Question

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:

• A. $\left[\frac{1}{8}, \frac{1}{5}\right]$
• B. $\left[\frac{1}{7}, \frac{1}{6}\right]$
• C. $\left[\frac{1}{7}, \frac{1}{5}\right]$
• D. $\left[\frac{1}{8}, \frac{1}{6}\right]$

Answer 1

The Correct Option is D

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:
\[\left[\frac{1}{8}, \frac{1}{6}\right].\]

Answer 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 \(\left[\frac{1}{8}, \frac{1}{6}\right]\).