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一．填空题  One. Fill-in-the-blank questions

1．函数 y = 1 − x 2 + x + 2 y = 1 − x 2 + x + 2 y=(1)/(sqrt(-x^(2)+x+2))y=\frac{1}{\sqrt{-x^{2}+x+2}} 的定义域是 qquad\qquad
1 qquad\qquad . y = 1 − x 2 + x + 2 y = 1 − x 2 + x + 2 y=(1)/(sqrt(-x^(2)+x+2))y=\frac{1}{\sqrt{-x^{2}+x+2}} The domain of the function is
2．函数 y = x x − 4 y = x x − 4 y=sqrt((x)/(x-4))y=\sqrt{\frac{x}{x-4}} 定义域 qquad\qquad ．
2. Function y = x x − 4 y = x x − 4 y=sqrt((x)/(x-4))y=\sqrt{\frac{x}{x-4}} Define the field qquad\qquad .
3．函数 y = 3 x 3 y = 3 x 3 y=3x^(3)y=3 x^{3} 的凸区间是 qquad\qquad ．
3. y = 3 x 3 y = 3 x 3 y=3x^(3)y=3 x^{3} The convex interval of the function is qquad\qquad .
4．函数 f ( x ) = x 3 − 3 x 2 − 9 x + 5 f ( x ) = x 3 − 3 x 2 − 9 x + 5 f(x)=x^(3)-3x^(2)-9x+5f(x)=x^{3}-3 x^{2}-9 x+5 的最大值是 qquad\qquad －．
4. The f ( x ) = x 3 − 3 x 2 − 9 x + 5 f ( x ) = x 3 − 3 x 2 − 9 x + 5 f(x)=x^(3)-3x^(2)-9x+5f(x)=x^{3}-3 x^{2}-9 x+5 maximum value of the function is qquad\qquad -.
5．假定隐函数 y = x + 1 2 cos ⁡ y y = x + 1 2 cos ⁡ y y=x+(1)/(2)cos yy=x+\frac{1}{2} \cos y ，那么 d y d x = d y d x = (dy)/(dx)=\frac{d y}{d x}= qquad\qquad ．
5. Assuming the implicit function y = x + 1 2 cos ⁡ y y = x + 1 2 cos ⁡ y y=x+(1)/(2)cos yy=x+\frac{1}{2} \cos y , then d y d x = d y d x = (dy)/(dx)=\frac{d y}{d x}= qquad\qquad .
6．假定 f ( x ) = x cos ⁡ x f ( x ) = x cos ⁡ x f(x)=x cos xf(x)=x \cos x ，那么 f ′ ′ ( x ) = f ′ ′ ( x ) = f^('')(x)=f^{\prime \prime}(x)= qquad\qquad ．
6. Suppose f ( x ) = x cos ⁡ x f ( x ) = x cos ⁡ x f(x)=x cos xf(x)=x \cos x that then qquad\qquad . f ′ ′ ( x ) = f ′ ′ ( x ) = f^('')(x)=f^{\prime \prime}(x)=
7．若 y = f ( cos ⁡ x ) y = f ( cos ⁡ x ) y=f(cos x)y=f(\cos x) 且 f ( u ) f ( u ) f(u)f(u) 是可导的，那么 d y d x = d y d x = (dy)/(dx)=\frac{d y}{d x}= qquad\qquad ．
7. If y = f ( cos ⁡ x ) y = f ( cos ⁡ x ) y=f(cos x)y=f(\cos x) and f ( u ) f ( u ) f(u)f(u) is derivable, then d y d x = d y d x = (dy)/(dx)=\frac{d y}{d x}= qquad\qquad .
8．若 y = f ( tan ⁡ x ) y = f ( tan ⁡ x ) y=f(tan x)y=f(\tan x) 且 f ( u ) f ( u ) f(u)f(u) 是可导的，那么 d y d x = d y d x = (dy)/(dx)=\frac{d y}{d x}= qquad\qquad .
8. If y = f ( tan ⁡ x ) y = f ( tan ⁡ x ) y=f(tan x)y=f(\tan x) and f ( u ) f ( u ) f(u)f(u) is derivable, then d y d x = d y d x = (dy)/(dx)=\frac{d y}{d x}= qquad\qquad .
9．若 ∑ n = 1 ∞ 1 n p ∑ n = 1 ∞   1 n p sum_(n=1)^(oo)(1)/(n^(p))\sum_{n=1}^{\infty} \frac{1}{n^{p}} 收玫，则 p p pp 的取值范围 qquad\qquad ．
9. If the rose ∑ n = 1 ∞ 1 n p ∑ n = 1 ∞   1 n p sum_(n=1)^(oo)(1)/(n^(p))\sum_{n=1}^{\infty} \frac{1}{n^{p}} is harvested, then p p pp the value range qquad\qquad of .
