true for all $x$ if $A eq 0$. 7. **Check the case $A = 0$:** $$m (1 - m) = 0 \\ m = 0 ext{ or } m = 1$$ - For $m=0$: $$0 < 0 + 2$$ Inequality becomes: $$0 < 2$$ True for all $x$. - For $m=1$: $$1 imes x (x - 2) < 1^2 x^2 + 2 \\ x^2 - 2 x < x^2 + 2 \\ -2 x < 2 \\ x > -1$$ This is not true for all $x$ (fails for $x o - ext{large}$). 8. **Conclusion for first inequality:** The inequality holds for all $x$ only if: $$m = 0$$ 9. **Analyze the second inequality:** $$2 x (2 x + 1) - 1 < m x (m x + 1)$$ Expand both sides: Left: $$4 x^2 + 2 x - 1$$ Right: $$m^2 x^2 + m x$$ Inequality: $$4 x^2 + 2 x - 1 < m^2 x^2 + m x$$ Bring all terms to one side: $$4 x^2 + 2 x - 1 - m^2 x^2 - m x < 0$$ Group terms: $$(4 - m^2) x^2 + (2 - m) x - 1 < 0$$ 10. **Define:** $$g(x) = (4 - m^2) x^2 + (2 - m) x - 1$$ We want $g(x) < 0$ for all real $x$. 11. **Conditions for $g(x) < 0$ for all $x$:** - Parabola opens downward: coefficient of $x^2$ negative: $$4 - m^2 < 0 \\ m^2 > 4 \\ m < -2 ext{ or } m > 2$$ - Discriminant $D_g < 0$: $$D_g = (2 - m)^2 - 4 (4 - m^2)(-1)$$ $$= (2 - m)^2 + 4 (4 - m^2)$$ Expand: $$(2 - m)^2 = m^2 - 4 m + 4$$ So: $$D_g = m^2 - 4 m + 4 + 16 - 4 m^2 = -3 m^2 - 4 m + 20$$ 12. **Solve $D_g < 0$:** $$-3 m^2 - 4 m + 20 < 0 \\ 3 m^2 + 4 m - 20 > 0$$ Find roots of $3 m^2 + 4 m - 20 = 0$: $$m = \frac{-4 \pm \sqrt{16 + 240}}{6} = \frac{-4 \pm 16}{6}$$ Roots: $$m_1 = \frac{-4 - 16}{6} = -\frac{20}{6} = -\frac{10}{3} \approx -3.33$$ $$m_2 = \frac{-4 + 16}{6} = \frac{12}{6} = 2$$ Since leading coefficient is positive, inequality $3 m^2 + 4 m - 20 > 0$ holds for: $$m < -\frac{10}{3} \text{ or } m > 2$$ 13. **Combine conditions for $g(x) < 0$ for all $x$:** - $m < -2$ or $m > 2$ - $m < -\frac{10}{3}$ or $m > 2$ Intersection: - For $m < -2$, must also have $m < -\frac{10}{3} \approx -3.33$, so $m < -3.33$ - For $m > 2$, both conditions hold. 14. **Final solution for second inequality:** $$m < -\frac{10}{3} \text{ or } m > 2$$ 15. **Summary:** - For inequality 1 to hold for all $x$: $$m = 0$$ - For inequality 2 to hold for all $x$: $$m < -\frac{10}{3} \text{ or } m > 2$$ **Answer:** $$\boxed{\text{Inequality 1: } m=0; \quad \text{Inequality 2: } m < -\frac{10}{3} \text{ or } m > 2}$$