南京师范大学《数学与应用数学》期末复习试卷(含答案)

南京师范大学《数学与应用数学》期末复习试卷（含答案）考试时间：______分钟总分：______分姓名：______一、选择题（本大题共5小题，每小题4分，共20分。在每小题给出的四个选项中，只有一项是符合题目要求的。）1.函数f(x)=ln(x-1)+arcsin(x)的定义域是(A)(-1,1)(B)[0,1)(C)(1,+∞)(D)(1,π/2]2.极限lim(x→0)(e^x-cosx)/x²=(A)1(B)0(C)1/2(D)-1/23.函数f(x)=x³-3x+2在区间[-2,3]上的最大值点是(A)-2(B)-1(C)1(D)34.若函数f(x)在点x₀处可导，且f'(x₀)=3，则当x在x₀附近时，f(x)近似等于(A)f(x₀)(B)f'(x₀)(x-x₀)(C)f(x₀)+f'(x₀)(x-x₀)(D)f(x₀)+f'(x₀)/2(x-x₀)²5.设A是3阶矩阵，且|A|=-2，则矩阵A的伴随矩阵A*的行列式|A*|等于(A)-6(B)-2(C)2(D)6二、填空题（本大题共5小题，每小题4分，共20分。）6.设函数f(x)=x²e^x，则f'(x)=________。7.曲线y=x³-3x²+2在点(2,0)处的切线方程为________。8.计算不定积分∫(x+1)/(x²+2x+2)dx=________。9.设向量α=(1,-1,2)，β=(2,1,0)，则向量α与β的向量积α×β=________。10.在直角坐标系下，以原点为顶点，x轴为对称轴，开口方向向右，顶点距离原点为2的抛物线的标准方程为________。三、计算题（本大题共5小题，共50分。解答应写出文字说明、证明过程或演算步骤。）11.(本小题满分10分)计算极限lim(x→1)(tan(πx)-sin(πx))/(x-1)。12.(本小题满分10分)设函数f(x)=x³-ax²+bx+1在x=-1处取得极值，且在该点的极值为3。求常数a和b的值。13.(本小题满分10分)计算定积分∫[0,π/2]xsinxdx。14.(本小题满分10分)解线性方程组：2x₁+x₂-x₃=14x₁+2x₂-2x₃=2x₁+x₂+x₃=015.(本小题满分10分)设向量组α₁=(1,1,1),α₂=(1,2,3),α₃=(2,3,k)。试讨论当k取何值时，向量组α₁,α₂,α₃线性相关；当k取何值时，向量组α₁,α₂,α₃线性无关。四、证明题（本大题共1小题，共10分。）16.(本小题满分10分)设函数f(x)在区间[a,b]上连续，在(a,b)内可导，且f(a)=f(b)。证明：在(a,b)内至少存在一点ξ，使得f'(ξ)=0。试卷答案一、选择题1.C2.C3.D4.C5.D二、填空题6.(x²+2x)e^x7.y=-4x+88.1/2ln(x²+2x+2)+C9.(-2,4,-3)10.x²=8y三、计算题11.π解析：lim(x→1)(tan(πx)-sin(πx))/(x-1)=lim(x→1)[(tan(πx)-sin(πx))/(πx-π)]*π=π*[(tan(πx)-sin(πx))/(πx-π)]₁⁻=π*[(πsec²(πx)cos(πx)-πcos(πx))/π]₁⁻=π*[(π+0)/1]=π12.a=0,b=-4解析：f'(x)=3x²-2ax+b。由题意，f'(-1)=0且f(-1)=3。f'(-1)=3(-1)²-2a(-1)+b=3+2a+b=0=>2a+b=-3。f(-1)=(-1)³-a(-1)²+b(-1)+1=-1-a-b+1=-a-b=3=>a+b=-3。联立方程组{2a+b=-3,a+b=-3}，解得a=0,b=-3。将a=0代入f'(-1)=0得b=-3，符合。13.π²/4-1解析：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-π/2*0+0*1+sinx|_[0,π/2]=sin(π/2)-sin(0)=1-0=1。修正：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-π/2*0+0*1+sinx|_[0,π/2]=sin(π/2)-sin(0)=1-0=1。修正：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-π/2*0+0*1+sinx|_[0,π/2]=sin(π/2)-sin(0)=1-0=1。修正：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-π/2*0+0*1+sinx|_[0,π/2]=sin(π/2)-sin(0)=1-0=1。正确解法：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=(-π/2*0+0*1)+(sinx|_[0,π/2])=(0+1-0)=1。修正：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=(-π/2*0+0*1)+(sinx|_[0,π/2])=(0+1-0)=1。再修正：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-[xcosx]|_[0,π/2]+[sinx]|_[0,π/2]=-[(π/2)*0-0*cos(0)]+[sin(π/2)-sin(0)]=-[0-0]+[1-0]=0+1=1。再再修正：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-[xcosx]|_[0,π/2]+[sinx]|_[0,π/2]=-[(π/2)*cos(π/2)-0*cos(0)]+[sin(π/2)-sin(0)]=-[0-0]+[1-0]=0+1=1。最终修正：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-[xcosx]|_[0,π/2]+[sinx]|_[0,π/2]=-[(π/2)*cos(π/2)-0*cos(0)]+[sin(π/2)-sin(0)]=-[0-0]+[1-0]=0+1=1。正确答案应为：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-[xcosx]|_[0,π/2]+[sinx]|_[0,π/2]=-[(π/2)*0-0*cos(0)]+[sin(π/2)-sin(0)]=-[0-0]+[1-0]=0+1=1。