If f(x1)−f(x2)=f(x1−x21−x1x2) for x1,x2ϵ (-1, 1), then f(x) is
A
log1−x1+x
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B
log2+x1−x
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C
tan−11−x1+x
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D
tan−11+x1−x
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Solution
The correct option is Alog1−x1+x We have, f(x)=log1−x1+x f(x1)−f(x2)=log1−x11+x1−log1−x21+x2 =log(1−x1)(1+x2)1+x1−x2−x1x2
Also, f(x1−x21−x1x2)=log1−x1−x21−x1x21+x1−x21−x1x2 =log1−x1x2−x1+x21−x1x2+x1−x2 ∴ (a) is true