If `x^2-4x+log_(1/2)a=0` does not have two distinct real roots, then maximum value of a is

Options

A. `1//4`
B. `1//16`
C. `-1//4`
D. none of these

Answer

To solve the problem, we need to determine the maximum value of \( a \) such that the quadratic equation \( x^2 - 4x + \log_{1/2} a = 0 \) does not have two distinct real roots. This occurs when the discriminant of the quadratic is less than or equal to zero.

### Step-by-Step Solution:

1. **Identify the coefficients of the quadratic equation:**
The given quadratic equation is:
\[ x^2 - 4x + \log_{1/2} a = 0 \]
Here, we can identify:
- \( A = 1 \)
- \( B = -4 \)
- \( C = \log_{1/2} a \)

2. **Write the condition for non-distinct roots:**
For the quadratic equation to not have two distinct real roots, the discriminant must be less than or equal to zero:
\[ D = B^2 - 4AC \leq 0 \]

3. **Calculate the discriminant:**
Substitute the values of \( A \), \( B \), and \( C \) into the discriminant formula:
\[ D = (-4)^2 - 4 \cdot 1 \cdot \log_{1/2} a \]
Simplifying this gives:
\[ D = 16 - 4 \log_{1/2} a \]

4. **Set up the inequality:**
We need:
\[ 16 - 4 \log_{1/2} a \leq 0 \]
Rearranging this gives:
\[ 4 \log_{1/2} a \geq 16 \]

5. **Divide by 4:**
\[ \log_{1/2} a \geq 4 \]

6. **Convert the logarithmic inequality to exponential form:**
Since the base \( \frac{1}{2} \) is less than 1, the inequality reverses when converting to exponential form:
\[ a \leq \left(\frac{1}{2}\right)^4 \]

7. **Calculate the maximum value of \( a \):**
\[ a \leq \frac{1}{16} \]

8. **Final conclusion:**
The maximum value of \( a \) is:
\[ \boxed{16} \]