The sum of all possible integral value of 'k' for which `5x ^(2) -2k x +1lt 0` has exactly one integral solution :

Options

A. 2
B. 3
C. 5
D. 9

Answer

To solve the problem, we need to find the sum of all possible integral values of \( k \) for which the quadratic inequality \( 5x^2 - 2kx + 1 < 0 \) has exactly one integral solution.

### Step-by-step Solution:

1. **Understanding the Quadratic Inequality**:
The quadratic inequality is given by:
\[ 5x^2 - 2kx + 1 < 0 \]
We need to find the values of \( k \) for which this inequality has exactly one integral solution.

2. **Finding the Roots of the Quadratic**:
The roots of the quadratic equation \( 5x^2 - 2kx + 1 = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5 \), \( b = -2k \), and \( c = 1 \). Thus, the roots are:
\[ x = \frac{2k \pm \sqrt{(-2k)^2 - 4 \cdot 5 \cdot 1}}{2 \cdot 5} \]
Simplifying this gives:
\[ x = \frac{2k \pm \sqrt{4k^2 - 20}}{10} \]
\[ x = \frac{k \pm \sqrt{k^2 - 5}}{5} \]

3. **Condition for Exactly One Integral Solution**:
For the quadratic to have exactly one integral solution, the difference between the two roots must be less than or equal to 2:
\[ \left( \frac{k + \sqrt{k^2 - 5}}{5} - \frac{k - \sqrt{k^2 - 5}}{5} \right) \leq 2 \]
Simplifying this gives:
\[ \frac{2\sqrt{k^2 - 5}}{5} \leq 2 \]
Multiplying both sides by 5:
\[ 2\sqrt{k^2 - 5} \leq 10 \]
Dividing by 2:
\[ \sqrt{k^2 - 5} \leq 5 \]
Squaring both sides:
\[ k^2 - 5 \leq 25 \implies k^2 \leq 30 \]

4. **Finding Lower Bound**:
Since \( k^2 - 5 \) must also be greater than or equal to 0 (as square roots cannot be negative):
\[ k^2 - 5 \geq 0 \implies k^2 \geq 5 \]

5. **Combining the Inequalities**:
We now have:
\[ 5 \leq k^2 \leq 30 \]
Taking square roots gives:
\[ \sqrt{5} \leq |k| \leq \sqrt{30} \]
Approximating the square roots:
\[ \sqrt{5} \approx 2.236 \quad \text{and} \quad \sqrt{30} \approx 5.477 \]
Thus, the integral values of \( k \) are:
\[ k = -5, -4, -3, -2, 2, 3, 4, 5 \]

6. **Finding Valid Values**:
We need to check which of these values yield exactly one integral solution.
- For \( k = 3 \): Roots are \( \frac{3 \pm \sqrt{4}}{5} = \frac{3 \pm 2}{5} \) which gives \( 1 \) and \( 0.2 \) (not valid).
- For \( k = 4 \): Roots are \( \frac{4 \pm \sqrt{11}}{5} \) which gives approximately \( 0.14 \) and \( 1.46 \) (valid).
- For \( k = 5 \): Roots are \( \frac{5 \pm \sqrt{20}}{5} \) which gives approximately \( 0.89 \) and \( 1.89 \) (valid).

Thus, valid integral values of \( k \) are \( 4 \) and \( 5 \).

7. **Calculating the Sum**:
The sum of all valid integral values of \( k \) is:
\[ 4 + 5 = 9 \]

### Final Answer:
The sum of all possible integral values of \( k \) is \( \boxed{9} \).