Let f : R → R be the function defined by f(x) = 1/(2 - cos x), ∀ x ∈…

The correct answer is Option (1) → $\left[ \frac{1}{3}, 1 \right]$ ##

Given function, $f(x) = \frac{1}{2 - \cos x}, \forall x \in R$

Let $y = \frac{1}{2 - \cos x}$

As the value of $\cos x$ lies in between $-1$ and $1$.

Therefore,

$\Rightarrow 2y - y\cos x = 1 \quad \Rightarrow y\cos x = 2y - 1$

$\Rightarrow \cos x = \frac{2y - 1}{y} = 2 - \frac{1}{y} \quad \Rightarrow \cos x = 2 - \frac{1}{y}$

$\Rightarrow -1 \le \cos x \le 1 \quad \Rightarrow -1 \le 2 - \frac{1}{y} \le 1$

$\Rightarrow -3 \le -\frac{1}{y} \le -1 \quad \Rightarrow 1 \le \frac{1}{y} \le 3$

$\Rightarrow \frac{1}{3} \le y \le 1$

So, range of $f$ is $\left[ \frac{1}{3}, 1 \right]$