Question

The range of the real valued function \( f(x) = \frac{15}{3 \sin x + 4 \cos x + 10} \) is:

• A. \([0, 3]\)
• B. \([-1, 3]\)
• C. \([1, 3]\)
• D. \([-1, 1]\)

Hint:

For functions involving sine and cosine, rewrite the sum \( a \sin x + b \cos x \) in the form \( R \sin(x + \alpha) \), where \( R = \sqrt{a^2 + b^2} \) to easily find the range.

Answer 1

The Correct Option is C

We are given the function \( f(x) = \frac{15}{3 \sin x + 4 \cos x + 10} \). To find the range of this function, we must first analyze the expression in the denominator: \[ g(x) = 3 \sin x + 4 \cos x + 10. \] Step 1:
The expression \( 3 \sin x + 4 \cos x \) can be rewritten in a more convenient form using the identity: \[ R \sin(x + \alpha) = 3 \sin x + 4 \cos x. \] Here, \( R = \sqrt{3^2 + 4^2} = 5 \), and the phase shift \( \alpha \) can be calculated as: \[ \tan \alpha = \frac{4}{3} \quad \Rightarrow \quad \alpha = \tan^{-1} \left( \frac{4}{3} \right). \] Thus, \[ g(x) = 5 \sin(x + \alpha) + 10. \] Step 2:
The range of \( 5 \sin(x + \alpha) \) is from \( -5 \) to \( 5 \), so the range of \( g(x) = 5 \sin(x + \alpha) + 10 \) is: \[ [10 - 5, 10 + 5] = [5, 15]. \] Step 3:
Now, we analyze the function \( f(x) = \frac{15}{g(x)} \). Since \( g(x) \) ranges from 5 to 15, the function \( f(x) \) will take values corresponding to: \[ f(x) = \frac{15}{g(x)} \quad \text{where} \quad g(x) \in [5, 15]. \] For \( g(x) = 5 \), \( f(x) = \frac{15}{5} = 3 \), and for \( g(x) = 15 \), \( f(x) = \frac{15}{15} = 1 \). Thus, the range of \( f(x) \) is \( [1, 3] \).