Question

The function $ f:R\to R $ defined by $ f(x)=\frac{{{e}^{|x|}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} $ is

• A. one-one and onto
• B. one-one but not onto
• C. not one-one but onto
• D. neither one-one nor onto

Answer 1

Answer: (D)

Given, $ f(x)=\frac{{{e}^{|x|}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} $
For $ x<0, $ $ f(x)=\frac{{{e}^{-x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}=0 $
Here, we see that for all negative values of x, we get always zero, it means it is not for one-one. For
$ x\ge 0, $
$ {{e}^{|x|}} > {{e}^{-x}} $
$ \therefore $ For $ x > 0,\,f(x) >0 $
and for $ x<0,\,f(x)=0 $
Hence, no negative real value of
$ f(x) $ exist, Hence, it is not onto.