Question:

Step 1: Analyze the given quadratic equation and conditions for its roots. The given quadratic equation is $x^2 - 2mx + m^2 - 1 = 0$. Let $f(x) = x^2 - 2mx + m^2 - 1$. For the roots of the equation to be real, the discriminant ($\Delta$) must be non-negative. The coefficient of $x^2$ is $a=1$, which is positive, meaning the parabola opens upwards. The conditions given are that both roots are greater than $-2$ and less than $4$. Let the roots be $\alpha$ and $\beta$. This implies $-2 < \alpha < 4$ and $-2 < \beta < 4$. For a quadratic $ax^2 + bx + c = 0$ with $a>0$, and both roots lying between $k_1$ and $k_2$, the following conditions must be satisfied: 1. $\Delta \ge 0$ (for real roots) 2. $a \cdot f(k_1) > 0$ 3. $a \cdot f(k_2) > 0$ 4. $k_1 < -\frac{b}{2a} < k_2$ (the vertex lies between $k_1$ and $k_2$) Step 2: Check the discriminant for real roots. The discriminant $\Delta$ for the quadratic equation $ax^2 + bx + c = 0$ is given by $b^2 - 4ac$. For $x^2 - 2mx + m^2 - 1 = 0$, we have $a=1$, $b=-2m$, $c=m^2-1$. $$ \Delta = (-2m)^2 - 4(1)(m^2 - 1) $$ $$ \Delta = 4m^2 - 4m^2 + 4 $$ $$ \Delta = 4 $$ Since $\Delta = 4 \ge 0$, the roots are always real for all values of $m$. This condition is always satisfied. Step 3: Apply the condition $f(-2) > 0$. Since $a=1 > 0$, for both roots to be greater than $-2$, the value of the quadratic function at $x=-2$ must be positive. $$ f(-2) = (-2)^2 - 2m(-2) + m^2 - 1 > 0 $$ $$ 4 + 4m + m^2 - 1 > 0 $$ $$ m^2 + 4m + 3 > 0 $$ Factorizing the quadratic expression: $$ (m+1)(m+3) > 0 $$ This inequality holds when $m < -3$ or $m > -1$. So, $m \in (-\infty, -3) \cup (-1, \infty)$. Step 4: Apply the condition $f(4) > 0$. Since $a=1 > 0$, for both roots to be less than $4$, the value of the quadratic function at $x=4$ must be positive. $$ f(4) = (4)^2 - 2m(4) + m^2 - 1 > 0 $$ $$ 16 - 8m + m^2 - 1 > 0 $$ $$ m^2 - 8m + 15 > 0 $$ Factorizing the quadratic expression: $$ (m-3)(m-5) > 0 $$ This inequality holds when $m < 3$ or $m > 5$. So, $m \in (-\infty, 3) \cup (5, \infty)$. Step 5: Apply the condition for the position of the vertex. The x-coordinate of the vertex of the parabola $ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. For $f(x) = x^2 - 2mx + m^2 - 1$, the vertex is at $x = -\frac{-2m}{2(1)} = m$. For both roots to lie between $-2$ and $4$, the x-coordinate of the vertex must also lie between $-2$ and $4$. $$ -2 < m < 4 $$ So, $m \in (-2, 4)$. Step 6: Find the intersection of all conditions. We need to find the values of $m$ that satisfy all three conditions derived: 1. From Step 3: $m \in (-\infty, -3) \cup (-1, \infty)$ 2. From Step 4: $m \in (-\infty, 3) \cup (5, \infty)$ 3. From Step 5: $m \in (-2, 4)$ Let's find the intersection of these intervals: First, intersect condition (1) and (3): $((-\infty, -3) \cup (-1, \infty)) \cap (-2, 4)$ The interval $(-\infty, -3)$ has no intersection with $(-2, 4)$. The interval $(-1, \infty)$ intersects with $(-2, 4)$ to give $(-1, 4)$. So, the intersection of (1) and (3) is $(-1, 4)$. Next, intersect this result with condition (2): $(-1, 4) \cap ((-\infty, 3) \cup (5, \infty))$ The interval $(-1, 4)$ intersects with $(-\infty, 3)$ to give $(-1, 3)$. The interval $(-1, 4)$ has no intersection with $(5, \infty)$. Therefore, the common interval for $m$ is $(-1, 3)$. The final answer is $\boxed{(-1 < m < 3)}$.