Let `f(x)` be a second degree polynomial function such that `f(-1)=f(1)` and `alpha,beta,gamma` are in A.P. Then, `f'(alpha),f'(beta),f'(gamma)` are in

Options

A. AP
B. GP
C. HP
D. none of these

Explanation

To solve the problem, we need to analyze the given conditions and derive the necessary relationships step by step.

### Step 1: Define the Polynomial Function
Let \( f(x) \) be a second-degree polynomial function, which can be expressed in the standard form:
\[ f(x) = ax^2 + bx + c \]

### Step 2: Differentiate the Polynomial
To find the derivative of the polynomial, we differentiate \( f(x) \):
\[ f'(x) = 2ax + b \]

### Step 3: Use the Given Condition \( f(-1) = f(1) \)
We know that \( f(-1) = f(1) \). Let's calculate both:
- For \( f(-1) \):
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]
- For \( f(1) \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]

Setting these equal gives us:
\[ a - b + c = a + b + c \]

### Step 4: Simplify the Equation
By simplifying the equation, we can eliminate \( a \) and \( c \):
\[ -b = b \implies 2b = 0 \implies b = 0 \]

### Step 5: Update the Derivative
Since \( b = 0 \), the derivative simplifies to:
\[ f'(x) = 2ax \]

### Step 6: Evaluate the Derivative at \( \alpha, \beta, \gamma \)
Now, we can evaluate the derivative at points \( \alpha, \beta, \gamma \):
\[ f'(\alpha) = 2a\alpha, \quad f'(\beta) = 2a\beta, \quad f'(\gamma) = 2a\gamma \]

### Step 7: Use the Condition that \( \alpha, \beta, \gamma \) are in A.P.
Since \( \alpha, \beta, \gamma \) are in arithmetic progression (A.P.), we have:
\[ 2\beta = \alpha + \gamma \]

### Step 8: Relate the Derivatives
Now, we can relate the derivatives:
\[ f'(\beta) = 2a\beta, \quad f'(\alpha) = 2a\alpha, \quad f'(\gamma) = 2a\gamma \]
Multiplying the A.P. condition by \( 2a \):
\[ 2f'(\beta) = f'(\alpha) + f'(\gamma) \]

### Step 9: Conclusion
From the relationship derived, we conclude that \( f'(\alpha), f'(\beta), f'(\gamma) \) are in A.P.

Thus, the final answer is that \( f'(\alpha), f'(\beta), f'(\gamma) \) are in A.P.

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