Question

If $f\left(x\right)+2f(1-x)=x^2+5\forall$ real values $x$ then $f\left(x\right)$ is given by

• A. $x^2-5$
• B. $2$
• C. $\frac{(x-2{)}^2+3}{3}$
• D. None of these

Answer 1

Answer: (C)

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

Given $f\left(x\right)+2f(1-x)=x^2+\mathrm{5...}\left(i\right)$

Replace
$x\to 1-x$

$\therefore f(1-x)+2f\left(x\right)=(1-x{)}^2+5...(ii)$

From $\left(i\right)$ and $\left(ii\right)$

$2\times \left(ii\right)-\left(i\right)\Rightarrow 2f(1-x)+4f\left(x\right)-f\left(x\right)-2f(1-x)=2(1-x{)}^2+(5\times 2)-x^2-5$

$\Rightarrow 3f\left(x\right)=2(x^2-2x+1)-x^2+5=x^2-4x+7$

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