Question

Let R be the set of all real numbers and let f be a function R to R such that $f\left(x\right)+\left(x+\frac{1}{2}\right)f(1-x)=1$, for all $xϵR$. Then $2f\left(0\right)+3f\left(1\right)$ is equal to.

• A. $2$
• B. $0$
• C. $-2$
• D. $-4$

Answer 1

Answer: (C)

$-2$

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

Let $x=0$, we have

$f\left(0\right)+\frac{1}{2}f\left(1\right)=1...\left(1\right)$

When $x=1$, we have

$f\left(1\right)+\frac{3}{2}f\left(0\right)=1$

Multiplying equation (1) by 2 and subtracting equation (2) from that, we have

$2f\left(0\right)+f\left(1\right)-f\left(1\right)-\frac{3}{2}f\left(0\right)=2-1=1$

$\Rightarrow \frac{1}{2}f\left(0\right)=1$

$\therefore f\left(0\right)=2$

$\therefore f\left(1\right)=-2$

$\Rightarrow 2f\left(0\right)+3f\left(1\right)=4-6=-2$