Question

The range of the function \( f(x) = \frac{1}{7 + 4\sin x + 3\cos x} \).

• A. \( \left[ \frac{1}{14}, \frac{1}{4} \right] \)
• B. \( \left[ \frac{1}{7}, \frac{1}{3} \right] \)
• C. \( \left[ \frac{1}{3}, \frac{1}{7} \right] \)
• D. \( \left[ \frac{1}{7}, 1 \right] \)

Hint:

When finding the range of a function involving sine and cosine, use the maximum and minimum values of the linear combination of sine and cosine to find the reciprocal range.

Answer 1

The Correct Option is A

Step 1: Expression for the range.
The function is of the form: \[ f(x) = \frac{1}{7 + 4\sin x + 3\cos x} \] The range of this function depends on the values of \( 7 + 4\sin x + 3\cos x \), which varies depending on \( x \).
Step 2: Find the minimum and maximum values of the denominator.
The expression \( 7 + 4\sin x + 3\cos x \) is a linear combination of sine and cosine functions. To find its range, we first find its maximum and minimum values. The maximum and minimum of \( a\sin x + b\cos x \) is given by: \[ \text{Max} = \sqrt{a^2 + b^2}, \quad \text{Min} = -\sqrt{a^2 + b^2} \] In our case, \( a = 4 \) and \( b = 3 \), so: \[ \text{Max} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5, \quad \text{Min} = -5 \] Thus, the range of \( 7 + 4\sin x + 3\cos x \) is: \[ [7 - 5, 7 + 5] = [2, 12] \]
Step 3: Range of \( f(x) \).
Since \( f(x) = \frac{1}{7 + 4\sin x + 3\cos x} \), the range of \( f(x) \) will be the reciprocal of the range of \( 7 + 4\sin x + 3\cos x \), which is: \[ \left[ \frac{1}{12}, \frac{1}{2} \right] \] Thus, the range of \( f(x) \) is \( \left[ \frac{1}{14}, \frac{1}{4} \right] \).
Final Answer: \( \left[ \frac{1}{14}, \frac{1}{4} \right] \).