🏖.2021汤家凤高数强化测试10套卷第一套强化测试卷第一.(3)题 

(3)已知f(x)再点(0,0)处的曲率圆为(x-1)²+(y-1)²=2,则\(\lim\limits_{x\to0}\frac{f(x)+x}{(1+x)^x-1}\)=()
A.2
B.\(\frac{1}{2}\)
C.1
D.-1
C

C,因为曲率圆(x-1)²+(y-1)²=2在(0,0)处切线斜率为-1,所以f(0)=0,f'(0)=-1,显然曲率半径R=\(\sqrt{2}\),

由曲率公式得

\(k=\frac{\left|f''(0)\right|}{[1+f'^2(0)]^\frac{3}{2}}=\frac{f''(0)}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\)

解得f''(0)=2,

故\(\lim\limits_{x\to0}\frac{f(x)+x}{(1+x)^x-1}\)=\(\lim\limits_{x\to0}\frac{f(x)+x}{e^{xln(1+x)}-1}\)=\(\lim\limits_{x\to0}\frac{f(x)+x}{xln(1+x)}\)=\(\lim\limits_{x\to0}\frac{f(x)+x}{x^2}\)=\(\lim\limits_{x\to0}\frac{f'(x)+1}{2x}\)=\(\frac{1}{2}\lim\limits_{x\to0}\frac{f'(x)-f(0)}{x}\)=\(\frac{1}{2}f''(0)=1\)

应选C

🏖.2021汤家凤高数强化测试10套卷第二套强化测试卷第一.(2)题 

(2)方程\(ae^x=x^2-x+1\)有三个实根,则a的取值范围为()
A.0<a<\(\frac{3}{e^2}\)
B.\(\frac{1}{e}<a<\frac{3}{e^2}\)
C.\(a<\frac{1}{e^2}\)                    
D.\(a>\frac{3}{e^2}\)
B

解:\(ae^x=x^2-x+1\)等价于\((x^2-x+1)e^{-x}-a=0\),

设\(f(x)=(x^2-x+1)e^{-x}-a(-\infty<x<+\infty)\),

令\(f'(x)=-(x-1)(x-2)e^{-x}=0,\)得x=1,x=2,

当x<1时,f'(x)<0,当1<x<2时,f'(x)>0,当x>2时,f'(x)<0,

则x=1为极小值,极小值点为\(f(1)=\frac{1}{e}-a\);x=2为极大值点,极大值为\(f(2)=\frac{3}{e^2}-a\),又\(\lim\limits_{x\to-\infty}f(x)=+\infty\),\(\lim\limits_{x\to+\infty}f(x)=-a\),

当\(\frac{1}{e}-a<0,\frac{3}{e^2}-a>0,-a<0,\)即,\(\frac{1}{e}<a<\frac{3}{e^2}\)时,f(x)恰有三个零点,从而原方程恰有三个根,应选B.

🏖.2021汤家凤高数强化测试10套卷第二套强化测试卷第二.(12)题 

12.设\(F(x,y,z)=x+y+z\),其中F连续可偏导且\((F_x-1)\cdot(F_y-1)\cdot(F_z-1)\not=0,\)则\(\frac{\partial x}{\partial y}+\frac{\partial y}{\partial z}+\frac{\partial z}{\partial x}\)=_____

解:因为\(F_x\not=1,\)所以由\(F(x,y,z)=x+y+z\)可以确定x为y,z得函数,\(F(x,y,z)=x+y+z\)两边对y求偏导可得

\(F_x\cdot\frac{\partial x}{\partial y}+F_y=\frac{\partial x}{\partial y}+1,\)

解得\(\frac{\partial x}{\partial y}=\frac{1-F_y}{F_x-1}\),同理可得\(\frac{\partial y}{\partial z}=\frac{1-F_z}{F_y-1},\frac{\partial z}{\partial x}=\frac{1-F_x}{F_z-1}\),则

原式=\(\frac{1-F_y}{F_x-1}+\frac{1-F_z}{F_y-1}+\frac{1-F_x}{F_z-1}\)

 🏖.2021汤家凤高数强化测试10套卷第二套强化测试卷第三.(20)题 

设\(f'(x)\in C[0,2]\),证明\(|f(x)|\le\left|\frac{1}{2}\int_{0}^{2}f(x)dx\right|+\int_{0}^{2}|f'(x)|dx\)

由积分中值定理,存在\(C\in[0,2]\),使得\(\int_{0}^{2}f(x)dx=2f(c)\)即\(\frac{1}{2}\int_{0}^{2}f(x)dx=f(c)\);

由牛顿莱布尼兹公式得

\(f(2)-f(c)=\int_{c}^{2}f(x)dx,\)或\(f(2)=f(c)+\int_{c}^{2}f'(x)dx\),

故\(|f(2)|\le|f(c)|+\left|\int_{c}^{2}f'(x)dx\right|\le\left|\frac{1}{2}\int_{0}^{2}f(x)dx\right|+\int_{0}^{2}|f'(x)|dx\)

 🏖.2021汤家凤高数强化测试10套卷第三套强化测试卷第二(10)题  

10.设y=y(x)由\(\begin{cases} x=t^2+2t,\\ \frac{d^2y}{dt^2}-y=2e^t\\ \end{cases}\)确定,且\(y(0)=0,y'(0)=-1,\)则\(\frac{d^2y}{dx^2}|_{t=0}=\)___________

