Question

Let R be the set of all real number and let f be a function R to R such that
$f\left(x\right)+\left(\begin{array}{c}x+\frac{1}{2}\end{array}\right)f(1-x)=1$, 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

The correct option is C -2

Given $f\left(x\right)+(x+\frac{1}{2})f(1-x)=1$............1
but x =0
$f\left(0\right)+\frac{1}{2}f\left(1\right)=1$
$\Rightarrow 2f\left(0\right)+f\left(1\right)=2$............(2)
put $x=1$ in $\left(1\right)$
$\Rightarrow f\left(1\right)+\frac{3}{2}f\left(0\right)=1$
$\Rightarrow 2f\left(1\right)+3f\left(0\right)=2$............(3)
Solving (2) &(3) we have
F(0)=2 & f(1)=-2
$\therefore 2f\left(0\right)+f\left(1\right)=4-6=-2$