The function `f(x)=(asinx+b cosx)/(c sinx+d cos x)` is decreasing, if

Options

A. `ad-bc gt 0 `
B. `ad - bc lt 0 `
C. `ab-cd gt 0 `
D. `ab - cd lt 0 `

Answer

To determine when the function \( f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x} \) is decreasing, we need to find its derivative \( f'(x) \) and analyze when it is less than zero.

### Step-by-Step Solution:

1. **Identify the Function**:
\[ f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x} \]

2. **Differentiate Using the Quotient Rule**:
The quotient rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then:
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
Here, \( u = a \sin x + b \cos x \) and \( v = c \sin x + d \cos x \).

3. **Find \( u' \) and \( v' \)**:
- \( u' = a \cos x - b \sin x \)
- \( v' = c \cos x - d \sin x \)

4. **Apply the Quotient Rule**:
\[ f'(x) = \frac{(a \cos x - b \sin x)(c \sin x + d \cos x) - (a \sin x + b \cos x)(c \cos x - d \sin x)}{(c \sin x + d \cos x)^2} \]

5. **Set the Derivative Less Than Zero**:
Since \( (c \sin x + d \cos x)^2 > 0 \) (as it is a square), we need to focus on the numerator:
\[ (a \cos x - b \sin x)(c \sin x + d \cos x) - (a \sin x + b \cos x)(c \cos x - d \sin x) < 0 \]

6. **Expand the Numerator**:
Expanding both products:
\[ = ac \sin x \cos x + ad \cos^2 x - bc \sin^2 x - bd \sin x \cos x - (ac \sin x \cos x - ad \sin^2 x + bc \cos^2 x - bd \sin x \cos x) < 0 \]
Simplifying this gives:
\[ ad \cos^2 x + ad \sin^2 x - bc \sin^2 x - bc \cos^2 x < 0 \]

7. **Combine Like Terms**:
\[ ad(\cos^2 x + \sin^2 x) - bc(\sin^2 x + \cos^2 x) < 0 \]
Since \( \cos^2 x + \sin^2 x = 1 \):
\[ ad - bc < 0 \]

8. **Final Condition**:
Therefore, the function \( f(x) \) is decreasing if:
\[ ad - bc < 0 \]

### Conclusion:
The function \( f(x) \) is decreasing if \( ad < bc \).