Relations & Functions
Functional Equations
Grade 12
Question:
<p>For <math>x \in \mathbb{R}</math>, the function <math>f(x)</math> satisfies <math>2f(x) + f(1-x) = x^2</math>. The value of <math>f(4)</math> is equal to</p>
<p>(a) <math>\frac{13}{3}</math></p>
<p>(b) <math>\frac{43}{3}</math></p>
<p>(c) <math>\frac{23}{3}</math></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Replace the variable with its complement to create a system of equations, then solve simultaneously to find the function.
<p><strong>Step 1:</strong> We have <math>2f(x) + f(1-x) = x^2</math> ... (i)</p><p><strong>Step 2:</strong> Replace <math>x</math> by <math>(1-x)</math> in equation (i):</p><p><math>2f(1-x) + f(x) = (1-x)^2</math> ... (ii)</p><p><strong>Step 3:</strong> Solve equations (i) and (ii):</p><p>From (i): <math>2f(x) + f(1-x) = x^2</math></p><p>Multiply by 2: <math>4f(x) + 2f(1-x) = 2x^2</math></p><p>Subtract (ii): <math>4f(x) + 2f(1-x) - 2f(1-x) - f(x) = 2x^2 - (1-x)^2</math></p><p><math>3f(x) = 2x^2 - (1-2x+x^2) = x^2 + 2x - 1</math></p><p><strong>Step 4:</strong> Therefore <math>f(x) = \frac{x^2 + 2x - 1}{3}</math></p><p><strong>Step 5:</strong> <math>f(4) = \frac{16 + 8 - 1}{3} = \frac{23}{3}</math></p><p>∴ Answer is (c) <math>\frac{23}{3}</math></p>
Correct Answer: C