Following questions are based on the given information for the following functions `f(x)`
`f(x)=2bx+f(-x), if x lt0`
`f(x)=a if x =0`
`f(x)=b+c-2cx+f(x-1), if x gt0`
`f(-19)` equals :

Options

A. A. `a-19b+361c`
B. B. `a+19(b-19c)`
C. C. `a-19(b+19c)`
D. D. none of these

Answer

To find \( f(-19) \) based on the given functions, we will follow these steps:

### Step 1: Identify the function for \( f(x) \) when \( x < 0 \)
Since \( -19 < 0 \), we will use the function:
\[ f(x) = 2bx + f(-x) \]
Thus, we can express \( f(-19) \) as:
\[ f(-19) = 2b(-19) + f(19) = -38b + f(19) \]

### Step 2: Find \( f(19) \)
Since \( 19 > 0 \), we will use the function:
\[ f(x) = b + c - 2cx + f(x-1) \]
We need to calculate \( f(19) \) by finding \( f(18), f(17), \ldots \) down to \( f(0) \).

### Step 3: Calculate \( f(0) \)
From the information given:
\[ f(0) = a \]

### Step 4: Calculate \( f(1) \)
Using the function for \( x = 1 \):
\[ f(1) = b + c - 2c(1) + f(0) = b + c - 2c + a = b - c + a \]

### Step 5: Calculate \( f(2) \)
Using the function for \( x = 2 \):
\[ f(2) = b + c - 2c(2) + f(1) = b + c - 4c + (b - c + a) = 2b - 4c + a \]

### Step 6: Calculate \( f(3) \)
Using the function for \( x = 3 \):
\[ f(3) = b + c - 2c(3) + f(2) = b + c - 6c + (2b - 4c + a) = 3b - 9c + a \]

### Step 7: Generalize \( f(n) \)
From the pattern observed, we can hypothesize:
\[ f(n) = nb - \frac{n(n+1)}{2}c + a \]
This can be verified by induction or by continuing the calculations.

### Step 8: Calculate \( f(19) \)
Using our generalized formula:
\[ f(19) = 19b - \frac{19(20)}{2}c + a = 19b - 190c + a \]

### Step 9: Substitute \( f(19) \) back into \( f(-19) \)
Now substituting \( f(19) \) back into our equation for \( f(-19) \):
\[ f(-19) = -38b + (19b - 190c + a) = -19b - 190c + a \]

### Final Result
Thus, we have:
\[ f(-19) = a - 19b - 190c \]