1．设$x = 0.03,y = 2\ln 1.01,z = \ln 1.1$，则x，y，z的大小关系为（ ）
A．$z > x > y$ B．$x > y > z$
C．$x > z > y$ D．$z > y > x$
解 由$x = \frac{3}{100}$，$y = 2\ln(1 + \frac{1}{100})$，$z = \ln(1 + \frac{1}{10})$，
若$t = \frac{1}{10}$，则$x = 3t^{2}$，$y = 2\ln(1 + t^{2})$，$z = \ln(1 + t)$，
令$f(t) = y - x = 2\ln(1 + t^{2}) - 3t^{2}$且$0 < t < 1$，则$f'(t) = \frac{4t}{1 + t^{2}} - 6t = \frac{- 2t(1 + 3t^{2})}{1 + t^{2}} < 0$，
所以$f(t)$在$(0,1)$上递减，故$f(t) < f(0) = 0$，即$y < x$，
令$g(t) = z - x = \ln(1 + t) - 3t^{2}$且$0 < t < 1$，则$g'(t) = \frac{1}{1 + t} - 6t$在$(0,1)$上递减，
若$g'(t) = 0$，则$\frac{1}{1 + t} = 6t$，可得$t = \frac{\sqrt{15} - 3}{6}$，故$(0,\frac{\sqrt{15} - 3}{6})$上$g'(t) > 0$，$g(t)$递增，
而$0 < \frac{1}{10} < \frac{\sqrt{15} - 3}{6}$，且在$(0,\frac{\sqrt{15} - 3}{6})$上$g(t) > g(0) = 0$，
所以$z - x > 0$，即$z > x$，
综上，$z > x > y$. 故选：A
2．已知$a = \frac{2}{21}$，$b = \ln 1.1$，$c = \frac{21}{220}$，则（ ）
A．$b < a < c$ B．$a < b < c$
C．$c < b < a$ D．$c < a < b$
解 易知$a = \frac{2}{21} = 2 \times \frac{1.1 - 1}{1.1 + 1}$，$c = \frac{21}{220} = \frac{1}{2} \times \left( 1.1 - \frac{1}{1.1} \right)$，
设$f(x) = \ln x - \frac{2(x - 1)}{x + 1}(x > 1)$，则$f'(x) = \frac{(x - 1)^{2}}{x(x + 1)^{2}} > 0$，所以$f(x)$单调递增，
所以$f(x) > f(1) = 0$，故$f(1.1) = \ln 1.1 - \frac{2(1.1 - 1)}{1.1 + 1} > 0$，
所以$b > a$
设$g(x) = \ln x - \frac{1}{2}\left( x - \frac{1}{x} \right)(x > 1)$，则$g'(x) = \frac{- (x - 1)^{2}}{2x^{2}} < 0$，所以$g(x)$单调递减，
所以$g(x) < g(1) = 0$，即$g(1.1) = \ln 1.1 - \frac{1}{2}\left( 1.1 - \frac{1}{1.1} \right) < 0$，所以$b < c$，
综上，$a < b < c$.故选：B