A) \[f(x)=\sqrt{1+x+{{x}^{2}}}-\sqrt{1-x+{{x}^{2}}}\]
B) \[f(x)=log\left( \frac{1-x}{1+x} \right)\]
C) \[f(x)=log\left( x+\sqrt{1+{{x}^{2}}} \right)\]
D) \[f(x)=\frac{{{e}^{x}}+{{e}^{-x}}}{2}\]
Correct Answer: D
Solution :
[d] \[f(-x)=\sqrt{1+(-x)+{{(-x)}^{2}}}-\sqrt{1-(-x)+{{(-x)}^{2}}}\] \[=\sqrt{1-x+{{x}^{2}}}-\sqrt{1+x+{{x}^{2}}}=-f(x)\] Hence, \[f(x)\] is odd. [b] \[f(-x)=log\left\{ \frac{1-(-x)}{1+(-x)} \right\}=\log \left( \frac{1+x}{1-x} \right)=-f(x)\] Hence, f(x) is odd. [c] \[f(-x)=log\left( -x+\sqrt{1+{{(-x)}^{2}}} \right)\] \[=\log \left\{ \frac{(-x+\sqrt{1+{{x}^{2}}})(x+\sqrt{1+{{x}^{2}}})}{(x+\sqrt{1+{{x}^{2}}})} \right\}\] \[=\log \left( \frac{1}{x+\sqrt{1+{{x}^{2}}}} \right)=-f(x)\] Hence, f(x) is odd. [d] \[f(-x)=\frac{{{e}^{-x}}+{{e}^{-(-x)}}}{2}=\frac{{{e}^{-x}}+{{e}^{x}}}{2}=f(x)\] Hence, f(x) is even.
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