If f x -log1- x -1- x - then f 2 x -1- x 2 is equal toA- fx2B- f x 3C- 2 f x D- 3 f x

The correct option is C {2 f(x)} f(x) = log ( 1+x/1 x ) Then, f ( 2x/1+x^2 ) = log ( 1+2x/1+x^2/1 2x/1+x^2 ) =log ( 1+x^2+2x/1+x^2/1+x^2 2x/1+x^2 ) =log ( (1 ...

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Byju's Answer

Standard XII

Mathematics

General Tips for Choosing Function

If f x =log1+...

Question

If $f (x) = l o g (\frac{1 + x}{1 - x})$ , then $f (\frac{2 x}{1 + x^{2}})$ is equal to

${f (x)}^{2}$

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${f (x)}^{3}$

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${2 f (x)}$

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${3 f (x)}$

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Solution

The correct option is C
${2 f (x)}$

$f (x) = l o g (\frac{1 + x}{1 - x})$
Then, $f (\frac{2 x}{1 + x^{2}}) = l o g (\frac{1 + \frac{2 x}{1 + x^{2}}}{1 - \frac{2 x}{1 + x^{2}}})$
$= l o g ⎛ ⎜ ⎝ \frac{\frac{1 + x^{2} + 2 x}{1 + x^{2}}}{\frac{1 + x^{2} - 2 x}{1 + x^{2}}} ⎞ ⎟ ⎠$
$= l o g (\frac{(1 + x)^{2}}{(1 - x)^{2}})$
$= 2 l o g (\frac{1 + x}{1 - x})$
$= 2 (f (x))$

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If f(x)=log(1+x1−x), then f(2x1+x2) is equal to

The correct option is C {2f(x)} f(x)=log(1+x1−x) Then, f(2x1+x2)=log(1+2x1+x21−2x1+x2) =log⎛⎜⎝1+x2+2x1+x21+x2−2x1+x2⎞⎟⎠ =log((1+x)2(1−x)2) =2 log(1+x1−x) =2(f(x))

If $f (x) = l o g (\frac{1 + x}{1 - x})$ , then $f (\frac{2 x}{1 + x^{2}})$ is equal to