To the equation `2^2pi//cos^((-1)x)-(a+1/2)2^pi//cos^((-1)x)-a^2=0` has only one real root, then `1lt=alt=3` (b) `ageq1` `alt=-3` (d) `ageq3`

Options

A. `1 le a le 3`
B. `a ge 1`
C. `a le -3`
D. `a ge 3`

Explanation

To solve the equation \(2^{\frac{2\pi}{\cos^{-1} x}} - \left(a + \frac{1}{2}\right)2^{\frac{\pi}{\cos^{-1} x}} - a^2 = 0\) such that it has only one real root, we can follow these steps:

### Step 1: Substitute and Rearrange
Let \( t = 2^{\frac{\pi}{\cos^{-1} x}} \). Then the equation can be rewritten as:
\[ t^2 - \left(a + \frac{1}{2}\right)t - a^2 = 0 \]

### Step 2: Apply the Quadratic Formula
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we have:
- \( a = 1 \)
- \( b = -\left(a + \frac{1}{2}\right) \)
- \( c = -a^2 \)

Substituting these values into the formula gives:
\[ t = \frac{a + \frac{1}{2} \pm \sqrt{\left(a + \frac{1}{2}\right)^2 - 4 \cdot 1 \cdot (-a^2)}}{2 \cdot 1} \]

### Step 3: Simplify the Discriminant
The discriminant \( D \) must be zero for the equation to have only one real root:
\[ D = \left(a + \frac{1}{2}\right)^2 + 4a^2 \]
Setting \( D = 0 \):
\[ \left(a + \frac{1}{2}\right)^2 + 4a^2 = 0 \]

### Step 4: Solve the Discriminant Equation
Expanding gives:
\[ a^2 + a + \frac{1}{4} + 4a^2 = 0 \implies 5a^2 + a + \frac{1}{4} = 0 \]
Using the quadratic formula again:
\[ a = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 5 \cdot \frac{1}{4}}}{2 \cdot 5} = \frac{-1 \pm \sqrt{1 - 5}}{10} = \frac{-1 \pm \sqrt{-4}}{10} \]
This indicates that \( a \) must be real, hence we need to analyze the conditions for the roots.

### Step 5: Analyze the Range of \( t \)
The range of \( t = 2^{\frac{\pi}{\cos^{-1} x}} \) is \( [2, \infty) \) since \( \cos^{-1} x \) ranges from \( 0 \) to \( \pi \).

### Step 6: Set Conditions for \( a \)
For the quadratic to have only one real root, we need:
\[ \left(a + \frac{1}{2}\right)^2 - 4(-a^2) = 0 \implies a + \frac{1}{2} = \pm \sqrt{4a^2} \]
This leads to two cases:
1. \( a + \frac{1}{2} = 2a \)
2. \( a + \frac{1}{2} = -2a \)

### Step 7: Solve Each Case
1. From \( a + \frac{1}{2} = 2a \):
\[ \frac{1}{2} = a \implies a = \frac{1}{2} \]
2. From \( a + \frac{1}{2} = -2a \):
\[ 3a = -\frac{1}{2} \implies a = -\frac{1}{6} \]

### Step 8: Check Conditions
We need to check which of the options satisfy \( 1 \leq a \leq 3 \) or \( a \geq 1 \) or \( a \leq -3 \) or \( a \geq 3 \).

### Conclusion
The valid conditions for \( a \) based on the analysis are:
- \( a \geq 1 \)
- \( a \leq -3 \)

Thus, the correct options are:
- (b) \( a \geq 1 \)
- (c) \( a \leq -3 \)