考研数学二(D)a=0,b<0 \frac {1}{x-1})的斜渐近线方程是(A)y=x+(B)y=x+\frac {1}{e}(C)y=x(D)y=x- \frac {1}{e} 【答案】B(2)f(x)= \cases { \frac {1}{ \sqrt {1+x^{2}}},x \le 0 \cr(x+1)\cos x,x>} 一个原函数为(A)F(x)= \cases { \ln(\sqrt {1+x^{2}}-x),x \le 0 \cr(x+1)\cos x- \sin x,x>0}(B)F(x)= \cases { \ln(\sqrt {1+x^{2}}-x)+1,x \le 0 \cr(x+1)\cos x- \sin x,x>0}(C)F(x)= \cases { \ln(\sqrt {1+x^{2}}-x),x \le 0 \cr(x+1)\sin x+\cos x,x>0} 【答案】D(D)F(x)= \cases { \ln(\sqrt {1+x^{2}}+x)+1,x \le 0 \cr(x+1)\sin x+\cos x,x>0}(3)设数列\{ x_{n} \} , \{ y_{n} \} 满足x_{1}=y_{1}= \frac {1}{2},x_{n+1}= \sin x_{n},y_{n+1}=y_{n}^{2},(A)x,是y,的高阶无穷小(B)y,是x,的高阶无穷小(C)x,是y,的等价无穷小(D)x,是y,的同阶但非等价无穷小【答案】B(4)微分方程;y''+ay'+by=0)的解在((- \infty ,+\infty)有界,则a,b的取值范围为()(A)a<0,b>0(B)a>0,b>0(C)a=0,b>0 【答案】(C)(5)设函数y=f(x)\ln \cases {x=2t+\mid t \mid \cr y= \mid t \mid \sin t} 6.则((A)f(x)连续,f'(0)不存在(B)f'(0)存在,f'(x)在x=0 处不连续(C)f'(x)连续.f"(0)和(D)f"(0)存在, f''(x)在x=0 处连续f''(x)【答案】C(6)若函数f(a)= \int _{2}^{+\infty } \frac {1}{