The curves `y = x^3 + x+ 1,2y = x^3 +5x ,"at" (1,3)` are

Options

A. touching each other
B. intersecting orthogonally
C. not intersecting
D. None of these

Answer

To determine whether the curves \( y = x^3 + x + 1 \) and \( 2y = x^3 + 5x \) touch each other at the point \( (1, 3) \), we need to follow these steps:

### Step 1: Verify the point lies on both curves
First, we need to check if the point \( (1, 3) \) lies on both curves.

For the first curve:
\[ y = x^3 + x + 1 \]
Substituting \( x = 1 \):
\[ y = 1^3 + 1 + 1 = 1 + 1 + 1 = 3 \]
So, the point \( (1, 3) \) lies on the first curve.

For the second curve:
\[ 2y = x^3 + 5x \]
Substituting \( x = 1 \):
\[ 2y = 1^3 + 5 \cdot 1 = 1 + 5 = 6 \implies y = \frac{6}{2} = 3 \]
So, the point \( (1, 3) \) also lies on the second curve.

### Step 2: Find the derivatives of both curves
Next, we need to find the derivatives of both curves to determine their slopes at the point \( (1, 3) \).

For the first curve:
\[ y = x^3 + x + 1 \]
The derivative is:
\[ \frac{dy}{dx} = 3x^2 + 1 \]
Now, substituting \( x = 1 \):
\[ \frac{dy}{dx} = 3(1)^2 + 1 = 3 + 1 = 4 \]
Thus, the slope \( m_1 = 4 \).

For the second curve:
\[ 2y = x^3 + 5x \implies y = \frac{x^3 + 5x}{2} \]
The derivative is:
\[ \frac{dy}{dx} = \frac{1}{2}(3x^2 + 5) \]
Now, substituting \( x = 1 \):
\[ \frac{dy}{dx} = \frac{1}{2}(3(1)^2 + 5) = \frac{1}{2}(3 + 5) = \frac{8}{2} = 4 \]
Thus, the slope \( m_2 = 4 \).

### Step 3: Compare the slopes
Since both slopes are equal:
\[ m_1 = m_2 = 4 \]
This indicates that the curves touch each other at the point \( (1, 3) \).

### Conclusion
The curves \( y = x^3 + x + 1 \) and \( 2y = x^3 + 5x \) touch each other at the point \( (1, 3) \).

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