If $f(x)+2f(1-x)=x^2+1\, ∀\, x ∈ R$ then f(x) is

$\frac{1}{3}(x^2+4x-3)$ $\frac{2}{3}(x^2+4x-3)$ $\frac{1}{3}(x^2-4x+3)$ $\frac{2}{3}(x^2-4x+3)$

$\frac{1}{3}(x^2-4x+3)$

$f(x) + 2f(1 − x) = x^2 + 1$ Replacing x → (1 − x) we get $f(1 − x)+2f(x)=(1-x)^2+1$  $⇒f(1 − x)+2f(x)=2+x^2-2x$  From these equations we get $f(x)=\frac{1}{3}(x^2-4x+3)$