given $2f(x) + f(1-x) = x^2$ find $f(-5)$

Question

I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.

A function $f$ has property that $2f(x)+ f(1-x) = x^2$ for any number of x. What is the number for $f(-5)$?

(A) $\frac {15}4$ (B) $\frac {21}5$ (C) $\frac {25}3$ (D) $\frac {14}3$ (E) $\frac {34}3$

It doesn't give us what $f(x)$ is equal to.

Tried to do comparison

\begin{align} 2f(-5) + f(1-(-5)) &= (-5)^2 \\ 2f(-5) + f(6) &= 25 \quad(1) \\ 2f(6) + f(1-(6)) &= (6)^2 \\ 2f(6) + f(-5) &= 36 \quad(2) \\ \end{align}

so we have the $f(-5)$ and $f(6)$ in both functions, we can add them together \begin{align} (1)+(2) \\ [2f(-5) + f(6)] + [2f(6) + f(-5)] &= 25+36 \\ 2f(-5) + f(-5) + f(6) + 2f(6) &= 61 \\ 3f(-5) + 3f(6) &= 61 \\ f(6) &= \frac {61}3 - f(-5) \quad(3) \\ \end{align}

replacing $f(6)$, at equation $\quad(3)$ to the first equation, $\quad(1)$

\begin{align} 2f(-5) + [\frac{61}{3} - f(-5)] &= 25 \\ 2f(-5) - f(-5) &= 25 - \frac {61}3 \\ f(-5) &= \frac{14}{3} \end{align}

EDIT: Well, I am blind, saw + to minus

Answer 1

Hint

If (this was the original post) $$2f(x) - f(1-x) = x^2$$ then $$2f(1-x) - f(x) = (1-x)^2$$ So, multiplying the first by $2$ and adding to the second, we have $$3f(x)=2x^2+(1-x)^2$$

Edit

By the way, the answer for $x=-5$ is not in the list of choices and your answer is perfectly correct.

You would get $\frac {34}3$ for $f(-3)$. All other cases correspond to $x$ expressed with radicals. Probably one more typo.

You changed the post (from $-$ to $+$) but the same method applies.

Answer 2

We have, $$2f(x)-f(1-x)=x^2$$ $$\implies 2f(6)-f(1-6)=6^2$$ $$\implies 2f(6)-f(-5)=36\tag 1$$ $$ 2f(-5)-f(1-(-5))=(-5)^2$$ $$\implies 2f(-5)-f(6)=25$$ $$\implies 4f(-5)-2f(6)=50\tag 2$$ Now, adding (1) & (2), we get $$3f(-5)=36+50=86$$ $$\implies f(-5)=\frac{86}{3}$$ Your answer is correct. There may be some printing mistake in the options provided in your book.