Let $f(x) = \frac{1}{7 - \sin 5x}$ be a function defined on $\mathbb{R}$. Then the range of the function $f(x)$ is equal to:
- $\left[\frac{1}{8}, \frac{1}{5}\right]$
- $\left[\frac{1}{7}, \frac{1}{6}\right]$
- $\left[\frac{1}{7}, \frac{1}{5}\right]$
- $\left[\frac{1}{8}, \frac{1}{6}\right]$
The Correct Option is D
Approach Solution - 1
Since \( \sin 5x \in [-1, 1] \), we have:
\[-\sin 5x \in [-1, 1]\]
Therefore:
\[7 - \sin 5x \in [6, 8]\]
Thus, the function \( f(x) = \frac{1}{7 - \sin 5x} \) takes values in the interval:
\[\frac{1}{7 - \sin 5x} \in \left[\frac{1}{8}, \frac{1}{6}\right]\]
Therefore, the range of \( f(x) \) is:
\[\left[\frac{1}{8}, \frac{1}{6}\right].\]
Approach Solution -2
To determine the range of the function \( f(x) = \frac{1}{7 - \sin 5x} \), we first need to analyze the possible values of \( \sin 5x \).
The sine function, \( \sin \theta \), has a range of \([-1, 1]\). Therefore, for \( \sin 5x \), we also have:
- \(-1 \leq \sin 5x \leq 1\)
Next, we substitute this range into the expression \( 7 - \sin 5x \) to find its range. Performing the calculations:
- At its maximum: \( \sin 5x = -1 \Rightarrow 7 - \sin 5x = 7 - (-1) = 8 \)
- At its minimum: \( \sin 5x = 1 \Rightarrow 7 - \sin 5x = 7 - 1 = 6 \)
Thus, the expression \( 7 - \sin 5x \) takes values in the range \([6, 8]\).
The function \( f(x) = \frac{1}{7 - \sin 5x} \) is the reciprocal of \( 7 - \sin 5x \). The range of the reciprocal function, when its input range is \([6, 8]\), will be \([\frac{1}{8}, \frac{1}{6}]\). This is because the reciprocal function inverts the order due to its decreasing nature.
Therefore, the range of the function \( f(x) \) is:
- \(\left[\frac{1}{8}, \frac{1}{6}\right]\)
Hence, the correct answer is \(\left[\frac{1}{8}, \frac{1}{6}\right]\).
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