Question

If \(f(x)+2f(1-x)=x^{2}+5\) for all real \(x\), then \(f(x)\) is equal to...

• A. \(\frac{(x-2)^{2}+3}{3}\)
• B. \(\frac{(x-2)^{2}-3}{3}\)
• C. \(\frac{(x-3)^{2}-2}{2}\)
• D. \(\frac{(x-3)^{2}+2}{2}\)

Hint:

Replace x by 1-x to get a second equation, then eliminate f(1-x).

Answer 1

The Correct Option is A

Step 1: Substitute 1-x
Replacing x with $1-x$ gives $f(1-x)+2f(x)=x^{2}-2x+6$.

Step 2: Solve the Pair
Let $a=f(x)$ and $b=f(1-x)$: $a+2b=x^{2}+5$ and $2a+b=x^{2}-2x+6$.
Then $3a=2(x^{2}-2x+6)-(x^{2}+5)=x^{2}-4x+7$.

Step 3: Check the Options
$x^{2}-4x+7=(x-2)^{2}+3$, so $f(x)=\dfrac{(x-2)^{2}+3}{3}$, option (A).

Final Answer:
The function is $\dfrac{(x-2)^{2}+3}{3}$, option (A). \[ \boxed{\text{(A) }\dfrac{(x-2)^{2}+3}{3}} \]