Q.Let $f : \mathbb{R} \to \mathbb{R}$ be the function defined by $f(x) = \dfrac{1}{2 - \cos x}$, $\forall\, x \in \mathbb{R}$. Then, find the range of $f$.

Domain and range of a function. The range is the set of all output values. Here $f(x) = \dfrac{1}{2 - \cos x}$ depends only on $\cos x$, so we bound the denominator and then take reciprocals.

Step 1: For all $x \in \mathbb{R}$, $-1 \leq \cos x \leq 1$. Subtracting from $2$ (which reverses the inequality) gives the denominator bounds:

Step 2: The denominator is always positive, and $t \mapsto \dfrac{1}{t}$ is decreasing for $t > 0$, so reciprocating flips the inequality:

$\frac{1}{3} \leq \frac{1}{2 - \cos x} \leq 1.$ …

Let’s understand what’s really going on here. The function $f(x) = \frac{1}{2 - \cos x}$ depends only on $\cos x$, which is periodic and not one-to-one over $\mathbb{R}$. So $f$ itself cannot be one-one globally — many different $x$ values give the same $\cos x$, hence the same $f(x)$. But the question asks for the range, not injectivity. The range is simply the set of all possible output values as $x$ runs over all real numbers.

The key idea: since $\cos x$ oscillates between $-1$ and $1$, the denominator $2 - \cos x$ oscillates between $2 - 1 = 1$ and $2 - (-1) = 3$. The reciprocal then gives the range.

Let’s work through it step by step.

1. Identify the bounds of $\cos x$.

   For all $x \in \mathbb{R}$, we know $-1 \le \cos x \le 1$.

2. Transform to the denominator $d(x) = 2 - \cos x$.

   Since subtracting $\cos x$ reverses the inequality direction:

which gives

3. Take reciprocals carefully. The function $g(t) = 1/t$ is strictly decreasing for $t > 0$. Here $d(x)$ is always positive (minimum 1), so the inequality flips when we take reciprocals:

Hence

4. Check that every value in $[1/3, 1]$ is actually attained. …

Method: Range of a bounded trigonometric rational function

For $f(x)=\dfrac{1}{k-\cos x}$ type functions, bound the inner function, transform through the denominator, then reciprocate.

Steps

Step 1: Bound the inner trig function

Use $-1\le\cos x\le 1$ as the starting inequality.

Step 2: Transform to the denominator

Subtracting $\cos x$ reverses the inequality: from $-1\le\cos x\le1$,

Step 3: Reciprocate, flipping the inequality

For positive quantities, $t\mapsto \tfrac1t$ is decreasing, so the direction flips:

Step 4: Confirm every value is attained …

Common Mistakes

Mistake 1: Not flipping the inequality when reciprocating

Why it's wrong: writing $1\le f(x)\le\tfrac13$ is backwards and impossible; reciprocals of positive numbers reverse order. Correct approach: from $1\le 2-\cos x\le3$ deduce $\tfrac13\le f(x)\le1$.

Mistake 2: Forgetting the inequality reverses when subtracting $\cos x$

Why it's wrong: $2-\cos x$ is largest when $\cos x$ is smallest. Correct approach: track the direction: $\cos x=-1\Rightarrow$ denominator $=3$. …