Which of the following is not true?

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(C) Let $f(x) = x - \log(1 + x)$. For $x > 0$,$f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x} > 0$. Thus,$f(x) > f(0) = 0$,so $\log(1 + x) < x$ is true.Let

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$e^{-x}  0$

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(C) Let $f(x) = x - \log(1 + x)$. For $x > 0$,$f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x} > 0$. Thus,$f(x) > f(0) = 0$,so $\log(1 + x) Let $g(x) = \log(1 + x) - \frac{x}{1+x}$. For $x > 0$,$g'(x) = \frac{1}{1+x} - \frac{(1+x) - x}{(1+x)^2} = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{x}{(1+x)^2} > 0$. Thus,$g(x) > g(0) = 0$,so $\frac{x}{1+x} Let $h(x) = e^{-x} - (1 - x)$. For $x > 0$,$h'(x) = -e^{-x} + 1$. Since $e^{-x}  0$,$h'(x) > 0$. Thus,$h(x) > h(0) = 0$,which implies $e^{-x} > 1 - x$. Therefore,the statement $e^{-x} < 1 - x$ is false.

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