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Let $f: R \rightarrow R$ be the function defined by $f(x)$ $=\frac{1}{2-\cos x}, \forall x \in R$. Then, find the range of $f$.
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Verified Answer
Here, $f(x)=\frac{1}{2-\cos x}, \forall x \in R$
Let $y=\frac{1}{2-\cos x}$
$$
\begin{aligned}
&\Rightarrow \cos x=2-\frac{1}{y} \\
&\Rightarrow-1 \leq \cos x \leq 1 \Rightarrow-1 \leq 2-\frac{1}{y} \leq 1
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow-3 \leq-\frac{1}{y} \leq-1 \Rightarrow 1 \leq \frac{1}{y} \leq 3 \\
&\Rightarrow \frac{1}{3} \leq y \leq 1
\end{aligned}
$$
So, Rauge of $\mathrm{y}$ is $\left[\frac{1}{3}, 1\right]$
Let $y=\frac{1}{2-\cos x}$
$$
\begin{aligned}
&\Rightarrow \cos x=2-\frac{1}{y} \\
&\Rightarrow-1 \leq \cos x \leq 1 \Rightarrow-1 \leq 2-\frac{1}{y} \leq 1
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow-3 \leq-\frac{1}{y} \leq-1 \Rightarrow 1 \leq \frac{1}{y} \leq 3 \\
&\Rightarrow \frac{1}{3} \leq y \leq 1
\end{aligned}
$$
So, Rauge of $\mathrm{y}$ is $\left[\frac{1}{3}, 1\right]$
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