Question

If \(f(x)+3f(1-x)=x+4\), then \(f(x)=\)

• A. \(\frac{19-4x}{8}\)
• B. \(\frac{11-4x}{8}\)
• C. \(\frac{11-2x}{8}\)
• D. \(\frac{11-2x}{9}\)
• E. \(\frac{11-4}{9}\)

Hint:

In functional equations involving \(f(x)\) and \(f(1-x)\), always substitute \(x \to 1-x\) to create a system and solve simultaneously.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a functional equation. The equation relates the value of the function at \(x\) to its value at \(1-x\). The trick to solving this type of equation is to create a second equation by substituting \(1-x\) for \(x\), and then solve the resulting system of two linear equations for \(f(x)\).
Step 2: Key Formula or Approach:
1. Start with the given equation: \(f(x) + 3f(1-x) = x + 4 \quad \cdots (1)\) 2. Replace \(x\) with \(1-x\) everywhere in the equation to get a second equation. 3. Solve the system of equations (1) and (2) for \(f(x)\).
Step 3: Detailed Explanation:
The given equation is: \[ f(x) + 3f(1-x) = x + 4 \quad \cdots (1) \] Now, replace \(x\) by \(1-x\): \[ f(1-x) + 3f(1-(1-x)) = (1-x) + 4 \] \[ f(1-x) + 3f(x) = 5 - x \quad \cdots (2) \] We now have a system of two equations with two "unknowns," \(f(x)\) and \(f(1-x)\). \begin{align} f(x) + 3f(1-x) &= x + 4 \quad &(1)
3f(x) + f(1-x) &= 5 - x \quad &(2) \end{align} To eliminate \(f(1-x)\), let's multiply equation (2) by 3 and subtract equation (1) from the result.
Multiplying equation (2) by 3 gives: \[ 9f(x) + 3f(1-x) = 3(5-x) = 15 - 3x \quad \cdots (3) \] Now, subtract equation (1) from equation (3): \[ (9f(x) + 3f(1-x)) - (f(x) + 3f(1-x)) = (15 - 3x) - (x + 4) \] \[ 8f(x) = 15 - 3x - x - 4 \] \[ 8f(x) = 11 - 4x \] Finally, solve for \(f(x)\): \[ f(x) = \frac{11 - 4x}{8} \] Step 4: Final Answer:
The function \(f(x)\) is \(\frac{11-4x}{8}\). This corresponds to option (B).