- Home
- Standard 12
- Mathematics
Let $R$ be the set of all real numbers and let $f$ be a function from $R$ to $R$ such that $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1$, for all $x \in R$. Then $2 f(0)+3 f(1)$ is equal to
$2$
$0$
$-2$
$-4$
Solution
(c)
Given,
$\Rightarrow f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1$
$\text { Put } x=1-x \text {, we get }$
$f(1-x)+\left(1-x+\frac{1}{2}\right) f(1-(1-x)=1$
$\Rightarrow f(1-x)+\left(\frac{3}{2}-x\right) f(x)=1$
Eq.$(ii)$ multiply by $\left(x+\frac{1}{2}\right)$ we get
$\left(\frac{3}{2}-x\right)\left(x+\frac{1}{2}\right) f(x)+\left(x+\frac{1}{2}\right)$
$f(1-x)=x+\frac{1}{2}$
On subtracting Eq. $(iii)$ from Eq.$(i)$, we get
$f(x)\left[1-\left(\frac{3}{4}+x-x^2\right)\right]=1-x-\frac{1}{2}$
$\Rightarrow f(x)=\frac{\frac{1}{2}-x}{x^2-x+\frac{1}{4}} \Rightarrow f(0)=2 \text { and } f(1)=-2$
$\because 2 f(0)+3 f(1)-2(2)+3(-2)=4-6=-2$
