Key Concepts and Formulas

• Root Location for Quadratic Equations: For a quadratic $f(x) = ax^2 + bx + c$ with $a > 0$, exactly one root lies in $(p, q)$ if $f(p)f(q) < 0$, or if $f(p)=0$ and the other root is in $(p,q)$, or if $f(q)=0$ and the other root is in $(p,q)$. A combined condition is $f(p)f(q) \le 0$, with careful checking of boundary cases where $f(p)$ or $f(q)$ are zero.
• Quadratic Formula: The roots of the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Solving Quadratic Inequalities: To solve $ax^2 + bx + c \le 0$ (or $\ge 0$), find the roots of $ax^2 + bx + c = 0$. If $a > 0$, the inequality is satisfied between the roots. If $a < 0$, the inequality is satisfied outside the roots.

Step-by-Step Solution

Step 1: Define the Quadratic Function and Analyze the Leading Coefficient

Let $f(x) = (\lambda^2 + 1)x^2 - 4\lambda x + 2$. We observe that the coefficient of the $x^2$ term is $\lambda^2 + 1$. Since $\lambda^2 \ge 0$ for all real $\lambda$, we have $\lambda^2 + 1 \ge 1 > 0$. This means the parabola opens upwards. Why: Identifying that the parabola opens upwards is crucial because the conditions for one root in an interval depend on the direction the parabola opens.

Step 2: Apply the Root Location Condition

For exactly one root to lie in the interval $(0, 1)$, we need $f(0)f(1) \le 0$. We will evaluate $f(0)$ and $f(1)$. Why: This condition ensures that the function changes sign within the interval, guaranteeing at least one root. We use $\le$ to include the possibility of a root at an endpoint.

Step 3: Evaluate f(0) and f(1)

• $f(0) = (\lambda^2 + 1)(0)^2 - 4\lambda(0) + 2 = 2$.
• $f(1) = (\lambda^2 + 1)(1)^2 - 4\lambda(1) + 2 = \lambda^2 + 1 - 4\lambda + 2 = \lambda^2 - 4\lambda + 3$. Why: These are the values of the quadratic at the endpoints of the interval, needed for the root location condition.

Step 4: Set up and Solve the Inequality

We have $f(0)f(1) \le 0$, so $2(\lambda^2 - 4\lambda + 3) \le 0$. Dividing by 2, we get $\lambda^2 - 4\lambda + 3 \le 0$. Factoring the quadratic, we have $(\lambda - 1)(\lambda - 3) \le 0$. This inequality is satisfied when $1 \le \lambda \le 3$. Why: Solving the inequality provides the initial range of possible values for $\lambda$.

Step 5: Check the Boundary Cases: $\lambda = 1$

If $\lambda = 1$, the quadratic equation becomes $(1^2 + 1)x^2 - 4(1)x + 2 = 0$, which simplifies to $2x^2 - 4x + 2 = 0$. Dividing by 2, we get $x^2 - 2x + 1 = 0$, or $(x - 1)^2 = 0$. This has a double root at $x = 1$, which is not in the interval $(0, 1)$. Therefore, $\lambda = 1$ is not a valid solution. Why: We must exclude $\lambda = 1$ because the root $x=1$ is not inside the open interval $(0, 1)$.

Step 6: Check the Boundary Cases: $\lambda = 3$

If $\lambda = 3$, the quadratic equation becomes $(3^2 + 1)x^2 - 4(3)x + 2 = 0$, which simplifies to $10x^2 - 12x + 2 = 0$. Dividing by 2, we get $5x^2 - 6x + 1 = 0$. Factoring, we have $(5x - 1)(x - 1) = 0$, so the roots are $x = \frac{1}{5}$ and $x = 1$. The root $x = \frac{1}{5}$ is in the interval $(0, 1)$, while $x = 1$ is not. Therefore, $\lambda = 3$ is a valid solution. Why: We include $\lambda = 3$ because exactly one root, $x = 1/5$, lies within the open interval $(0, 1)$.

Step 7: State the Final Interval for $\lambda$

Since $\lambda = 1$ is excluded and $\lambda = 3$ is included, the set of all real values of $\lambda$ is $(1, 3]$.

Common Mistakes & Tips

• Forgetting to Check Endpoints: Failing to check the boundary cases where $f(0) = 0$ or $f(1) = 0$ can lead to an incorrect interval. Remember that the problem asks for exactly one root in the open interval.
• Incorrectly Solving the Inequality: Make sure to correctly factor and solve the quadratic inequality.
• Not Considering the Leading Coefficient: Always consider whether the parabola opens upwards or downwards, as the conditions for root location depend on this.

Summary

We found the range of $\lambda$ by using the condition $f(0)f(1) \le 0$, where $f(x)$ is the given quadratic. We then checked the boundary values of the resulting interval to ensure that there was exactly one root within the open interval $(0, 1)$. This led us to exclude $\lambda = 1$ and include $\lambda = 3$, giving the final interval $(1, 3]$.

Final Answer The final answer is \boxed{(1, 3]}, which corresponds to option (D).