Discussion
\(f(x) = \frac{1}{7 - \cos}, x \in \mathbb{R}\) . Then the range of \(f\) is
Answer: E
The range is \(\left[\frac{1}{8}, \frac{1}{6}\right]\).
Reading the denominator as \(7 - \cos x\) (the \(x\) is missing in the paper): since \(-1 \le \cos x \le 1\),
\[ \begin{aligned} 6 \le 7 - \cos x &\le 8 \\ \frac{1}{8} \le f(x) &\le \frac{1}{6} \end{aligned} \]
Both end values are attained (at \(\cos x = -1\) and \(\cos x = 1\)), so the interval is closed. The negative intervals are impossible because the denominator is always positive, and intervals near 1 would need a denominator near 1.
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