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If ƒ(x) = loge (1 - x/1 + x), |x| < 1, , then f(2x/1 + x^2) is equal to

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If ƒ(x) = loge (1 - x/1 + x), |x| < 1, , then f(2x/1 + x2)  is equal to

(1)  2ƒ(x)   (2) 2ƒ(x2)  (3)  (ƒ(x))2   (4) – 2ƒ(x)

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Correct option  (1)  2ƒ(x)

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