Question:medium

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

Show Hint

For a quadratic equation to have real roots, always use discriminant \(D\geq 0\).
  • \(\pm 1\)
  • \(\frac{1}{2}\)
  • \(1\)
  • \(\pm \frac{1}{2}\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
This question connects the algebraic expression $x + \frac{1}{x}$ with the trigonometric function $\cos\theta$. The key is to know the range of values that both of these expressions can take.
Step 2: Key Formula or Approach:
We need to use two important range properties: 1. For any non-zero real number $x$, the value of $x + \frac{1}{x}$ is always in the interval $(-\infty, -2] \cup [2, \infty)$. This means $|x + \frac{1}{x}| \ge 2$. 2. The range of the cosine function is $[-1, 1]$. This means $|\cos\theta| \le 1$.
Step 3: Detailed Explanation:
We are given the equation: \[ x + \frac{1}{x} = 2 \cos\theta \] From the property of the expression on the left side, we know that: \[ |x + \frac{1}{x}| \ge 2 \] Substituting the given equation into this inequality: \[ |2 \cos\theta| \ge 2 \] \[ 2 |\cos\theta| \ge 2 \] \[ |\cos\theta| \ge 1 \] Now, from the property of the cosine function, we know that: \[ |\cos\theta| \le 1 \] We have two conditions for $|\cos\theta|$: it must be greater than or equal to 1, and it must be less than or equal to 1. The only value that satisfies both conditions simultaneously is: \[ |\cos\theta| = 1 \] This implies that: \[ \cos\theta = 1 \quad \text{or} \quad \cos\theta = -1 \] So, $\cos\theta = \pm 1$.
Step 4: Final Answer:
The only possible values for $\cos\theta$ are $\pm 1$. Therefore, option (A) is the correct answer.
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