The abscissas of point `Pa n dQ` on the curve `y=e^x+e^(-x)` such that tangents at `Pa n dQ` make `60^0` with the x-axis are. )a) `1n((sqrt(3)+sqrt(7))/7)a n d1n((sqrt(3)+sqrt(5))/2)` (b) `1n((sqrt(3)+sqrt(7))/2)` (c) `1n((sqrt(7)-sqrt(3))/2)` (d) `+-1n((sqrt(3)+sqrt(7))/2)`

Options

A. `ln((sqrt3+sqrt7)/(7))and ln((sqrt3+sqrt5)/(2))`
B. `((sqrt3+sqrt7)/(2))`
C. `ln((sqrt7-sqrt3)/(2))`
D. `+-ln((sqrt3+sqrt7)/(2))`

Answer

To find the abscissas of points \( P \) and \( Q \) on the curve \( y = e^x + e^{-x} \) such that the tangents at these points make an angle of \( 60^\circ \) with the x-axis, we will follow these steps:

### Step 1: Determine the slope of the tangent line
The slope \( m \) of the tangent line at a point on the curve can be found using the derivative of the function. The angle \( \theta \) that the tangent makes with the x-axis is given as \( 60^\circ \). Therefore, the slope of the tangent line is:
\[ m = \tan(60^\circ) = \sqrt{3} \]

### Step 2: Find the derivative of the function
To find the slope of the tangent line at any point on the curve, we need to differentiate the function \( y = e^x + e^{-x} \):
\[ \frac{dy}{dx} = e^x - e^{-x} \]

### Step 3: Set the derivative equal to the slope
We set the derivative equal to the slope we found in Step 1:
\[ e^x - e^{-x} = \sqrt{3} \]

### Step 4: Multiply through by \( e^x \)
To eliminate the negative exponent, we multiply both sides by \( e^x \):
\[ (e^x)^2 - \sqrt{3} e^x - 1 = 0 \]
Let \( t = e^x \). Then the equation becomes:
\[ t^2 - \sqrt{3} t - 1 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ t = \frac{\sqrt{3} \pm \sqrt{3 + 4}}{2} \]
\[ t = \frac{\sqrt{3} \pm \sqrt{7}}{2} \]

### Step 6: Convert back to \( x \)
Since \( t = e^x \), we have:
\[ e^x = \frac{\sqrt{3} + \sqrt{7}}{2} \quad \text{and} \quad e^x = \frac{\sqrt{3} - \sqrt{7}}{2} \]
Taking the natural logarithm of both sides gives us:
\[ x = \ln\left(\frac{\sqrt{3} + \sqrt{7}}{2}\right) \quad \text{and} \quad x = \ln\left(\frac{\sqrt{3} - \sqrt{7}}{2}\right) \]

### Step 7: Determine valid solutions
Since \( \sqrt{7} > \sqrt{3} \), the term \( \frac{\sqrt{3} - \sqrt{7}}{2} \) is negative, and we discard it because the logarithm of a negative number is undefined. Thus, the valid abscissa is:
\[ x = \ln\left(\frac{\sqrt{3} + \sqrt{7}}{2}\right) \]

### Final Answer
The abscissas of points \( P \) and \( Q \) are:
\[ x = \ln\left(\frac{\sqrt{3} + \sqrt{7}}{2}\right) \]