Question

Let $f(x) = \frac{10}{7 + 4\sin x + 3\cos x}, x \in \mathbb{R}$. Then the range of the function $f$ is

• A. $[\frac{5}{7}, 5]$
• B. $[\frac{5}{7}, \frac{10}{7}]$
• C. $[\frac{5}{6}, 5]$
• D. $[\frac{5}{3}, 5]$
• E. $[\frac{5}{3}, \frac{10}{3}]$

Hint:

For \( a \sin x + b \cos x \), remember the Pythagorean triplets like (3, 4, 5). This allows you to immediately identify the range as $\pm 5$ without lengthy calculations.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To find the range of \(f(x)\), we need the minimum and maximum values of its denominator.
The denominator contains a linear combination of sine and cosine.
Step 2: Key Formula or Approach:
The expression \(a\sin x + b\cos x\) has a minimum value of \(-\sqrt{a^2 + b^2}\) and a maximum value of \(\sqrt{a^2 + b^2}\).
We apply this to \(4\sin x + 3\cos x\).
Step 3: Detailed Explanation:
Calculate the extreme values of \(4\sin x + 3\cos x\):
\[ R = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5 \] So, \(-5 \leq 4\sin x + 3\cos x \leq 5\).
Now substitute this into the denominator \(D(x) = 7 + 4\sin x + 3\cos x\):
Minimum of \(D(x) = 7 - 5 = 2\).
Maximum of \(D(x) = 7 + 5 = 12\).
The function is \(f(x) = \frac{10}{D(x)}\).
When \(D(x)\) is maximum (12), \(f(x)\) is minimum: \(\frac{10}{12} = \frac{5}{6}\).
When \(D(x)\) is minimum (2), \(f(x)\) is maximum: \(\frac{10}{2} = 5\).
Therefore, the range is \([\frac{5}{6}, 5]\).
Step 4: Final Answer:
The range of the function is \([\frac{5}{6}, 5]\).