For the curve `sinx+siny=1` lying in first quadrant. If `lim_(xrarr0) x^(alpha)(d^(2)y)/(dx^(2))` exists and non-zero than `2alpha=`

Options

A. 3
B. 4
C. 5
D. 1

Explanation

To solve the problem, we need to analyze the given equation of the curve and find the second derivative of \( y \) with respect to \( x \). The steps are outlined below:

### Step 1: Start with the given equation
The equation of the curve is:
\[ \sin x + \sin y = 1 \]

### Step 2: Differentiate the equation with respect to \( x \)
Differentiating both sides with respect to \( x \):
\[ \cos x + \cos y \frac{dy}{dx} = 0 \]
From this, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{\cos x}{\cos y} \]

### Step 3: Differentiate again to find \( \frac{d^2y}{dx^2} \)
Now we differentiate \( \frac{dy}{dx} \) again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{\cos x}{\cos y}\right) \]
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = -\frac{\sin x \cos y + \cos x \sin y \frac{dy}{dx}}{\cos^2 y} \]
Substituting \( \frac{dy}{dx} = -\frac{\cos x}{\cos y} \):
\[ \frac{d^2y}{dx^2} = -\frac{\sin x \cos y - \cos x \sin y \frac{\cos x}{\cos y}}{\cos^2 y} \]
Simplifying:
\[ \frac{d^2y}{dx^2} = -\frac{\sin x \cos y - \frac{\cos^2 x \sin y}{\cos y}}{\cos^2 y} \]
\[ = -\frac{\sin x \cos y - \cos^2 x (1 - \sin x)}{\cos^2 y} \]
\[ = -\frac{\sin x \cos y - \cos^2 x + \cos^2 x \sin x}{\cos^2 y} \]
\[ = -\frac{\sin x \cos y + \cos^2 x \sin x - \cos^2 x}{\cos^2 y} \]

### Step 4: Analyze the limit
We need to find:
\[ \lim_{x \to 0} x^{\alpha} \frac{d^2y}{dx^2} \]
To ensure this limit exists and is non-zero, we need to analyze the behavior of \( \frac{d^2y}{dx^2} \) as \( x \to 0 \).

### Step 5: Use Taylor series expansion
Using the Taylor expansion around \( x = 0 \):
\[ \sin x \approx x \quad \text{and} \quad \cos x \approx 1 \]
Thus:
\[ \sin y \approx 1 - x \quad \text{and} \quad \cos y \approx 1 \]
Substituting these approximations into \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} \approx -\frac{x(1) + 1 \cdot (1 - x)(-1)}{1} = -\frac{x + 1 - x}{1} = -1 \]

### Step 6: Find the limit
Now substituting back into the limit:
\[ \lim_{x \to 0} x^{\alpha} \cdot (-1) = -x^{\alpha} \]
For this limit to exist and be non-zero, we require:
\[ \alpha = 0 \]

### Step 7: Calculate \( 2\alpha \)
Thus:
\[ 2\alpha = 2 \cdot 0 = 0 \]

### Final Answer
The value of \( 2\alpha \) is:
\[ \boxed{0} \]