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If $f(x)$ satisfies the relation $2 f(x)+f(1-x)=x^{2}$ for all real $x$, then $f(x)$ is
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The correct answer is:
$\frac{x^{2}+2 x-1}{3}$
Given, $2 f(x)+f(1-x)=x^{2}$
Replacing $x$ by $(1-x)$, we get
$\quad 2 f(1-x)+f(x)=(1-x)^{2}$ $\Rightarrow \quad 2 f(1-x)+f(x)=1+x^{2}-2 x$ Multiplying Eq. (i) by 2 and subtracting Eq. (ii) from Eq. (i), we get $$ 3 f(x)=x^{2}+2 x-1 \Rightarrow f(x)=\frac{x^{2}+2 x-1}{3} $$ Eq. (i), we get
Replacing $x$ by $(1-x)$, we get
$\quad 2 f(1-x)+f(x)=(1-x)^{2}$ $\Rightarrow \quad 2 f(1-x)+f(x)=1+x^{2}-2 x$ Multiplying Eq. (i) by 2 and subtracting Eq. (ii) from Eq. (i), we get $$ 3 f(x)=x^{2}+2 x-1 \Rightarrow f(x)=\frac{x^{2}+2 x-1}{3} $$ Eq. (i), we get
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