`x+y-ln(x+y)=2x+5` has a vertical tangent at the point `(alpha,beta)` then `alpha+beta` is equal to

Options

A. `-1`
B. 1
C. 2
D. `-2`

Answer

To solve the problem, we need to find the point \((\alpha, \beta)\) where the equation \(x + y - \ln(x + y) = 2x + 5\) has a vertical tangent. A vertical tangent implies that the derivative \(\frac{dy}{dx}\) is infinite, which occurs when the denominator of the derivative is zero.

### Step-by-step Solution:

1. **Differentiate the given equation**:
We start with the equation:
\[ x + y - \ln(x + y) = 2x + 5 \]
Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x + y) - \frac{d}{dx}(\ln(x + y)) = \frac{d}{dx}(2x + 5) \]
Using the chain rule for the logarithm, we have:
\[ 1 + \frac{dy}{dx} - \frac{1}{x + y}\left(1 + \frac{dy}{dx}\right) = 2 \]

2. **Rearranging the equation**:
Rearranging gives:
\[ 1 + \frac{dy}{dx} - \frac{1 + \frac{dy}{dx}}{x + y} = 2 \]
Simplifying this, we get:
\[ \frac{dy}{dx} - \frac{1 + \frac{dy}{dx}}{x + y} = 1 \]

3. **Finding \(\frac{dy}{dx}\)**:
Multiply through by \(x + y\) to eliminate the fraction:
\[ (x + y)\frac{dy}{dx} - (1 + \frac{dy}{dx}) = (x + y) \]
This simplifies to:
\[ (x + y)\frac{dy}{dx} - 1 - \frac{dy}{dx} = x + y \]
Rearranging gives:
\[ (x + y - 1)\frac{dy}{dx} = x + y + 1 \]
Thus,
\[ \frac{dy}{dx} = \frac{x + y + 1}{x + y - 1} \]

4. **Setting the derivative to be infinite**:
For the tangent to be vertical, we set the denominator to zero:
\[ x + y - 1 = 0 \]
This implies:
\[ x + y = 1 \]

5. **Substituting back into the original equation**:
Substitute \(y = 1 - x\) into the original equation:
\[ x + (1 - x) - \ln(1) = 2x + 5 \]
This simplifies to:
\[ 1 = 2x + 5 \]
Rearranging gives:
\[ 2x = 1 - 5 \implies 2x = -4 \implies x = -2 \]
Then substituting back to find \(y\):
\[ y = 1 - (-2) = 3 \]

6. **Finding \(\alpha + \beta\)**:
We have found \((\alpha, \beta) = (-2, 3)\). Therefore:
\[ \alpha + \beta = -2 + 3 = 1 \]

### Final Answer:
\[ \alpha + \beta = 1 \]