Question

Let $f\left(x\right)=\frac{x^2-4}{x^2+4}$ for $|x|>2$, then the function $f:(-\infty ,-2)\cup [2,\infty )\to (-1,1)$ is

• A. One-one into
• B. One-one onto
• C. Many one into
• D. Many one onto

Answer 1

The correct option is B One-one onto

$\begin{array}{c}f\left(x\right)=\frac{x^2-4}{x^2+4}\\ letf\left(x\right)=y\\ \Rightarrow y=\frac{x^2-4}{x^2+4}\\ \Rightarrow x^{2}y+4y-x^2+4=0\\ \Rightarrow x^2\left(y-1\right)+4\left(y+1\right)=0\\ \Rightarrow x^2=\frac{-4\left(y+1\right)}{y-1}\\ \Rightarrow x=\frac{\sqrt{4\left(y+1\right)}}{\left(1-y\right)}\\ \Rightarrow \frac{4\left(y+1\right)}{1-y}0\\ \Rightarrow y+,0\\ \Rightarrow y-1\\ \Rightarrow 1-y0\\ \Rightarrow y1\\ Hence,rangeof{f}^{b}is\left(-1,1whichisequaltoCodomain\right)\\ \therefore f^{h}isone-one-onto1mu\end{array}$