Question

The range of the function $f\left(x\right)=\frac{x^2-x}{x^2+2x}$ is
(a) R
(b) R − {1}
(c) R − {−1/2, 1}
(d) None of these

Answer 1

(c) R − {-1/2,1}

$f\left(x\right)=\frac{x^2-x}{x^2+2x}$
$$\begin{gathered} \mathrm{Let}y=\frac{x^2-x}{x^2+2x}\left[\mathrm{Also},x\ne 0\right] \\ \Rightarrow y=\frac{x(x-1)}{x(x+2)} \\ \Rightarrow y=\frac{(x-1)}{(x+2)} \\ \Rightarrow xy+2y=x-1 \\ \Rightarrow x=\frac{2y+1}{1-y} \\ \text{Here},1-y\ne 0. \\ \mathrm{or},y\ne 1. \\ \mathrm{Also},x\ne 0 \\ \Rightarrow \frac{2y+1}{1-y}\ne 0 \\ \Rightarrow y\ne -\frac{1}{2} \\ \mathrm{Thus},\mathrm{range}\left(f\right)=\mathrm{R}-\{-\frac{1}{2},1\}. \end{gathered}$$