Key Concepts and Formulas

• Location of Roots: For a quadratic $f(x) = ax^2 + bx + c$ with roots $\alpha$ and $\beta$, if $a > 0$ and $f(p) < 0$, then $\alpha < p < \beta$. If $a > 0$ and $f(p) > 0$, then $p < \alpha$ or $p > \beta$.
• Quadratic Formula and Discriminant: The roots of $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. The discriminant is $\Delta = b^2 - 4ac$.
• Interval Notation: $(a, b)$ denotes the open interval between $a$ and $b$, excluding $a$ and $b$.

Step-by-Step Solution

Step 1: Define the quadratic function and analyze the given information.

We are given the quadratic equation $2x^2 - 8x + k = 0$. Let $f(x) = 2x^2 - 8x + k$. We know that one root, $\alpha$, lies in the interval $(1, 2)$, and the other root, $\beta$, lies in the interval $(2, 3)$. Therefore, $1 < \alpha < 2$ and $2 < \beta < 3$. Since the coefficient of $x^2$ is $2 > 0$, the parabola opens upwards.

Step 2: Apply the location of roots concept at x = 1.

Since $1 < \alpha < 2 < \beta < 3$, we have $1 < \alpha$. Because the parabola opens upwards, $f(1) > 0$. $f(1) = 2(1)^2 - 8(1) + k = 2 - 8 + k = k - 6$. Therefore, $k - 6 > 0$, which implies $k > 6$.

Step 3: Apply the location of roots concept at x = 2.

Since $1 < \alpha < 2 < \beta < 3$, we have $\alpha < 2 < \beta$. Because the parabola opens upwards, $f(2) < 0$. $f(2) = 2(2)^2 - 8(2) + k = 8 - 16 + k = k - 8$. Therefore, $k - 8 < 0$, which implies $k < 8$.

Step 4: Apply the location of roots concept at x = 3.

Since $1 < \alpha < 2 < \beta < 3$, we have $\beta < 3$. Because the parabola opens upwards, $f(3) > 0$. $f(3) = 2(3)^2 - 8(3) + k = 18 - 24 + k = k - 6$. Therefore, $k - 6 > 0$, which implies $k > 6$.

Step 5: Determine the range of k.

Combining the inequalities, we have $k > 6$ and $k < 8$. This can be written as $6 < k < 8$.

Step 6: Find the integral values of k.

We are looking for integer values of $k$ that satisfy $6 < k < 8$. The only integer that satisfies this inequality is $k = 7$.

Step 7: Count the number of integral values.

There is only one integer value of $k$ that satisfies the given conditions, which is $k=7$.

Common Mistakes & Tips

• Sign of 'a': Always pay attention to the sign of the leading coefficient 'a'. It dictates the direction of the parabola and thus the signs of $f(x)$ relative to the roots.
• Strict Inequalities: Since the roots lie strictly within the given intervals, use strict inequalities ($>, <$). If the roots could lie at the endpoints, use non-strict inequalities ($\ge, \le$).
• Discriminant Check (Not Needed Here): In some problems, after finding the range of $k$, you might need to check if the discriminant is non-negative to ensure the roots are real. In this case, the condition $f(2) < 0$ already guarantees real and distinct roots.

Summary

The problem requires finding the integral values of $k$ for which the roots of $2x^2 - 8x + k = 0$ lie in the intervals $(1, 2)$ and $(2, 3)$. By applying the location of roots theorem with the conditions $f(1) > 0$, $f(2) < 0$, and $f(3) > 0$, we found that $6 < k < 8$. Therefore, the only integer value of $k$ is $7$, and the number of such values is 1.

Final Answer

The final answer is \boxed{1}, which corresponds to option (C).