The angle at which the curve y = `(a+1)e^(ax)` intersects y-axis is

Options

A. `tan^(-1)(a^(2)+a)`
B. `sin^(-1)((1)/(sqrt(1+a^(4))))`
C. `cot^(-1)(a^(2)+a)`
D. `sec^(-1)sqrt(1+a^(4))`

Answer

To find the angle at which the curve \( y = (a+1)e^{ax} \) intersects the y-axis, we will follow these steps:

### Step 1: Differentiate the function
The first step is to find the derivative of the function to determine the slope of the curve at the point where it intersects the y-axis.

Given the function:
\[ y = (a + 1)e^{ax} \]

We differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = (a + 1) \cdot \frac{d}{dx}(e^{ax}) = (a + 1) \cdot ae^{ax} \]

### Step 2: Evaluate the derivative at the y-axis
The curve intersects the y-axis at \( x = 0 \). We will substitute \( x = 0 \) into the derivative to find the slope at this point.

\[ \frac{dy}{dx} \bigg|_{x=0} = (a + 1) \cdot a \cdot e^{a \cdot 0} = (a + 1) \cdot a \cdot 1 = a(a + 1) \]

### Step 3: Relate the slope to the angle
The slope \( m \) of the curve at the point of intersection with the y-axis is given by:
\[ m = a(a + 1) \]

The angle \( \theta \) that the curve makes with the positive direction of the x-axis can be related to the slope by the formula:
\[ m = \tan(\theta) \]

Thus, we have:
\[ \tan(\theta) = a(a + 1) \]

### Step 4: Find the angle
To find \( \theta \), we take the arctangent of both sides:
\[ \theta = \tan^{-1}(a(a + 1)) \]

### Step 5: Find the angle with respect to the y-axis
Since the angle \( \alpha \) between the slope and the x-axis is related to the angle \( \theta \) we found, we can use the relationship:
\[ \alpha + \theta = \frac{\pi}{2} \]
This means:
\[ \alpha = \frac{\pi}{2} - \theta \]
Thus, we can express \( \theta \) in terms of \( \alpha \):
\[ \theta = \frac{\pi}{2} - \tan^{-1}(a(a + 1)) \]

### Final Result
The angle at which the curve intersects the y-axis is:
\[ \theta = \cot^{-1}(a(a + 1)) \]