A) \[(-\infty ,\infty )\]
B) \[[0,1)\]
C) \[(-1,0]\]
D) \[(-1,1)\]
Correct Answer: C
Solution :
| [c] \[f(x)=\frac{{{e}^{x}}-{{e}^{\left| x \right|}}}{{{e}^{x}}+{{e}^{\left| x \right|}}}=\left\{ \begin{matrix} 0, \\ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}, \\ \end{matrix} \right.\,\,\,\begin{matrix} x\ge 0 \\ x<0 \\ \end{matrix}\] |
| Clearly, \[f(x)\] is identically zero if \[x\ge 0\] (1) |
| If \[x<0,\] let \[y=f(x)=\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}\] or \[{{e}^{2x}}=\frac{1+y}{1-y}\] |
| \[\because \,\,\,\,x<0;\,\,{{e}^{2x}}<1\] or \[0<{{e}^{2x}}<1\] |
| \[\therefore 0<\frac{1+y}{1-y}<1\] or \[\frac{1+y}{1-y}>0\] and \[\frac{1+y}{1-y}<1\] |
| or \[(y+1)(y-1)<0\] and \[\frac{2y}{1-y}<0\] |
| i.e., \[-1<y<1\] and \[y<0\] or \[y>1\] |
| or \[-1<y<0\] (2) |
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