If m be the slope of the tangent to the curve `e^(2y) = 1+4x^(2)`, then

Options

A. `m lt 1`
B. `|m| le 1`
C. `|m| ge 1`
D. none of these

Answer

To find the slope of the tangent to the curve given by the equation \( e^{2y} = 1 + 4x^2 \), we will differentiate the equation with respect to \( x \) and simplify the result.

### Step-by-Step Solution:

1. **Differentiate the Equation**:
Start with the equation:
\[ e^{2y} = 1 + 4x^2 \]
Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(e^{2y}) = \frac{d}{dx}(1 + 4x^2) \]

2. **Apply the Chain Rule**:
For the left side, use the chain rule:
\[ e^{2y} \cdot \frac{d}{dx}(2y) = e^{2y} \cdot 2 \frac{dy}{dx} \]
For the right side:
\[ \frac{d}{dx}(1) + \frac{d}{dx}(4x^2) = 0 + 8x \]

So, we have:
\[ e^{2y} \cdot 2 \frac{dy}{dx} = 8x \]

3. **Solve for \(\frac{dy}{dx}\)**:
Rearranging the equation gives:
\[ 2 e^{2y} \frac{dy}{dx} = 8x \]
Dividing both sides by \( 2 e^{2y} \):
\[ \frac{dy}{dx} = \frac{8x}{2 e^{2y}} = \frac{4x}{e^{2y}} \]

4. **Substitute \( e^{2y} \)**:
From the original equation, we know \( e^{2y} = 1 + 4x^2 \). Substitute this into the derivative:
\[ \frac{dy}{dx} = \frac{4x}{1 + 4x^2} \]

5. **Identify the Slope \( m \)**:
Thus, the slope of the tangent \( m \) is:
\[ m = \frac{4x}{1 + 4x^2} \]

6. **Express \( m \) in a Different Form**:
We can rewrite \( m \) as:
\[ m = \frac{2(2x)}{1 + (2x)^2} \]

7. **Find the Modulus of \( m \)**:
To find the modulus:
\[ |m| = \left| \frac{2(2x)}{1 + (2x)^2} \right| = \frac{2|2x|}{1 + (2x)^2} \]

8. **Apply the AM-GM Inequality**:
Since \( 1 + (2x)^2 \) is always positive, we can apply the AM-GM inequality:
\[ 1 + (2x)^2 \geq 2|2x| \]
This implies:
\[ |m| \leq 1 \]

### Conclusion:
Thus, we conclude that:
\[ |m| \leq 1 \]