A) Increasing
B) Decreasing
C) Even
D) None of these
Correct Answer: A
Solution :
\[f(x)=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}\] Þ(i)\[f(-x)=\frac{{{e}^{-2x}}-1}{{{e}^{-2x}}+1}=\frac{1-{{e}^{2x}}}{1+{{e}^{2x}}}\]Þ\[f(x)=-\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}=-f(x)\] \[f(x)\] is an odd function. Again \[f(x)=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}\Rightarrow {f}'(x)=\frac{4{{e}^{2x}}}{{{(1+{{e}^{2x}})}^{2}}}>0\,\forall \,n\in R\] Þ \[f(x)\] is an increasing function.You need to login to perform this action.
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