10．若 ∑ n = 1 ∞ 1 n p ∑ n = 1 ∞   1 n p sum_(n=1)^(oo)(1)/(n^(p))\sum_{n=1}^{\infty} \frac{1}{n^{p}} 是发散的，则 p p pp 的取值范围 qquad\qquad ．
10. If it ∑ n = 1 ∞ 1 n p ∑ n = 1 ∞   1 n p sum_(n=1)^(oo)(1)/(n^(p))\sum_{n=1}^{\infty} \frac{1}{n^{p}} is divergent, then the range p p pp qquad\qquad of values of .
11．函数 1 2 + x 1 2 + x (1)/(2+x)\frac{1}{2+x} 在 x = 0 x = 0 x=0x=0 处可幂级数展开 则 1 2 + x = 1 2 + x = (1)/(2+x)=\frac{1}{2+x}= qquad\qquad .
11. The function 1 2 + x 1 2 + x (1)/(2+x)\frac{1}{2+x} can be expanded x = 0 x = 0 x=0x=0 by a power series 1 2 + x = 1 2 + x = (1)/(2+x)=\frac{1}{2+x}= qquad\qquad at .
12．函数 1 1 + x 1 1 + x (1)/(1+x)\frac{1}{1+x} 在 x = 0 x = 0 x=0x=0 处可幂级数展开．那么 1 1 + x = 1 1 + x = (1)/(1+x)=\frac{1}{1+x}= qquad\qquad ．
12. The function 1 1 + x 1 1 + x (1)/(1+x)\frac{1}{1+x} can be expanded x = 0 x = 0 x=0x=0 by a power series at . 1 1 + x = 1 1 + x = (1)/(1+x)=\frac{1}{1+x}= Then qquad\qquad .

二．选择题  Two. Multiple choice questions

1．假定 f ( x ) = { ( x − 1 ) 2 − 1 < x ≤ 0 x + 1 , 0 < x ≤ 1 f ( x ) = ( x − 1 ) 2      − 1 < x ≤ 0 x + 1 ,      0 < x ≤ 1 f(x)={[(x-1)^(2),-1 < x <= 0],[x+1",",0 < x <= 1]:}f(x)=\left\{\begin{array}{ll}(x-1)^{2} & -1<x \leq 0 \\ x+1, & 0<x \leq 1\end{array}\right. ，那么 lim x → 0 f ( x ) = ( ) lim x → 0   f ( x ) = ( ) lim_(x rarr0)f(x)=(quad)\lim _{x \rightarrow 0} f(x)=(\quad) ．
1. Suppose f ( x ) = { ( x − 1 ) 2 − 1 < x ≤ 0 x + 1 , 0 < x ≤ 1 f ( x ) = ( x − 1 ) 2      − 1 < x ≤ 0 x + 1 ,      0 < x ≤ 1 f(x)={[(x-1)^(2),-1 < x <= 0],[x+1",",0 < x <= 1]:}f(x)=\left\{\begin{array}{ll}(x-1)^{2} & -1<x \leq 0 \\ x+1, & 0<x \leq 1\end{array}\right. that then lim x → 0 f ( x ) = ( ) lim x → 0   f ( x ) = ( ) lim_(x rarr0)f(x)=(quad)\lim _{x \rightarrow 0} f(x)=(\quad) .
A． 1
B． 0
C．-1
D．不存在  D. Non-existent
2．函数 f ( x ) = | x − 1 | ( ) f ( x ) = | x − 1 | ( ) f(x)=|x-1|quad(quad)f(x)=|x-1| \quad(\quad) ．  2. Functions f ( x ) = | x − 1 | ( ) f ( x ) = | x − 1 | ( ) f(x)=|x-1|quad(quad)f(x)=|x-1| \quad(\quad) .
