Question

If $f\left(x\right)-2f(1-x)=x^2+2$, then what is $f\left(x\right)$?

• A. $f\left(x\right)=-x^2+\frac{4}{3}x-\frac{3}{8}$
• B. $f\left(x\right)=-x^2+\frac{4}{3}x-\frac{8}{3}$
• C. $f\left(x\right)=-x^2+\frac{8}{3}x-\frac{4}{3}$
• D. $f\left(x\right)=-x^2+\frac{3}{8}x-\frac{3}{4}$

Answer 1

Answer: (B)

$f\left(x\right)=-x^2+\frac{4}{3}x-\frac{8}{3}$

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

Setting $x=1-x$ then we get

$f(1-x)-2f(1-1+x)={(1-x)}^2+2$

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

$2f(1-x)-4f\left(x\right)=2x^2-4x+6$ .......$\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get

$-3f\left(x\right)=3x^2-4x+8$

$\therefore f\left(x\right)=-x^2+\frac{4}{3}x-\frac{8}{3}$