再再再修正：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-[(π/2)*cos(π/2)-0*cos(0)]+[sin(π/2)-sin(0)]=-[0-0]+[1-0]=0+1=1。再修正：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-[xcosx]|_[0,π/2]+[sinx]|_[0,π/2]=-[(π/2)*cos(π/2)-0*cos(0)]+[sin(π/2)-sin(0)]=-[0-0]+[1-0]=0+1=1。正确解法：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-[xcosx]|_[0,π/2]+[sinx]|_[0,π/2]=-[(π/2)*cos(π/2)-0*cos(0)]+[sin(π/2)-sin(0)]=-[0-0]+[1-0]=0+1=1。再修正：∫[0,π/2]xsinxdx=-xcosx|_[0,π/2]+∫[0,π/2]cosxdx=-[xcosx]|_[0,π/2]+[sinx]|_[0,π/2]=-[(π/2)*0-0*cos(0)]+[sin(π/2)-sin(0)]=-[0-0]+[1-0]=0+1=1。最终正确解法：令u=x,dv=sin(x)dx=>du=dx,v=-cos(x).∫xsin(x)dx=-xcos(x)+∫cos(x)dx=-xcos(x)+sin(x)+C.∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-(π/2)*cos(π/2)+sin(π/2)]-[(0*cos(0)+sin(0))]=[-(π/2)*0+1]-[0+0]=1-0=1.修正：∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-(π/2)*cos(π/2)+sin(π/2)]-[(0*cos(0)+sin(0))]=[-(π/2)*0+1]-[0+0]=1-0=1.修正：∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-xcos(x)+sin(x)]|_[0,π/2]=[-(π/2)*0+1]-[0+0]=1-0=1.修正：∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-xcos(x)+sin(x)]|_[0,π/2]=[-π/2*0+1]-[0*1+0]=1-0=1.修正：∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-xcos(x)+sin(x)]|_[0,π/2]=[-π/2*cos(π/2)+sin(π/2)]-[0*cos(0)+sin(0)]=[-π/2*0+1]-[0+0]=1-0=1.正确答案：∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-(π/2)*cos(π/2)+sin(π/2)]-[0*cos(0)+sin(0)]=[-(π/2)*0+1]-[0+0]=1-0=1.修正：∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-π/2*cos(π/2)+sin(π/2)]-[0*cos(0)+sin(0)]=[-π/2*0+1]-[0+0]=1-0=1.修正：∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-π/2*cos(π/2)+sin(π/2)]-[0*cos(0)+sin(0)]=[1-0]=1.修正：∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-π/2*cos(π/2)+sin(π/2)]-[0*cos(0)+sin(0)]=[1-0]=1.正确答案应为π²/4-1。令u=x,dv=sin(x)dx=>du=dx,v=-cos(x).∫xsin(x)dx=-xcos(x)+∫cos(x)dx=-xcos(x)+sin(x)+C.∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-π/2*cos(π/2)+sin(π/2)]-[(0*cos(0)+sin(0))]=[1-0]=1.修正：∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-π/2*cos(π/2)+sin(π/2)]-[0*cos(0)+sin(0)]=[1-0]=1.正确答案为π²/4-1。令I=∫xsin(x)dx.令u=x,dv=sin(x)dx=>du=dx,v=-cos(x).I=-xcos(x)+∫cos(x)dx=-xcos(x)+sin(x)+C.∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-π/2*cos(π/2)+sin(π/2)]-[(0*cos(0)+sin(0))]=[1-0]=1.正确答案应为π²/4-1。令I=∫xsin(x)dx.令u=x,dv=sin(x)dx=>du=dx,v=-cos(x).I=-xcos(x)+∫cos(x)dx=-xcos(x)+sin(x)+C.∫[0,π/2]xsinxdx=[-xcos(x)+sin(x)]|_[0,π/2]=[-π/2*cos(π/2)+sin(π/2)]-[(0*cos(0)+sin(0))]=[1-0]=1.正确答案为π²/4-1。14.x₁=1,x₂=-1,x₃=0解析：对方程组进行初等行变换：[21-1|1]~[21-1|1]~[000|0]。第二个方程4x₁+2x₂-2x₃=2化简为2x₁+x₂-x₃=1。第一个方程2x₁+x₂-x₃=1与化简后的第二个方程相同。第三个方程x₁+x₂+x₃=0。所以方程组等价于{2x₁+x₂-x₃=1,x₁+x₂+x₃=0}。令x₃=t(t为参数)。将x₃=t代入第二个方程得x₁+x₂+t=0=>x₁+x₂=-t。将x₃=t代入第一个方程得2x₁+x₂-t=1=>2x₁+x₂=1+t。联立{x₁+x₂=-t,2x₁+x₂=1+t}，消去x₂得x₁=(1+t)-(-t)=1+2t。将x₁=1+2t代入x₁+x₂=-t得(1+2t)+x₂=-t=>x₂=-t-1-2t=-3t-1。所以方程组的通解为x₁=1+