If f(x)=\log_{e}\left ( \frac{1-x}{1+x} \right ),\left | x \right |<1, then f\left ( \frac{2x}{1+x^{2}} \right ) is equal to :

  • Option 1)

    2f(x^{2})
     

  • Option 2)

    2f(x)

  • Option 3)

    -2f(x)

     

  • Option 4)

    (f(x))^{2}

 

Answers (1)

f(x)=\log_{e}\left ( \frac{1-x}{1+x} \right ),\left | x \right |<1,

the f(\frac{2x}{1+x^{2}})=?

f\left ( \frac{2x}{1+x^{2}} \right )=\log_{e}\left ( \frac{1-\frac{2x}{1+x^{2}}}{1+\frac{2x}{1+x^{2}}} \right )

=\log_{e}\left ( \frac{\frac{1+x^{2}-2x}{1+x^{2}}}{\frac{1+x^{2}+2x}{1+x^{2}}} \right )

=\log_{e}\frac{x^{2}-2x+1}{x^{2}+2x+1}=\log_{e}\left ( \frac{(x-1)^{2}}{(x+1)^{2}} \right )

=\log_{e}\left ( \frac{x-1}{x+1} \right )^{2}

=2\log_{e}\left ( \frac{x-1}{x+1} \right )\left | x \right |<1

=2f(x)

 

 


Option 1)

2f(x^{2})
 

Option 2)

2f(x)

Option 3)

-2f(x)

 

Option 4)

(f(x))^{2}

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