Question

For real x and if \(x + \frac{1}{x} = 2 \cos \theta\) then \(\cos \theta\) is

• A. \(\pm 1\)
• B. 1/2
• C. 1
• D. \(\pm 1/2\)

Hint:

This is a classic problem that tests your knowledge of the ranges of functions. Whenever you see the expression \(x + \frac{1}{x}\) for a real number \(x\), immediately recall its range: \((-\infty, -2] \cup [2, \infty)\). Comparing this with the bounded range of a trigonometric function like cosine or sine will quickly lead you to the solution at the boundary points.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given an equation that relates a real number \(x\) to a trigonometric function \(\cos \theta\). We need to find the possible values of \(\cos \theta\).

Step 2: Key Formula or Approach:
The problem requires understanding the range of the function \(f(x) = x + \frac{1}{x}\) for real \(x\), and the range of the function \(g(\theta) = 2 \cos \theta\). The equality can only hold where their ranges overlap.
The range of \(f(x) = x + \frac{1}{x}\) can be found using the AM-GM inequality or calculus.
For \(x \gt 0\), by AM-GM, \(\frac{x + 1/x}{2} \ge \sqrt{x \cdot \frac{1}{x}} \implies x + \frac{1}{x} \ge 2\).
For \(x \lt 0\), let \(x = -y\) where \(y \gt 0\). Then \(x + \frac{1}{x} = -y - \frac{1}{y} = -(y + \frac{1}{y})\). Since \(y + \frac{1}{y} \ge 2\), we have \( -(y + \frac{1}{y}) \le -2\).
So, the range of \(x + \frac{1}{x}\) is \((-\infty, -2] \cup [2, \infty)\).
The range of \(\cos \theta\) is \([-1, 1]\). Therefore, the range of \(2 \cos \theta\) is \([-2, 2]\).

Step 3: Detailed Explanation:
We have the equation:
\[ x + \frac{1}{x} = 2 \cos \theta \]
Let's analyze the possible values for both sides of the equation.
The Left-Hand Side (LHS): \(y_1 = x + \frac{1}{x}\). As established, for any real \(x \neq 0\), the value of \(y_1\) must satisfy \(y_1 \ge 2\) or \(y_1 \le -2\). So, \(|y_1| \ge 2\).
The Right-Hand Side (RHS): \(y_2 = 2 \cos \theta\). Since the range of \(\cos \theta\) is \([-1, 1]\), the range of \(y_2\) is \([-2, 2]\). So, \(|y_2| \le 2\).
For the equation to hold, we must have \(y_1 = y_2\). The only values that are in both the range of \(y_1\) and the range of \(y_2\) are the boundary points where \(|y_1| \ge 2\) and \(|y_2| \le 2\) meet.
This can only happen when the value is exactly 2 or -2.
So, we must have:
\[ 2 \cos \theta = 2 \quad \text{or} \quad 2 \cos \theta = -2 \]
Solving for \(\cos \theta\):
\[ \cos \theta = 1 \quad \text{or} \quad \cos \theta = -1 \]
This can be written concisely as \(\cos \theta = \pm 1\).

Step 4: Final Answer:
The only possible values for \(\cos \theta\) are 1 and -1.