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Question

If f(x)=loge(1x1+x),|x|<1, then f(2x1+x2) is equal to:

A
2f(x2)
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B
2f(x)
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C
2f(x)
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D
(f(x))2
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Solution

The correct option is C 2f(x)
As, f(x)=loge(1x1+x)
f(2x1+x2)=loge⎜ ⎜ ⎜12x1+x21+2x1+x2⎟ ⎟ ⎟=loge(1+x22x1+x2+2x)=ln(1x1+x)2=2f(x)

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