Question

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

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

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

Solving (2) & (3) we have

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

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