Q.Range of $f(x) = \dfrac{1}{1 - 2\cos x}$ is
(A) $\left[\dfrac{1}{3},\ 1\right]$
(B) $\left[-1,\ \dfrac{1}{3}\right]$
(C) $(-\infty,\ -1] \cup \left[\dfrac{1}{3},\ \infty\right)$
(D) $\left[-\dfrac{1}{3},\ 1\right]$

Concept: Rational Function Domain — the denominator must not be zero, and the range of $\cos x$ restricts the possible outputs.

Step 1: Denominator $1 - 2\cos x \neq 0 \Rightarrow \cos x \neq \frac12 \Rightarrow x \neq \pm\frac{\pi}{3} + 2n\pi$.

Step 2: $\cos x \in [-1, 1]$, so $1 - 2\cos x \in [1 - 2(1),\; 1 - 2(-1)] = [-1, 3]$, but excluding $0$ at $\cos x = \frac12$. …

Concept and Intuition

When finding the range of a rational function involving a trigonometric expression, the key is to work from the inside out. Here, the function is $f(x) = \frac{1}{g(x)}$ where $g(x) = 1 - 2\cos x$.

The critical insight: the range of $1/g(x)$ is not simply the reciprocal of the range of $g(x)$ — because if $g(x)$ can take both positive and negative values, the reciprocal function will have a break (a vertical asymptote) where $g(x) = 0$. The reciprocal of a set that crosses zero gives two disjoint intervals stretching to infinity.

So the plan is:

1. Find the range of $g(x) = 1 - 2\cos x$.
2. Identify where $g(x) = 0$ (the asymptote).
3. Apply the reciprocal transformation separately on the positive and negative parts of the range of $g(x)$.

Step-by-Step Solution

1. Find the range of $\cos x$.

We know that for all real $x$:

2. Find the range of $2\cos x$.

Multiplying by 2 (a positive constant) preserves the inequality direction:

3. Find the range of $1 - 2\cos x$.

Subtract $2\cos x$ from 1. To get the new bounds, we substitute the extreme values:

• When $\cos x = -1$: $1 - 2(-1) = 1 + 2 = 3$
• When $\cos x = 1$: $1 - 2(1) = 1 - 2 = -1$

Since $1 - 2\cos x$ is a continuous function of $\cos x$, and $\cos x$ varies continuously, the range is:

So $g(x) \in [-1, 3]$.

4. Check if $g(x)$ can be zero.

This is crucial. Set $1 - 2\cos x = 0$:

Since $\frac{1}{2}$ lies within $[-1, 1]$, this equation has solutions (e.g., $x = \frac{\pi}{3}$). Therefore, $g(x) = 0$ is attained, and $f(x)$ has a vertical asymptote there — the function is undefined at those points, and near them $f(x)$ tends to $\pm\infty$.

5. Split the range of $g(x)$ at zero.

The denominator $g(x)$ takes values in $[-1, 3]$, and it crosses zero. So we split:

• Negative part: $g(x) \in [-1, 0)$
• Positive part: $g(x) \in (0, 3]$ …