1. **Problem statement:** Find all real values of $\lambda$ such that the quadratic equation $$(\lambda^2 + 1)x^2 - 4\lambda x + 2 = 0$$ has exactly one root in the interval $(0,1)$. 2. **Recall the quadratic formula:** For $ax^2 + bx + c = 0$, roots are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$ The discriminant $\Delta = b^2 - 4ac$ determines the nature of roots. 3. **Identify coefficients:** Here, $$a = \lambda^2 + 1 > 0, \quad b = -4\lambda, \quad c = 2.$$ Since $a > 0$ for all real $\lambda$, the parabola opens upwards. 4. **Condition for roots:** The quadratic can have 0, 1, or 2 real roots depending on $\Delta$. Calculate discriminant: $$\Delta = (-4\lambda)^2 - 4(\lambda^2 + 1)(2) = 16\lambda^2 - 8(\lambda^2 + 1) = 16\lambda^2 - 8\lambda^2 - 8 = 8\lambda^2 - 8 = 8(\lambda^2 - 1).$$ 5. **Number of roots:** - If $\Delta < 0$, no real roots. - If $\Delta = 0$, one real root (double root). - If $\Delta > 0$, two distinct real roots. 6. **We want exactly one root in $(0,1)$:** This can happen in two ways: **Case 1: Double root inside $(0,1)$** - $\Delta = 0 \implies 8(\lambda^2 - 1) = 0 \implies \lambda^2 = 1 \implies \lambda = \pm 1.$ Find the root for $\lambda = 1$: $$a = 1^2 + 1 = 2, \quad b = -4(1) = -4, \quad c = 2.$$ Root: $$x = \frac{-b}{2a} = \frac{4}{4} = 1.$$ Root is at $x=1$, which is not inside $(0,1)$. For $\lambda = -1$: $$a = (-1)^2 + 1 = 2, \quad b = -4(-1) = 4, \quad c = 2.$$ Root: $$x = \frac{-b}{2a} = \frac{-4}{4} = -1,$$ not in $(0,1)$. So no double root inside $(0,1)$. **Case 2: Two distinct roots, exactly one in $(0,1)$** Since $a>0$, parabola opens upward. The quadratic is continuous and crosses the x-axis at two points if $\Delta > 0$. 7. **Find roots explicitly:** $$x = \frac{4\lambda \pm \sqrt{8(\lambda^2 - 1)}}{2(\lambda^2 + 1)} = \frac{4\lambda \pm 2\sqrt{2}\sqrt{\lambda^2 - 1}}{2(\lambda^2 + 1)} = \frac{2\lambda \pm \sqrt{2}\sqrt{\lambda^2 - 1}}{\lambda^2 + 1}.$$ 8. **Domain for real roots:** $\lambda^2 - 1 \geq 0 \implies |\lambda| \geq 1$. 9. **We want exactly one root in $(0,1)$:** Let the roots be: $$x_1 = \frac{2\lambda - \sqrt{2}\sqrt{\lambda^2 - 1}}{\lambda^2 + 1}, \quad x_2 = \frac{2\lambda + \sqrt{2}\sqrt{\lambda^2 - 1}}{\lambda^2 + 1}.$$ Since $a>0$, $x_1 < x_2$. 10. **Check intervals for $\lambda \geq 1$:** - For $\lambda = 1$, roots coincide at $x=1$ (excluded). - For $\lambda > 1$, analyze roots: Check if $x_1 < 0$ and $0 < x_2 < 1$ to have exactly one root in $(0,1)$. - $x_1 < 0$: $$2\lambda - \sqrt{2}\sqrt{\lambda^2 - 1} < 0 \implies 2\lambda < \sqrt{2}\sqrt{\lambda^2 - 1}.$$ Square both sides: $$4\lambda^2 < 2(\lambda^2 - 1) \implies 4\lambda^2 < 2\lambda^2 - 2 \implies 2\lambda^2 < -2,$$ which is impossible for real $\lambda$. So $x_1$ is never negative for $\lambda > 1$. Try $x_1$ in $(0,1)$: - $x_1 > 0$ and $x_1 < 1$? Similarly for $x_2$. 11. **Try $\lambda = 2$ as example:** $$x_1 = \frac{4 - \sqrt{2}\sqrt{4 - 1}}{5} = \frac{4 - \sqrt{2}\sqrt{3}}{5} \approx \frac{4 - 2.45}{5} = \frac{1.55}{5} = 0.31,$$ $$x_2 = \frac{4 + 2.45}{5} = \frac{6.45}{5} = 1.29 > 1.$$ So one root in $(0,1)$. 12. **Try $\lambda = -2$:** $$x_1 = \frac{-4 - \sqrt{2}\sqrt{4 - 1}}{5} = \frac{-4 - 2.45}{5} = \frac{-6.45}{5} = -1.29 < 0,$$ $$x_2 = \frac{-4 + 2.45}{5} = \frac{-1.55}{5} = -0.31 < 0,$$ No roots in $(0,1)$. 13. **Summary:** For $\lambda > 1$, exactly one root lies in $(0,1)$. For $\lambda < -1$, no roots in $(0,1)$. For $|\lambda| < 1$, no real roots. For $\lambda = \pm 1$, double root at $x=1$ or $x=-1$ (not in $(0,1)$). 14. **Final answer:** $$\boxed{\lambda \in (1, \infty)}.$$