A) one-one and onto
B) one-one but not onto
C) not one-one but onto
D) neither one-one nor onto
Correct Answer: D
Solution :
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.
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