Q.If for real values of $x$, $\cos\theta = x + \dfrac{1}{x}$, then
(A) $\theta$ is an acute angle
(B) $\theta$ is right angle
(C) $\theta$ is an obtuse angle
(D) No value of $\theta$ is possible

Concept: Range of $\cos\theta$ and AM–GM inequality.

For real $x$, the expression $x + \frac{1}{x}$ has a minimum magnitude of 2 when $x>0$ and a maximum magnitude of $-2$ when $x<0$.

Specifically, by AM–GM, for $x>0$: $x + \frac{1}{x} \ge 2$; for $x<0$: $x + \frac{1}{x} \le -2$.

Thus $\cos\theta = x + \frac{1}{x}$ implies $\cos\theta \ge 2$ or $\cos\theta \le -2$. …

Concept and Intuition

This problem is a classic trap: it looks like a trigonometric equation, but the real constraint comes from the algebraic expression $x + \frac{1}{x}$. For real $x$, this sum has a well-known range — it is either $\ge 2$ or $\le -2$, never between $-2$ and $2$. Meanwhile, $\cos\theta$ is always between $-1$ and $1$. So the equation $\cos\theta = x + \frac{1}{x}$ asks us to equate two quantities whose ranges do not overlap at all. That immediately tells us no real $\theta$ can satisfy it.

Let’s verify this step by step.

Step-by-Step Solution

1. Recall the range of $\cos\theta$ For any real angle $\theta$, the cosine function satisfies:

This is a fundamental property — no real $\theta$ can make $\cos\theta$ go outside this interval.

2. Analyze the expression $x + \frac{1}{x}$ for real $x$ Consider two cases for real $x$:
   • If $x > 0$, then by AM–GM inequality:

 Equality occurs when $x = 1$.

• If $x < 0$, let $x = -t$ where $t > 0$. Then:

 Equality occurs when $t = 1$, i.e., $x = -1$.

• If $x = 0$, the expression is undefined (division by zero), so $x = 0$ is not allowed.

Therefore, for all real $x \neq 0$:

3. Compare the two ranges …