Read ihe following information carefully and answer the questions given below: Let `f(x) =ax^(2) + bx + c` such that `f(1) =f(-1)` and a. b. c are in arithmetic progression.
`f''(a).f''(b)` and `f''( c)` are:

Options

A. In AP only
B. In GP only
C. In both AP and GP
D. Neither in AP nor in GP

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the function and conditions
We are given a quadratic function:
\[ f(x) = ax^2 + bx + c \]
with the condition that:
\[ f(1) = f(-1) \]
and that \( a, b, c \) are in arithmetic progression.

### Step 2: Find the condition from \( f(1) = f(-1) \)
Calculate \( f(1) \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]

Calculate \( f(-1) \):
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]

Set these equal to each other:
\[ a + b + c = a - b + c \]

### Step 3: Simplify the equation
Subtract \( a + c \) from both sides:
\[ b = -b \]

This implies:
\[ 2b = 0 \]
So, we find:
\[ b = 0 \]

### Step 4: Use the condition of arithmetic progression
Since \( a, b, c \) are in arithmetic progression and we found \( b = 0 \), we can express \( c \) in terms of \( a \):
If \( b = 0 \), then \( a, 0, c \) are in arithmetic progression, which means:
\[ c = -a \]

### Step 5: Find the second derivative of \( f(x) \)
Now, we need to find the second derivative \( f''(x) \):
First, find the first derivative:
\[ f'(x) = 2ax + b = 2ax \] (since \( b = 0 \))

Now, find the second derivative:
\[ f''(x) = 2a \]

### Step 6: Evaluate \( f''(a), f''(b), f''(c) \)
Since \( f''(x) = 2a \) is a constant, we have:
\[ f''(a) = 2a \]
\[ f''(b) = 2a \]
\[ f''(c) = 2a \]

### Step 7: Determine the relationship between \( f''(a), f''(b), f''(c) \)
Since \( f''(a) = f''(b) = f''(c) = 2a \), they are all equal. This means they are in both arithmetic progression (AP) and geometric progression (GP).

### Final Conclusion
Thus, we can conclude that:
\[ f''(a), f''(b), f''(c) \text{ are in both AP and GP.} \]