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.

List-$I$	List-$II$
$(A) \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$	$(I) \cos x-\sin x$
$(B) \tan ^{-1}\left(\frac{1-x}{1+x}\right)$	$(II) \frac{-1}{1+x^2}$
$(C) e^{\log (\sin x+\cos x)}$	$(III) \frac{2}{1+x^2}$
$(D) \sqrt{1-\sin 2 x} \text{ for } (0 < x < \frac{\pi}{4})$	$(IV) \cos x+\sin x$
	$(V) -\sin x-\cos x$

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