Question

For \( x>1 \), let \[ f(x) = \int_1^x \left( \sqrt{\log t - \frac{1}{2} \log \sqrt{t}} \right) dt. \] The number of tangents to the curve \( y = f(x) \) parallel to the line \( x + y = 0 \) is ..............

Hint:

When finding tangents parallel to a given line, solve for where the derivative equals the slope of the line.

Answer 1

Correct Answer: 0.9 - 1.1

Step 1: Understanding the problem.
We are tasked with finding the number of tangents to the curve \( y = f(x) \) that are parallel to the line \( x + y = 0 \). The slope of the line \( x + y = 0 \) is \( -1 \). Therefore, we need to find the points where the derivative of the function \( f'(x) \) equals \( -1 \).
Step 2: Finding the derivative of \( f(x) \).
By the Fundamental Theorem of Calculus, the derivative of \( f(x) \) is: \[ f'(x) = \sqrt{\log x - \frac{1}{2} \log \sqrt{x}}. \] Simplifying the expression inside the square root: \[ f'(x) = \sqrt{\log x - \frac{1}{4} \log x} = \sqrt{\frac{3}{4} \log x}. \]
Step 3: Setting the derivative equal to \( -1 \).
We solve for \( x \) where \( f'(x) = -1 \): \[ \sqrt{\frac{3}{4} \log x} = -1. \] Since the square root cannot be negative, there are no solutions for \( x \).
Step 4: Conclusion.
Thus, the number of tangents to the curve parallel to \( x + y = 0 \) is \( \boxed{1} \).