The value of 'a' for which the equation `4cosec^(2)pi(a+x)+a^(2)-4a=0` has a real solution is

Options

A. a=1
B. a=2
C. a=10
D. None of these

Answer

To solve the equation \( 4 \csc^2(\pi(a+x)) + a^2 - 4a = 0 \) for the value of \( a \) that allows for a real solution, we will follow these steps:

### Step 1: Rewrite the Equation
Start by isolating the trigonometric part of the equation:
\[ 4 \csc^2(\pi(a+x)) = 4a - a^2 \]
This can be rewritten as:
\[ \csc^2(\pi(a+x)) = \frac{4a - a^2}{4} \]

### Step 2: Set Conditions for Cosecant
Recall that \( \csc^2(\theta) \) is always greater than or equal to 1 for real \( \theta \). Therefore, we set the following inequality:
\[ \frac{4a - a^2}{4} \geq 1 \]

### Step 3: Multiply by 4
To eliminate the fraction, multiply both sides by 4:
\[ 4a - a^2 \geq 4 \]

### Step 4: Rearrange the Inequality
Rearranging gives us:
\[ -a^2 + 4a - 4 \geq 0 \]
or
\[ -a^2 + 4a - 4 = 0 \]

### Step 5: Factor the Quadratic
We can factor the quadratic equation:
\[ -(a^2 - 4a + 4) \geq 0 \]
This simplifies to:
\[ -(a-2)^2 \geq 0 \]

### Step 6: Analyze the Inequality
The expression \(-(a-2)^2\) is non-positive, which implies:
\[ (a-2)^2 = 0 \]
This means:
\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \]

### Conclusion
The value of \( a \) for which the equation has a real solution is:
\[ \boxed{2} \]