Which of the following is not true?

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(D) Let $f(x) = \log(1 + x) - x$. Then $f'(x) = \frac{1}{1 + x} - 1 = \frac{-x}{1 + x}$.For $x > 0$,$f'(x) < 0$,so $f(x)$ is strictly decreasing.Since

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$e^x  0$

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(D) Let $f(x) = \log(1 + x) - x$. Then $f'(x) = \frac{1}{1 + x} - 1 = \frac{-x}{1 + x}$.For $x > 0$,$f'(x) Since $f(0) = 0$,$f(x)  0$,hence $\log(1 + x) Let $g(x) = \log(1 + x) - \frac{x}{1 + x}$. Then $g'(x) = \frac{1}{1 + x} - \frac{(1 + x) - x}{(1 + x)^2} = \frac{1}{1 + x} - \frac{1}{(1 + x)^2} = \frac{x}{(1 + x)^2}$.For $x > 0$,$g'(x) > 0$,so $g(x)$ is strictly increasing.Since $g(0) = 0$,$g(x) > 0$ for $x > 0$,hence $\frac{x}{1 + x} Using the Taylor series expansion,$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$.For $x > 0$,all terms $\frac{x^n}{n!}$ are positive,so $e^x > 1 + x$ is true.For $x > 0$,$e^x > 1$ and $1 - x  1 - x$ is always true.Therefore,the statement $e^x < 1 - x$ is not true.

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