If `2^(2pi//sin^((-1)x))-2(a+2)^(pi//sin^((-1)x))+8a<0` for at least one real `x ,` then

Options

A. `(1)/(8) le a le 2`
B. `a le 2`
C. `a in R - {2}`
D. `a in [0, (1)/(8)) uu (2, oo)`

Explanation

To solve the inequality \( 2^{\frac{2\pi}{\sin^{-1} x}} - 2(a + 2)^{\frac{\pi}{\sin^{-1} x}} + 8a < 0 \) for at least one real \( x \), we can follow these steps:

### Step 1: Substitute \( t = 2^{\frac{2\pi}{\sin^{-1} x}} \)

Let \( t = 2^{\frac{2\pi}{\sin^{-1} x}} \). Since \( \sin^{-1} x \) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), the value of \( t \) will range as follows:
- When \( \sin^{-1} x \) approaches \( 0 \), \( t \) approaches \( 2^{\infty} \) (which is infinity).
- When \( \sin^{-1} x \) approaches \( \frac{\pi}{2} \), \( t \) approaches \( 2^{\frac{2\pi}{\frac{\pi}{2}}} = 2^4 = 16 \).

Thus, \( t \) can take values in the intervals \( (0, 16) \) and \( (16, \infty) \).

### Step 2: Rewrite the inequality in terms of \( t \)

Substituting \( t \) into the inequality gives:
\[ t - 2(a + 2)^{\frac{\pi}{\sin^{-1} x}} + 8a < 0 \]

### Step 3: Express \( (a + 2)^{\frac{\pi}{\sin^{-1} x}} \) in terms of \( t \)

We can express \( (a + 2)^{\frac{\pi}{\sin^{-1} x}} \) as:
\[ (a + 2)^{\frac{\pi}{\sin^{-1} x}} = (a + 2)^{\frac{\pi}{\frac{2\pi}{\log_2 t}}} = (a + 2)^{\frac{\log_2 t}{2}} \]
This implies:
\[ (a + 2)^{\frac{\pi}{sin^{-1} x}} = \sqrt{(a + 2)^{\log_2 t}} = (a + 2)^{\log_2 t} = t^{\log_2(a + 2)} \]

### Step 4: Substitute back into the inequality

Now substituting back, we have:
\[ t - 2t^{\log_2(a + 2)} + 8a < 0 \]

### Step 5: Analyze the quadratic inequality

This can be rearranged into a standard quadratic form:
\[ -2t^{\log_2(a + 2)} + t + 8a < 0 \]

### Step 6: Find the roots of the quadratic

The roots of the quadratic \( -2t^{\log_2(a + 2)} + t + 8a = 0 \) can be found using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = -2, b = 1, c = 8a \).

### Step 7: Determine the conditions for the inequality to hold

For the quadratic to be less than zero for at least one real \( x \), the discriminant must be non-negative:
\[ 1^2 - 4(-2)(8a) \geq 0 \]
This simplifies to:
\[ 1 + 64a \geq 0 \implies a \geq -\frac{1}{64} \]

### Step 8: Check the intervals for \( a \)

Since we also need to ensure that \( t \) can take values in the ranges derived earlier, we analyze the values of \( a \) further.

### Conclusion

The final condition we derived is:
\[ a \geq -\frac{1}{64} \]
This means that for at least one real \( x \), the inequality holds true if \( a \) lies within the derived intervals.