A．在 x = 1 x = 1 x=1x=1 处连续且可导  A. Continuous and directable x = 1 x = 1 x=1x=1 at
B．在 x = 1 x = 1 x=1x=1 处不连续  B. Discontinuity x = 1 x = 1 x=1x=1 at
C．在 x = 0 x = 0 x=0x=0 处连续且可导  C. Continuous and conductible x = 0 x = 0 x=0x=0 at
D．在 x = 0 x = 0 x=0x=0 处不连续  D. Discontinuity x = 0 x = 0 x=0x=0 at
3 下列哪个函数在［－1，1］满足罗尔定理．（ ）
3 Which of the following functions satisfies Roll's theorem at [-1,1].( )
A． f ( x ) = 1 x 2 f ( x ) = 1 x 2 f(x)=(1)/(x^(2))f(x)=\frac{1}{x^{2}}
B． f ( x ) = | x | f ( x ) = | x | f(x)=|x|f(x)=|x|
C． f ( x ) = 1 − x 2 f ( x ) = 1 − x 2 f(x)=1-x^(2)f(x)=1-x^{2}
D． f ( x ) = x 2 − 2 x − 1 f ( x ) = x 2 − 2 x − 1 f(x)=x^(2)-2x-1f(x)=x^{2}-2 x-1
4．当 x → ∞ x → ∞ x rarr oox \rightarrow \infty ，下列有极限的是？（）
4. When x → ∞ x → ∞ x rarr oox \rightarrow \infty , what are the following limits? （）
A， x sin ⁡ 1 x x sin ⁡ 1 x x sin((1)/(x))x \sin \frac{1}{x}
B， sin ⁡ 1 x 2 sin ⁡ 1 x 2 sin((1)/(x^(2)))\sin \frac{1}{x^{2}}
C， x + 1 x 2 − x x + 1 x 2 − x (x+1)/(x^(2)-x)\frac{x+1}{x^{2}-x}
D， cot ⁡ x cot ⁡ x cot x\cot x
5． ∫ 0 + ∞ 1 1 + x 2 d x = ( ) ∫ 0 + ∞   1 1 + x 2 d x = ( ) int_(0)^(+oo)(1)/(1+x^(2))dx=(quad)\int_{0}^{+\infty} \frac{1}{1+x^{2}} d x=(\quad) ．
A． 1 π 1 π (1)/(pi)\frac{1}{\pi}
B． π 2 π 2 (pi)/(2)\frac{\pi}{2}
C． 1
D． 4 π 4 π (4)/(pi)\frac{4}{\pi}
6． ∫ 1 1 − 2 x d x = ( ) ∫ 1 1 − 2 x d x = ( ) int(1)/(1-2x)dx=(quad)\int \frac{1}{1-2 x} d x=(\quad) ．
A． − 1 2 In | 1 − 2 x | + C − 1 2 In | 1 − 2 x | + C -(1)/(2)In|1-2x|+C-\frac{1}{2} \operatorname{In}|1-2 x|+C
B．In | 1 − 2 x | + C | 1 − 2 x | + C |1-2x|+C|1-2 x|+C  B． In | 1 − 2 x | + C | 1 − 2 x | + C |1-2x|+C|1-2 x|+C
C． − 1 2 In | 1 − x | + C − 1 2 In | 1 − x | + C -(1)/(2)In|1-x|+C-\frac{1}{2} \operatorname{In}|1-x|+C
D． − 1 2 In | 1 − 2 x | − 1 2 In | 1 − 2 x | -(1)/(2)In|1-2x|-\frac{1}{2} \operatorname{In}|1-2 x|
7． ∫ x e − x 2 d x = ( ) ∫ x e − x 2 d x = ( ) int xe^(-x^(2))dx=(quad)\int x e^{-x^{2}} d x=(\quad) ．
A． − 1 2 e − x 2 + C − 1 2 e − x 2 + C -(1)/(2)e^(-x^(2))+C-\frac{1}{2} e^{-x^{2}}+C
B． − 2 e − x 2 + C − 2 e − x 2 + C -2e^(-x^(2))+C-2 e^{-x^{2}}+C
C． 1 2 e − x 2 + C 1 2 e − x 2 + C (1)/(2)e^(-x^(2))+C\frac{1}{2} e^{-x^{2}}+C
D． e − x 2 + C e − x 2 + C e^(-x^(2))+Ce^{-x^{2}}+C
8．下列哪个级数是收敛（ ）  8. Which of the following series is convergence ( )
A． ∑ n = 1 ∞ 2 1 n ∑ n = 1 ∞   2 1 n sum_(n=1)^(oo)2^((1)/(n))\sum_{n=1}^{\infty} 2^{\frac{1}{n}}
B． ∑ n = 1 ∞ ( n + 1 − n ) ∑ n = 1 ∞   ( n + 1 − n ) sum_(n=1)^(oo)(sqrt(n+1)-sqrtn)\sum_{n=1}^{\infty}(\sqrt{n+1}-\sqrt{n})
C． ∑ n = 1 ∞ [ ( 3 4 ) n + 1 n + 1 ] ∑ n = 1 ∞   3 4 n + 1 n + 1 sum_(n=1)^(oo)[((3)/(4))^(n)+(1)/(sqrt(n+1))]\sum_{n=1}^{\infty}\left[\left(\frac{3}{4}\right)^{n}+\frac{1}{\sqrt{n+1}}\right]
D． ∑ n = 1 ∞ 1 3 n 3 ∑ n = 1 ∞   1 3 n 3 sum_(n=1)^(oo)(1)/(3n^(3))\sum_{n=1}^{\infty} \frac{1}{3 n^{3}}
9．下列哪个级数是发散的（ ）。  9. Which of the following series is divergent ( ).
