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`log_(e)(e-1)``log_(e)(e+1)``log_(e)e`none of these

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Clearly, f(x) is continuous on [0,1] and differentiable on (0,1). Therefore, therer exists `c in (0,1)` such that <br> `f'(c)=(f(1)-f(0))/(1-0)rArr e^(c)=e-1rArr c=log_(e) (e-1)`**Revision of limits**

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