Question

If \( f(x) + 2f(1 - x) = x^2 + 5 \) for all real values of \( x \), then \( f(x) \) is given by:

• A. \( x^2 - 5 \)
• B. \( 2 \)
• C. \( \frac{(x - 2)^2 + 3}{3} \)
• D. None of these

Hint:

In functional equations involving \( f(x) \) and \( f(1 - x) \), always consider substituting \( x \to 1 - x \) to form a system of equations. Solving them simultaneously often reveals the explicit form of the function.

Answer 1

The Correct Option is C

Step 1: Given equation is
\[ f(x) + 2f(1 - x) = x^2 + 5 \quad \text{(1)} \] Step 2: Replace \( x \) with \( 1 - x \) in equation (1)
\[ f(1 - x) + 2f(x) = (1 - x)^2 + 5 = x^2 - 2x + 1 + 5 = x^2 - 2x + 6 \quad \text{(2)} \] Step 3: Solve equations (1) and (2) simultaneously
Equation (1): \( f(x) + 2f(1 - x) = x^2 + 5 \)
Equation (2): \( 2f(x) + f(1 - x) = x^2 - 2x + 6 \) Multiply equation (1) by 2:
\[ 2f(x) + 4f(1 - x) = 2x^2 + 10 \quad \text{(3)} \] Now subtract (2) from (3):
\[ (2f(x) + 4f(1 - x)) - (2f(x) + f(1 - x)) = (2x^2 + 10) - (x^2 - 2x + 6) \] \[ 3f(1 - x) = x^2 + 2x + 4 \Rightarrow f(1 - x) = \frac{x^2 + 2x + 4}{3} \] Put this value back into equation (1):
\[ f(x) + 2 \cdot \frac{x^2 + 2x + 4}{3} = x^2 + 5 \] \[ f(x) + \frac{2(x^2 + 2x + 4)}{3} = x^2 + 5 \Rightarrow f(x) = x^2 + 5 - \frac{2(x^2 + 2x + 4)}{3} \] \[ f(x) = \frac{3(x^2 + 5) - 2(x^2 + 2x + 4)}{3} = \frac{3x^2 + 15 - 2x^2 - 4x - 8}{3} = \frac{x^2 - 4x + 7}{3} \] \[ f(x) = \frac{(x - 2)^2 + 3}{3} \]