Let a denotes the number of non-negative values of p for which the equation `p.2^x +2^(-x)=5` possess a unique solution. Then find the value of 'a'.

Options

A. `0`
B. `2`
C. `1`
D. None of these

Answer

To solve the problem, we need to determine the number of non-negative values of \( p \) for which the equation

\[ p \cdot 2^x + 2^{-x} = 5 \]

has a unique solution.

### Step-by-step Solution:

1. **Rewrite the Equation**:
Start by rewriting the equation in terms of \( t = 2^x \). Thus, we have:
\[ p \cdot t + \frac{1}{t} = 5 \]
Multiplying through by \( t \) (assuming \( t > 0 \)):
\[ p t^2 - 5t + 1 = 0 \]

2. **Identify Conditions for Unique Solutions**:
A quadratic equation \( at^2 + bt + c = 0 \) has a unique solution when its discriminant \( D \) is zero. The discriminant is given by:
\[ D = b^2 - 4ac \]
For our equation \( p t^2 - 5t + 1 = 0 \), we have:
- \( a = p \)
- \( b = -5 \)
- \( c = 1 \)

Thus, the discriminant becomes:
\[ D = (-5)^2 - 4 \cdot p \cdot 1 = 25 - 4p \]

3. **Set the Discriminant to Zero**:
For the quadratic to have a unique solution, set the discriminant equal to zero:
\[ 25 - 4p = 0 \]
Solving for \( p \):
\[ 4p = 25 \implies p = \frac{25}{4} \]

4. **Check for the Case When \( p = 0 \)**:
Additionally, we need to consider the case when \( p = 0 \):
\[ 0 \cdot t^2 - 5t + 1 = 0 \implies -5t + 1 = 0 \implies t = \frac{1}{5} \]
This gives us a valid solution for \( p = 0 \).

5. **Count Non-Negative Values of \( p \)**:
We have found two non-negative values of \( p \):
- \( p = 0 \)
- \( p = \frac{25}{4} \)

Therefore, the total number of non-negative values of \( p \) for which the equation has a unique solution is:
\[ a = 2 \]

### Final Answer:
The value of \( a \) is \( 2 \).