Question

Give that $f\left(x\right)+f\left(\frac{1}{1-x}\right)=\frac{2(1-2x)}{x(1-x)},x\in R,x\ne 1$ then the given $f\left(x\right)$ is

• A. One-one
• B. Many-one
• C. $f\left(0\right)=1$
• D. none of these

Answer 1

Answer: (A)

One-one

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

Put $x=\frac{1}{1-x}$

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

Put $x=\frac{x-1}{x}$ in $\left(1\right)$

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

Consider $\left(1\right)-\left(2\right)+\left(3\right)$

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

$⟹f\left(x\right)=1$ which does not lie in the domain