If `f(x) = ( 10 cos x + 5 cos 3x + cos 5x)/(cos 6x + 6cos 4x + 15 cos 2x +10)` then `f(0)+f'(0) + f''(0)=`

Options

A. `- ( 1)/( 2)`
B. 0
C. `(1)/(2)`
D. 1

Answer

To solve the problem, we need to evaluate \( f(0) + f'(0) + f''(0) \) for the function

\[ f(x) = \frac{10 \cos x + 5 \cos 3x + \cos 5x}{\cos 6x + 6 \cos 4x + 15 \cos 2x + 10} \]

### Step 1: Evaluate \( f(0) \)

First, we calculate \( f(0) \):

\[ f(0) = \frac{10 \cos(0) + 5 \cos(0) + \cos(0)}{\cos(0) + 6 \cos(0) + 15 \cos(0) + 10} \]

Since \( \cos(0) = 1 \):

\[ f(0) = \frac{10 \cdot 1 + 5 \cdot 1 + 1}{1 + 6 \cdot 1 + 15 \cdot 1 + 10} = \frac{10 + 5 + 1}{1 + 6 + 15 + 10} = \frac{16}{32} = \frac{1}{2} \]

### Step 2: Differentiate \( f(x) \) to find \( f'(x) \)

Using the quotient rule:

\[ f'(x) = \frac{(g(x) \cdot f'(x) - f(x) \cdot g'(x))}{(g(x))^2} \]

where \( f(x) = 10 \cos x + 5 \cos 3x + \cos 5x \) and \( g(x) = \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 \).

#### Calculate \( f'(0) \)

1. Differentiate \( f(x) \):
- \( f'(x) = -10 \sin x - 15 \sin 3x - 5 \sin 5x \)
- \( g'(x) = -6 \sin 6x - 24 \sin 4x - 30 \sin 2x \)

2. Evaluate \( f'(0) \):
- \( f'(0) = -10 \sin(0) - 15 \sin(0) - 5 \sin(0) = 0 \)
- \( g'(0) = -6 \sin(0) - 24 \sin(0) - 30 \sin(0) = 0 \)

Using L'Hôpital's Rule since both \( f(0) \) and \( g(0) \) are 0:

\[ f'(0) = \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \frac{0}{0} \]

### Step 3: Differentiate again to find \( f''(x) \)

1. Differentiate \( f'(x) \) to find \( f''(x) \):
- \( f''(x) = -10 \cos x - 45 \cos 3x - 25 \cos 5x \)

2. Evaluate \( f''(0) \):
- \( f''(0) = -10 \cdot 1 - 45 \cdot 1 - 25 \cdot 1 = -10 - 45 - 25 = -80 \)

### Step 4: Combine results

Now, we combine \( f(0) \), \( f'(0) \), and \( f''(0) \):

\[ f(0) + f'(0) + f''(0) = \frac{1}{2} + 0 - 80 = \frac{1}{2} - 80 = \frac{1 - 160}{2} = \frac{-159}{2} \]

### Final Answer

Thus,

\[ f(0) + f'(0) + f''(0) = \frac{-159}{2} \]