Cosine of the angle of intersection of curves `y = 3^(x-1) logx and y= x^(x)-1` at (1,0) is

Options

A. 1
B. `1/2`
C. 0
D. `1/3`

Answer

To find the cosine of the angle of intersection of the curves \( y = 3^{(x-1)} \log x \) and \( y = x^x - 1 \) at the point \( (1, 0) \), we will follow these steps:

### Step 1: Find the slope of the tangent to the first curve
The first curve is given by:
\[ y = 3^{(x-1)} \log x \]
We need to find the derivative \( \frac{dy}{dx} \) at the point \( (1, 0) \). We will use the product rule for differentiation.

Let:
- \( u = 3^{(x-1)} \)
- \( v = \log x \)

Using the product rule \( \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \):

1. Calculate \( \frac{du}{dx} \):
\[ \frac{du}{dx} = 3^{(x-1)} \log 3 \]

2. Calculate \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = \frac{1}{x} \]

Now, applying the product rule:
\[ \frac{dy}{dx} = 3^{(x-1)} \cdot \frac{1}{x} + \log x \cdot 3^{(x-1)} \log 3 \]

Now, substituting \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 3^{(1-1)} \cdot 1 + 0 \cdot 3^{(1-1)} \log 3 = 1 + 0 = 1 \]

Thus, the slope \( m_1 \) of the tangent to the first curve at \( (1, 0) \) is:
\[ m_1 = 1 \]

### Step 2: Find the slope of the tangent to the second curve
The second curve is given by:
\[ y = x^x - 1 \]
We need to find \( \frac{dy}{dx} \) at the point \( (1, 0) \).

Using logarithmic differentiation:
\[ \log y = x \log x \]
Differentiating both sides:
\[ \frac{1}{y} \frac{dy}{dx} = \log x + 1 \]
Thus,
\[ \frac{dy}{dx} = y \cdot (\log x + 1) \]
Substituting \( y = x^x \) back:
\[ \frac{dy}{dx} = x^x (\log x + 1) \]

Now, substituting \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 1^1 \cdot (0 + 1) = 1 \]

Thus, the slope \( m_2 \) of the tangent to the second curve at \( (1, 0) \) is:
\[ m_2 = 1 \]

### Step 3: Find the cosine of the angle between the tangents
The angle \( \theta \) between two curves can be found using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \( m_1 = 1 \) and \( m_2 = 1 \):
\[ \tan \theta = \left| \frac{1 - 1}{1 + 1 \cdot 1} \right| = \left| \frac{0}{2} \right| = 0 \]

Since \( \tan \theta = 0 \), it implies that \( \theta = 0 \).

Thus, the cosine of the angle is:
\[ \cos \theta = \cos 0 = 1 \]

### Final Answer
The cosine of the angle of intersection of the curves at the point \( (1, 0) \) is:
\[ \boxed{1} \]