Let `f (x) = log _e (sinx ), ( 0 lt x lt pi ) and g(x) = sin ^(-1) (e ^(-x)), (x ge 0)`. If `alpha` is a positive real number such that ` a = ( fog)' ( alpha ) and b = (fog ) ( alpha )`, then

Options

A. `a alpha^(2) + b alpha -a = -2 alpha^(2)`
B. `a alpha^(2) + b alpha + a = 0`
C. `a alpha^(2) - b alpha -a = 1`
D. `a alpha^(2) - b alpha - a = 0`

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) defined as follows:

1. \( a = (f \circ g)'(\alpha) \)
2. \( b = (f \circ g)(\alpha) \)

Where:
- \( f(x) = \log_e(\sin x) \) for \( 0 < x < \pi \)
- \( g(x) = \sin^{-1}(e^{-x}) \) for \( x \geq 0 \)

### Step 1: Find \( f(g(x)) \)

We start by substituting \( g(x) \) into \( f(x) \):

\[ f(g(x)) = f(\sin^{-1}(e^{-x})) = \log_e(\sin(\sin^{-1}(e^{-x}))) \]

Using the property that \( \sin(\sin^{-1}(y)) = y \), we have:

\[ f(g(x)) = \log_e(e^{-x}) \]

### Step 2: Simplify \( f(g(x)) \)

Now, we can simplify \( \log_e(e^{-x}) \):

\[ f(g(x)) = -x \]

### Step 3: Find \( b \)

Now we can find \( b \):

\[ b = f(g(\alpha)) = -\alpha \]

### Step 4: Differentiate \( f(g(x)) \) to find \( a \)

Next, we differentiate \( f(g(x)) \):

\[ (f \circ g)'(x) = \frac{d}{dx}(-x) = -1 \]

Thus, we find:

\[ a = (f \circ g)'(\alpha) = -1 \]

### Step 5: Summary of results

We have determined:

- \( a = -1 \)
- \( b = -\alpha \)

### Step 6: Check options

Now we need to check which of the given options are correct based on \( a \) and \( b \):

1. **Option 1:** \( -1 \cdot \alpha^2 + (-\alpha)(-\alpha) - (-1) = -2\alpha^2 \)
2. **Option 2:** \( -1 \cdot \alpha^2 + (-\alpha)(-\alpha) - 1 = -2\alpha^2 - 1 \)
3. **Option 3:** \( -1 \cdot \alpha^2 - (-\alpha)(-\alpha) - (-1) = 1 \)
4. **Option 4:** \( -1 \cdot \alpha^2 - (-\alpha)(-\alpha) = 0 \)

After evaluating each option:

- **Option 1:** Not equal.
- **Option 2:** Not equal.
- **Option 3:** Equal to 1, thus correct.
- **Option 4:** Not equal.

### Final Answer

The correct option is **Option 3**.

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