Question

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

• A. $2$
• B. $0$
• C. $-2$
• D. $-4$

Answer 1

Answer: (C)

$-2$

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

Substitute $x=0$

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

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

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

Substitute $x=1$

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

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

$2f\left(1\right)+3f\left(0\right)=2$ ............(2)

Multiply equation (1) with $2$ and subtract equation (2) from the result.

$2\left\{2f\right(0)+f(1\left)\right\}-2f\left(1\right)-3f\left(0\right)=4-2$

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

$f\left(0\right)=2$ ...............(3)

Now, from equation (1),

$2f\left(0\right)+f\left(1\right)=2$

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

$f\left(1\right)=2-4=-2$ ................(4)

Now,

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

$=4-6=-2$