The abscissa of the point on the curve `y=a[e^(x//a)+e^(-x//a)]` when the tangent is parallel to the x-axis is

Options

A. 0
B. 1
C. a
D. 2a

Answer

To find the abscissa of the point on the curve \( y = a \left( e^{\frac{x}{a}} + e^{-\frac{x}{a}} \right) \) when the tangent is parallel to the x-axis, we need to follow these steps:

### Step-by-Step Solution:

1. **Understand the condition for the tangent to be parallel to the x-axis**:
The tangent to the curve is parallel to the x-axis when the derivative \( \frac{dy}{dx} = 0 \).

2. **Differentiate the function**:
We start by differentiating \( y \) with respect to \( x \):
\[ y = a \left( e^{\frac{x}{a}} + e^{-\frac{x}{a}} \right) \]
Using the chain rule, we differentiate:
\[ \frac{dy}{dx} = a \left( \frac{1}{a} e^{\frac{x}{a}} - \frac{1}{a} e^{-\frac{x}{a}} \right) \]
Simplifying this gives:
\[ \frac{dy}{dx} = e^{\frac{x}{a}} - e^{-\frac{x}{a}} \]

3. **Set the derivative to zero**:
To find where the tangent is parallel to the x-axis, we set the derivative equal to zero:
\[ e^{\frac{x}{a}} - e^{-\frac{x}{a}} = 0 \]

4. **Solve for \( x \)**:
Rearranging the equation gives:
\[ e^{\frac{x}{a}} = e^{-\frac{x}{a}} \]
This implies:
\[ e^{\frac{x}{a}} \cdot e^{\frac{x}{a}} = 1 \quad \text{or} \quad e^{\frac{2x}{a}} = 1 \]
Taking the natural logarithm on both sides:
\[ \frac{2x}{a} = 0 \]
Therefore,
\[ x = 0 \]

5. **Conclusion**:
The abscissa of the point on the curve when the tangent is parallel to the x-axis is \( x = 0 \).

### Final Answer:
The abscissa is \( 0 \).