The number of integral values of k, for which one root of the equation `2x^2 –8x + k = 0` lies in the interval `(1, 2)` and its other root lies in the interval `(2, 3)` is

Options

A. `2`
B. `0`
C. `1`
D. `3`

Answer

To solve the problem, we need to find the number of integral values of \( k \) for which one root of the quadratic equation \( 2x^2 - 8x + k = 0 \) lies in the interval \( (1, 2) \) and the other root lies in the interval \( (2, 3) \).

### Step 1: Identify the quadratic function
The given quadratic equation is:
\[ f(x) = 2x^2 - 8x + k \]

### Step 2: Evaluate the function at the endpoints of the intervals
We need to evaluate \( f(1) \), \( f(2) \), and \( f(3) \):
1. Calculate \( f(1) \):
\[ f(1) = 2(1)^2 - 8(1) + k = 2 - 8 + k = k - 6 \]

2. Calculate \( f(2) \):
\[ f(2) = 2(2)^2 - 8(2) + k = 8 - 16 + k = k - 8 \]

3. Calculate \( f(3) \):
\[ f(3) = 2(3)^2 - 8(3) + k = 18 - 24 + k = k - 6 \]

### Step 3: Apply the Intermediate Value Theorem
For the roots to lie in the specified intervals, we require:
- One root in \( (1, 2) \) implies \( f(1) \cdot f(2) < 0 \):
\[ (k - 6)(k - 8) < 0 \]

- One root in \( (2, 3) \) implies \( f(2) \cdot f(3) < 0 \):
\[ (k - 8)(k - 6) < 0 \]

### Step 4: Solve the inequalities
1. From \( (k - 6)(k - 8) < 0 \):
- The critical points are \( k = 6 \) and \( k = 8 \).
- The solution to this inequality is \( 6 < k < 8 \).

2. From \( (k - 8)(k - 6) < 0 \):
- The critical points are the same \( k = 6 \) and \( k = 8 \).
- The solution is also \( 6 < k < 8 \).

### Step 5: Determine integral values of \( k \)
The only integral value of \( k \) that satisfies \( 6 < k < 8 \) is:
\[ k = 7 \]

### Conclusion
Thus, the number of integral values of \( k \) for which one root lies in \( (1, 2) \) and the other in \( (2, 3) \) is:
\[ \boxed{1} \]