The function f : R – {–1} → R – {1} defined by $f(x) = \frac{x}{x+1}$ is

Both 1-1 and onto

f : R – {–1} → R – {1}

$f(x) = \frac{x}{x+1} =y$  so  $y_1 = y_2 ⇒ \frac{x_1}{x_1+1} = \frac{x_2}{x_2+1}$

$\Rightarrow x_1\left(x_2+1\right)=x_2\left(x_1+1\right)$

$\Rightarrow x_2 x_1+x_1=x_2 x_1+x_2$

$\Rightarrow x_1=x_2$  ⇒  function is one one.

so  $y=\frac{x}{x+1} \Rightarrow xy + y = x  \Rightarrow  y = x(1-y)$

so  $x = \frac{y}{1-y}$   →   for every y atleast some x exist

⇒  it is onto

Option A