Question

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

• A. \(\pm 1\)
• B. \(\frac{1}{2}\)
• C. \(1\)
• D. \(\pm\frac{1}{2}\)

Hint:

For real \(x\), \(x+\frac{1}{x}\) cannot lie between \(-2\) and \(2\), except at the end values. Since \(2\cos\theta\) lies between \(-2\) and \(2\), only \(\pm2\) are possible.

Answer 1

The Correct Option is A

We are given \[ x+\frac{1}{x}=2\cos\theta. \] Here \(x\) is real and \(x\neq 0\). For real \(x\), we know that: \[ x+\frac{1}{x}\geq 2 \quad \text{if } x>0. \] Also, \[ x+\frac{1}{x}\leq -2 \quad \text{if } x<0. \] Therefore, for real \(x\), \[ x+\frac{1}{x}\in (-\infty,-2]\cup [2,\infty). \] But the right-hand side is \[ 2\cos\theta. \] Since \[ -1\leq \cos\theta \leq 1, \] we have \[ -2\leq 2\cos\theta \leq 2. \] Now both conditions must be satisfied: \[ x+\frac{1}{x}=2\cos\theta. \] The only possible values common to both ranges are \[ 2 \quad \text{and} \quad -2. \] Therefore, \[ 2\cos\theta=2 \] or \[ 2\cos\theta=-2. \] So, \[ \cos\theta=1 \] or \[ \cos\theta=-1. \] Hence, \[ \cos\theta=\pm 1. \]