For all x in (0, 1),which of the following in | vedclass

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(B) Consider the function $f(x) = x - \log_e(1 + x)$ for $x \in (0, 1)$.Taking the derivative,we get $f'(x) = 1 - \frac{1}{1 + x} = \frac{1 + x - 1}{1

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$\log_e(1 + x) < x$

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(B) Consider the function $f(x) = x - \log_e(1 + x)$ for $x \in (0, 1)$.Taking the derivative,we get $f'(x) = 1 - \frac{1}{1 + x} = \frac{1 + x - 1}{1 + x} = \frac{x}{1 + x}$.Since $x > 0$,$f'(x) > 0$ for all $x \in (0, 1)$.This implies that $f(x)$ is a strictly increasing function on $(0, 1)$.Since $f(0) = 0 - \log_e(1) = 0$,and $f(x)$ is strictly increasing,$f(x) > f(0)$ for all $x > 0$.Therefore,$x - \log_e(1 + x) > 0$,which means $\log_e(1 + x) Thus,option $(b)$ is correct.

For all $x \in (0, 1)$,which of the following inequalities is true?

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