Equation Analysis for Real Solutions

We are given the equation \(x \cos \theta = x^2 + p\). The goal is to find the condition on the parameter \(p\) such that this equation has at least one real solution for \(x\) for all values of \(\theta\) within the interval \(0 \le \theta \le \frac{\pi}{4}\).

Rearranging the Equation

First, we rewrite the equation to express \(p\) in terms of \(x\) and \(\theta\):

\( p = x \cos \theta - x^2 \)

This form implies that for a specific value of \(\theta\), a real solution \(x\) exists if \(p\) is less than or equal to the maximum value that the expression \(x \cos \theta - x^2\) can attain.

Finding the Maximum Value of \(p\) for a Fixed \(\theta\)

Consider the expression \(f(x, \theta) = x \cos \theta - x^2\). To find the maximum value achievable by \(p\) for a fixed \(\theta\), we find the maximum of \(f(x, \theta)\) with respect to \(x\). The function \(f(x, \theta)\) is a quadratic in \(x\) shaped like an inverted parabola.

The maximum value occurs at the vertex. The \(x\)-coordinate of the vertex is:

\( x_{vertex} = \frac{-(\cos \theta)}{2(-1)} = \frac{\cos \theta}{2} \)

Substituting this \(x\) value back into the expression gives the maximum value:

\( \text{Max value for fixed } \theta = f\left(\frac{\cos \theta}{2}, \theta\right) = \left(\frac{\cos \theta}{2}\right) \cos \theta - \left(\frac{\cos \theta}{2}\right)^2 \)

\( = \frac{\cos^2 \theta}{2} - \frac{\cos^2 \theta}{4} \)

\( = \frac{\cos^2 \theta}{4} \)

Therefore, for a real solution \(x\) to exist for a specific \(\theta\), \(p\) must satisfy \(p \le \frac{\cos^2 \theta}{4}\).

Determining the Overall Condition on \(p\)

The problem requires that a real solution exists for every \(\theta\) in the interval \(0 \le \theta \le \frac{\pi}{4}\). This implies that \(p\) must be compatible with the range of values that \(\frac{\cos^2 \theta}{4}\) can take within this interval. Specifically, for the condition \(p \le \frac{\cos^2 \theta}{4}\) to be potentially met across the interval, \(p\) must not exceed the maximum value that the expression \(\frac{\cos^2 \theta}{4}\) can attain.

So, we need to find the maximum value of \(\frac{\cos^2 \theta}{4}\) over the interval \(0 \le \theta \le \frac{\pi}{4}\):

\( p \le \max_{0 \le \theta \le \frac{\pi}{4}} \left( \frac{\cos^2 \theta}{4} \right) \)

Analyzing the Maximum Value in the Interval

Let's examine the behavior of \(\cos^2 \theta\) in the interval \(0 \le \theta \le \frac{\pi}{4}\):

• The function \(\cos \theta\) is decreasing throughout this interval. It ranges from \(\cos(0) = 1\) down to \(\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\).
• Consequently, \(\cos^2 \theta\) is also a decreasing function in this interval.
• The maximum value of \(\cos^2 \theta\) will occur at the smallest value of \(\theta\), which is \(\theta = 0\).

Calculating this maximum value:

\( \max_{0 \le \theta \le \frac{\pi}{4}} (\cos^2 \theta) = (\cos 0)^2 = 1^2 = 1 \)

Final Condition for \(p\)

Now we use this maximum value to determine the condition for \(p\):

\( p \le \frac{1}{4} \times (\text{maximum value of } \cos^2 \theta) \)

\( p \le \frac{1}{4} \times 1 \)

\( p \le \frac{1}{4} \)

Therefore, the condition required for the equation \(x \cos \theta = x^2 + p\) to have a real solution \(x\) for \(\theta\) in the range \(0 \le \theta \le \frac{\pi}{4}\) is \(p \le \frac{1}{4}\).