A． ∑ n = 1 ∞ ( 1 − cos ⁡ π n ) ∑ n = 1 ∞   1 − cos ⁡ π n sum_(n=1)^(oo)(1-cos((pi )/(n)))\sum_{n=1}^{\infty}\left(1-\cos \frac{\pi}{n}\right)
B． ∑ n = 1 ∞ 1 n 2 ∑ n = 1 ∞   1 n 2 sum_(n=1)^(oo)(1)/(n^(2))\sum_{n=1}^{\infty} \frac{1}{n^{2}}
C． ∑ n = 2 ∞ sin ⁡ 1 n ∑ n = 2 ∞   sin ⁡ 1 n sum_(n=2)^(oo)sin((1)/(sqrtn))\sum_{n=2}^{\infty} \sin \frac{1}{\sqrt{n}}
D． ∑ n = 1 ∞ 1 n ! ∑ n = 1 ∞   1 n ! sum_(n=1)^(oo)(1)/(n!)\sum_{n=1}^{\infty} \frac{1}{n!} ；
10．级数 ∑ n = 0 ∞ ( − 1 ) n 1 n + 1 ∑ n = 0 ∞   ( − 1 ) n 1 n + 1 sum_(n=0)^(oo)(-1)^(n)(1)/(n+1)\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{n+1} 是（ ）
10. Series ∑ n = 0 ∞ ( − 1 ) n 1 n + 1 ∑ n = 0 ∞   ( − 1 ) n 1 n + 1 sum_(n=0)^(oo)(-1)^(n)(1)/(n+1)\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{n+1} Yes ( )
A．绝对收玫  A. Absolute Harvest
B．条件收敛  B. Conditional convergence
C．发散  C. Divergence
D．收敛性不能确定  D. Convergence cannot be determined
1．计算 lim x → 1 x 2 − 1 x 2 + 2 x − 3 lim x → 1   x 2 − 1 x 2 + 2 x − 3 lim_(x rarr1)(x^(2)-1)/(x^(2)+2x-3)\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{2}+2 x-3} ．  1. Calculation lim x → 1 x 2 − 1 x 2 + 2 x − 3 lim x → 1   x 2 − 1 x 2 + 2 x − 3 lim_(x rarr1)(x^(2)-1)/(x^(2)+2x-3)\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{2}+2 x-3} .
2．计算 lim x → ∞ 2 x 3 + 3 x 2 + 5 7 x 3 + 4 x 2 − 1 lim x → ∞   2 x 3 + 3 x 2 + 5 7 x 3 + 4 x 2 − 1 lim_(x rarr oo)(2x^(3)+3x^(2)+5)/(7x^(3)+4x^(2)-1)\lim _{x \rightarrow \infty} \frac{2 x^{3}+3 x^{2}+5}{7 x^{3}+4 x^{2}-1} ．  2. Calculation lim x → ∞ 2 x 3 + 3 x 2 + 5 7 x 3 + 4 x 2 − 1 lim x → ∞   2 x 3 + 3 x 2 + 5 7 x 3 + 4 x 2 − 1 lim_(x rarr oo)(2x^(3)+3x^(2)+5)/(7x^(3)+4x^(2)-1)\lim _{x \rightarrow \infty} \frac{2 x^{3}+3 x^{2}+5}{7 x^{3}+4 x^{2}-1} .
3．若 y = e x cos ⁡ x + x 1 − x y = e x cos ⁡ x + x 1 − x y=e^(x)cos x+(x)/(1-x)y=e^{x} \cos x+\frac{x}{1-x} ，则 y ′ y ′ y^(')y^{\prime} ．
3. If y = e x cos ⁡ x + x 1 − x y = e x cos ⁡ x + x 1 − x y=e^(x)cos x+(x)/(1-x)y=e^{x} \cos x+\frac{x}{1-x} , then y ′ y ′ y^(')y^{\prime} .
4．若 y = e x cos ⁡ x + x 1 − x y = e x cos ⁡ x + x 1 − x y=e^(x)cos x+(x)/(1-x)y=e^{x} \cos x+\frac{x}{1-x}