A) \[f(x)=\frac{{{a}^{x}}+{{a}^{-x}}}{{{a}^{x}}-{{a}^{-x}}}\]
B) \[f(x)=\frac{{{a}^{x}}+1}{{{a}^{x}}-1}\]
C) \[f(x)=x\frac{{{a}^{x}}-1}{{{a}^{x}}+1}\]
D) \[f(x)={{\log }_{e}}(x+\sqrt{{{x}^{2}}+1})\]
Correct Answer: C
Solution :
If\[f(-x)=f(x),\]then\[f(x)\]is an even function \[\therefore \] Putting\[x=-x\]in \[x\frac{{{a}^{x}}-1}{{{a}^{x}}+1}\] \[\therefore \] \[f(-x)=-x.\frac{{{a}^{-x}}-1}{{{a}^{-x}}+1}\] \[=-x.\frac{{{a}^{-x}}(1-{{a}^{x}})}{{{a}^{-x}}(1+{{a}^{x}})}\] \[=-x\frac{(1-{{a}^{x}})}{(1+{{a}^{x}})}\] \[=x\left( \frac{{{a}^{x}}-1}{{{a}^{x}}+1} \right)=f(x)\] \[\therefore \] \[x\frac{{{a}^{x}}-1}{{{a}^{x}}+1}\]is an even function.
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