1．若$a = \ln 1.01,b = \frac{2}{201},c = \sqrt{1.02} - 1$，则（ ）
A．$a < b < c$ B．$b < a < c$
C．$\text{b} < \text{c} < \text{a}$ D．$\text{c} < \text{a} < \text{b}$
解 由于$a = \ln 1.01 = \ln(1 + 0.01),c = \sqrt{1.02} - 1 = \sqrt{1 + 0.01 \times 2} - 1$，
故设函数$f(x) = \ln(1 + x) - \sqrt{1 + 2x} + 1,(x > 0)$ ，
则$f'(x) = \frac{1}{1 + x} - \frac{1}{\sqrt{1 + 2x}} = \frac{\sqrt{1 + 2x} - (1 + x)}{(1 + x)\sqrt{1 + 2x}}$，$x > 0$，
由于${(\sqrt{1 + 2x})}^{2} - {(1 + x)}^{2} = - x^{2} < 0$，所以${(\sqrt{1 + 2x})}^{2} < {(1 + x)}^{2}$，
即$\sqrt{1 + 2x} - (1 + x) < 0$，即$f'(x) < 0$，
故$f(x) = \ln(1 + x) - \sqrt{1 + 2x} + 1,(x > 0)$为单调递减函数，
故$f(x) < f(0) = 0$，即$\ln(1 + x) < \sqrt{1 + 2x} - 1,(x > 0)$，
令$x = 0.01$，则$\ln(1 + 0.01) < \sqrt{1 + 2 \times 0.01} - 1$，即$a < c$；
又$a = \ln 1.01 = \ln(1 + 0.01),b = \frac{2}{201} = \frac{2 \times 0.01}{2 + 0.01}$，
令$g(x) = \ln(x + 1) - \frac{2x}{2 + x},(x > 0)$，
则$g'(x) = \frac{1}{x + 1} - \frac{4}{{(x + 2)}^{2}} = \frac{x^{2}}{(x + 1){(x + 2)}^{2}} > 0,(x > 0)$，
即$g(x) = \ln(x + 1) - \frac{2x}{2 + x},(x > 0)$为单调递增函数，
故$g(x) > g(0) = 0$，即$\ln(x + 1) > \frac{2x}{2 + x},(x > 0)$，
令$x = 0.01$，则$\ln 1.01 > \frac{2 \times 0.01}{2 + 0.01} = \frac{2}{201}$，即$a > b$，
故$b < a < c$，故选：B
2．已知$a = \frac{1}{11}$，$b = \sqrt{\frac{1}{10}}$，$c = \ln\frac{11}{10}$．则（ ）
A．$a > b > c$ B．$b > c > a$ C．$c > b > a$ D．$b > a > c$
解令$f(x) = x - \ln(x + 1)$，$0 < x < 1$，则$f'(x) = 1 - \frac{1}{x + 1} = \frac{x}{x + 1} > 0$，
$\therefore f(x)$在$(0,1)$上单调递增，$\therefore f(x) > f(0) = 0$，即$x > \ln(x + 1)$；
令$g(x) = \ln(x + 1) - 1 + \frac{1}{x + 1}$，$0 < x < 1$，则$g'(x) = \frac{1}{x + 1} - \frac{1}{(x + 1)^{2}} = \frac{x}{(x + 1)^{2}} > 0$，
$\therefore g(x)$在$(0,1)$上单调递增，$\therefore g(x) > g(0) = 0$，即$\ln(x + 1) > 1 - \frac{1}{x + 1}$；
又当$0 < x < 1$时，$\sqrt{x} > x$，$\therefore$当$0 < x < 1$时，$\sqrt{x} > x > \ln(x + 1) > 1 - \frac{1}{x + 1}$；
则当$x = \frac{1}{10}$时，$\sqrt{\frac{1}{10}} > \frac{1}{10} > \ln\frac{11}{10} > \frac{1}{11}$，即$b > c > a$.故选：B.