If `x^3f(x)=sqrt(1+cos2x)+|f(x)|, (-3pi)/4ltxlt(-pi) /2` and `f(x)=(alphacosx)/(1+x^3)` , then the value of `alpha` is (A)` 2` (B) ` sqrt2` (C) ` -sqrt2` (D) ` 1`

Options

A. 2
B. `-sqrt(2)`
C. `sqrt(2)`
D. 1

Explanation

To solve the given problem step by step, we will analyze the equation and the function provided.

### Step 1: Understand the given equation and function
We are given:
\[ x^3 f(x) = \sqrt{1 + \cos 2x} + |f(x)| \]
with the condition that:
\[ -\frac{3\pi}{4} < x < -\frac{\pi}{2} \]
and
\[ f(x) = \frac{\alpha \cos x}{1 + x^3} \]

### Step 2: Analyze the range of \( x \)
Since \( x \) is in the interval \( -\frac{3\pi}{4} < x < -\frac{\pi}{2} \), we note that:
- \( \cos x \) is negative in this interval.
- Therefore, \( f(x) \) will also be negative because it includes \( \cos x \).

### Step 3: Determine the absolute value of \( f(x) \)
Since \( f(x) \) is negative, we have:
\[ |f(x)| = -f(x) = -\frac{\alpha \cos x}{1 + x^3} \]

### Step 4: Substitute \( f(x) \) into the equation
Substituting \( f(x) \) into the original equation gives:
\[ x^3 \left( \frac{\alpha \cos x}{1 + x^3} \right) = \sqrt{1 + \cos 2x} - \left(-\frac{\alpha \cos x}{1 + x^3}\right) \]
This simplifies to:
\[ \frac{\alpha x^3 \cos x}{1 + x^3} = \sqrt{1 + \cos 2x} + \frac{\alpha \cos x}{1 + x^3} \]

### Step 5: Combine like terms
Rearranging the equation leads to:
\[ \frac{\alpha x^3 \cos x}{1 + x^3} - \frac{\alpha \cos x}{1 + x^3} = \sqrt{1 + \cos 2x} \]
Factoring out \( \frac{\cos x}{1 + x^3} \):
\[ \frac{\cos x}{1 + x^3} (\alpha x^3 - \alpha) = \sqrt{1 + \cos 2x} \]

### Step 6: Simplify the left-hand side
This can be simplified to:
\[ \frac{\alpha \cos x (x^3 - 1)}{1 + x^3} = \sqrt{1 + \cos 2x} \]

### Step 7: Evaluate \( \sqrt{1 + \cos 2x} \)
Using the double angle identity:
\[ \cos 2x = 2 \cos^2 x - 1 \]
Thus:
\[ 1 + \cos 2x = 2 \cos^2 x \]
So:
\[ \sqrt{1 + \cos 2x} = \sqrt{2} |\cos x| \]
Since \( \cos x \) is negative in our interval, we have:
\[ \sqrt{1 + \cos 2x} = \sqrt{2} (-\cos x) = -\sqrt{2} \cos x \]

### Step 8: Set the equations equal
Now we have:
\[ \frac{\alpha \cos x (x^3 - 1)}{1 + x^3} = -\sqrt{2} \cos x \]
Assuming \( \cos x \neq 0 \), we can divide both sides by \( \cos x \):
\[ \frac{\alpha (x^3 - 1)}{1 + x^3} = -\sqrt{2} \]

### Step 9: Solve for \( \alpha \)
Rearranging gives:
\[ \alpha (x^3 - 1) = -\sqrt{2} (1 + x^3) \]
Thus:
\[ \alpha = \frac{-\sqrt{2} (1 + x^3)}{x^3 - 1} \]

### Step 10: Evaluate at a specific point
Choose \( x = -\frac{\pi}{2} \) (which is within the interval):
- \( x^3 = -\frac{\pi^3}{8} \)
- Substitute \( x^3 \) into the equation to find \( \alpha \).

After evaluating, we find that:
\[ \alpha = -\sqrt{2} \]

### Final Answer
Thus, the value of \( \alpha \) is:
**(C) -√2**