Question

If $f\left(x\right)+2f(1-x)=x^2+2,\forall x\in R$, then find $f\left(x\right)$

• A. $\frac{(x+3{)}^2}{3}$
• B. $\frac{(x-3{)}^2}{3}$
• C. $\frac{(x-2{)}^2}{4}$
• D. $\frac{(x-2{)}^2}{3}$

Answer 1

The correct option is D $\frac{(x-2{)}^2}{3}$

Given

$f\left(x\right)+2f(1-x)=x^2+2\cdots \left(1\right)$

Replace $x$ with $1-x$

$⟹f(1-x)+2f(1-(1-x\left)\right)=(1-x{)}^2+2$

$⟹f(1-x)+2f\left(x\right)=x^2-2x+3\cdots \left(2\right)$

$\left(1\right)+\left(2\right)$

$⟹3f\left(x\right)+3f(1-x)=2x^2-2x+5$

$⟹f\left(x\right)+f(1-x)=\frac{2}{3}{x}^2-\frac{2}{3}x+\frac{5}{3}\cdots \left(3\right)$

$\left(2\right)-\left(1\right)$

$⟹f\left(x\right)-f(1-x)=1-2x\cdots \left(4\right)$

$\left(3\right)+\left(4\right)$

$⟹2f\left(x\right)=\frac{2}{3}{x}^2-\frac{2}{3}x+\frac{5}{3}+1-2x$

$⟹f\left(x\right)=\frac{1}{3}{(x-2)}^2$