Functions Question 384
Question: The range of the function $ f(x)=\frac{e^{x}-{e^{| x |}}}{e^{x}+{e^{| x |}}} $ is
Options:
A) $ (-\infty ,\infty ) $
B) $ [0,1) $
C) $ (-1,0] $
D) $ (-1,1) $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ f(x)=\frac{e^{x}-{e^{| x |}}}{e^{x}+{e^{| x |}}}= \begin{cases} 0, \\ \frac{e^{x}-{e^{-x}}}{e^{x}+{e^{-x}}}, \\ \end{cases} .,\begin{matrix} x\ge 0 \\ x<0 \\ \end{cases} $ 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)
BETA
