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Question

If $$f(x)=\log \left (\dfrac{1+x}{1-x} \right )$$, then $$f \left( \dfrac{2x}{1+x^2} \right )$$ is equal to


A
(f(x))2
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B
(f(x))3
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C
2f(x)
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D
3f(x)
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Solution

The correct option is D $$2f(x)$$
$$f\left( x \right) =\log { \left(\cfrac { 1+x }{ 1-x } \right) } $$

$$ f\left(\cfrac { 2x }{ 1+{ x }^{ 2 } } \right)=\log { \left(\cfrac { 1+\cfrac { 2x }{ 1+{ x }^{ 2 } }  }{ 1-\cfrac { 2x }{ 1+{ x }^{ 2 } }  } \right) } $$

                         $$ =\log { \left(\cfrac { \cfrac { 1+{ x }^{ 2 }+2x }{ 1+{ x }^{ 2 } }  }{ \cfrac { 1+{ x }^{ 2 }-2x }{ 1+{ x }^{ 2 } }  } \right) } $$

                         $$ =\log  { \left(\cfrac { 1+x }{ 1-x } \right) } ^2$$

                         $$ = 2(f\left( x \right)) $$

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