Question

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

• A. $\frac{2}{3}(x^2-4x+3)$
• B. $\frac{2}{3}(x^2+4x-3)$
• C. $\frac{1}{3}(x^2+4x-3)$
• D. $\frac{1}{3}(x^2-4x+3)$

Answer 1

The correct option is D

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

$f\left(x\right)+2f(1-x)=x^2+1$ .........(i)
Replacing $x$ by $1–x$
$f(1-x)+2f\left(x\right)=(1-x{)}^2+1$ .......(ii)
multiplying (ii) by 2 and subtracting it from (1), we get
$-3f\left(x\right)=x^2-2(1-x{)}^2-1$
$3f\left(x\right)=2(1-x{)}^2+1-x^2=(1-x\left)\right(2-2x+1+x)=(1-x\left)\right(3-x)=x^2-4x+3$
$\Rightarrow f\left(x\right)=\frac{1}{3}(x^2-4x+3)$