If f(x) + 2f (1 – x) = x² + 2 ∀ x ∈ R, then f(x) =

The correct option is 1/3 (x – 2)².

To find the function $f\left(x\right)$, we are given the functional equation:
$f\left(x\right)+2f(1-x)=x^2+2$         — (Equation 1)
for all real numbers $x$.

To solve this equation, we can substitute $1-x$ in place of $x$ throughout Equation 1. This gives:
$f(1-x)+2f(1-(1-x\left)\right)={(1-x)}^2+2$

Simplifying the terms inside the functions and expanding the right-hand side:
$f(1-x)+2f\left(x\right)=(1-2x+x^2)+2$
$2f\left(x\right)+f(1-x)=x^2-2x+3$         — (Equation 2)

Now we have a system of two linear equations in terms of $f\left(x\right)$ and $f(1-x)$:
1)   $f\left(x\right)+2f(1-x)=x^2+2$
2)   $2f\left(x\right)+f(1-x)=x^2-2x+3$

To eliminate the term $f(1-x)$, we multiply Equation 2 by 2:
$4f\left(x\right)+2f(1-x)=2x^2-4x+6$         — (Equation 3)

Now, subtract Equation 1 from Equation 3:
$\left[4f\right(x)+2f(1-x\left)\right]-\left[f\right(x)+2f(1-x\left)\right]=(2x^2-4x+6)-(x^2+2)$
$3f\left(x\right)=x^2-4x+4$

Recognizing that the right-hand side is a perfect square, we can rewrite it:
$3f\left(x\right)={(x-2)}^2$

Dividing both sides by 3, we obtain the expression for $f\left(x\right)$:
$f\left(x\right)=\frac{1}{3}{(x-2)}^2$