方程\(\frac{d^2y}{dt^2}-y=0\)得特征方程为\(\lambda^2-1=0\),特征值为\(\lambda_1=-1,\lambda_2=1,\)

方程\(\frac{d^2y}{dt^2}-y=0\)的通解为\(y=C_1e^{-t}+C_2e^t;\)

设方程\(\frac{d^2y}{dt^2}-y=2e^t\)的通解为\(y=ate^t,\)代入得a=1,

即\(y=te^t\)为特解,从而原方程通解为

\(y=C_1e^{-t}+C_2e^t+te^t\),

由\(y|_{t=0}=0,y'|_{t=0}=-1\)得\(C_1=1,C_2=-1,\)故\(y=e^{-t}-e^t+te^t\)

\(y'=\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-e^{-t}-e^t+(t+1)e^t}{2t+2}=\frac{-e^{-t}+te^t}{2t+2},\)

\(\frac{d^2y}{dx^2}=\frac{d\left(\frac{dy}{dx}\right)/dt}{dx/dt}=\frac{\left(\frac{-e^{-t}+te^t}{2t+2}\right)'}{2t+2}\)

\(=\frac{[e^{-t}+(t+1)e^t](2t+2)-2(-e^{-t}+te^t))}{(2t+2)^3}\)

故\(\frac{d^2y}{dx^2}|_{t=0}\)\(=\frac{[e^{-t}+(t+1)e^t](2t+2)-2(-e^{-t}+te^t))}{(2t+2)^3}|_{t=0}=\frac{3}{4}.\)

 

 🏖.2021汤家凤高数强化测试10套卷第四套强化测试卷第一(7)题  

7.设A为3阶可逆矩阵,将A的第1行的4倍加到第3行得B,下列结论正确得是()        
A.A*的第3列的4倍加到第1列得B*
B.A*得第3列得-4倍加到第1列得B*
C.A*得第1行得4倍加到第3行得B*
D.A*得第1行得-4倍加到第3行得B*
B

由\(B=\begin{pmatrix} 1 &0&0\\ 0&1&0\\ 4&0&1 \end{pmatrix}A\)得|B|=|A|,且

\(B^*=|B||B|^{-1}=|A|A^{-1} \begin{pmatrix} 1 &0&0\\ 0&1&0\\ 4&0&1 \end{pmatrix}\)

\(=A^*\begin{pmatrix} 1 &0&0\\ 0&1&0\\ -4&0&1 \end{pmatrix}\)

即A*得第3列得-4倍加到第1列得B*,应选B.

 🏖.2021汤家凤高数强化测试10套卷第四套强化测试卷第三(15)题  

(15)设函数y=f(x)二阶可导,且\(f''(x)>0,f(0)=0,f'(0)=0,\)求\(\lim_\limits{x\to0}\frac{x^3f(u)}{f(x)sin^3u},\)其中u是曲线\(y=f(x)\)在点\(P(x,f(x))\)处的切线在x轴上的截距

15.设点P处得到切线为\(Y-f(x)=f'(x)(X-x)\)

令Y=0,则切线在x轴上截距为\(u=x-\frac{f(x)}{f'(x)}\)

\(\lim_\limits{x\to0}\frac{u}{x}=1-\lim_\limits{x\to0}\frac{f(x)}{xf'(x)}\)

\(=1-\lim_\limits{x\to0}\frac{1}{\frac{f'(x)-f'(0)}{x}}\cdot\frac{f(x)}{x^2}\)

\(=1-\frac{1}{f''(0)}\lim_\limits{x\to0}\frac{f(x)}{x^2}=1-\frac{1}{2f''(0)}\lim_\limits{x\to0}\frac{f'(x)}{x}\)

\(=1-\frac{1}{2f''(0)}​​\lim_\limits{x\to0}\frac{f'(x)-f'(0)}{x}=\frac{1}{2},\)

\(\lim_\limits{x\to0}\frac{x^3f(u)}{f(x)sin^3u}=\lim_\limits{x\to0}\frac{x^3f(u)}{f(x)u^3}=\lim_\limits{x\to0}\frac{x}{u}\cdot\frac{\frac{f(u)}{u^2}}{\frac{f(x)}{x^2}}\)

因为\(\lim_\limits{x\to0}\frac{f(x)}{x^2}=\frac{1}{2}\lim_\limits{x\to0}\frac{f'(x)-f'(0)}{x}=\frac{1}{2}f''(0),\)

\(\lim_\limits{x\to0}\frac{f(u)}{u^2}=\frac{1}{2}\lim_\limits{u\to0}\frac{f'(u)-f'(0)}{u}=\frac{1}{2}f''(0),\)

所以\(\lim_\limits{x\to0}\frac{x^3f(u)}{f(x)sin^3u}=\lim_\limits{x\to0}\frac{x}{u}\cdot\frac{\frac{f(u)}{u^2}}{\frac{f(x)}{x^2}}=\lim_\limits{x\to0}\frac{x}{u}=2.\)

 

 🏖.2021汤家凤高数强化测试10套卷第五套强化测试卷第一(8)题  

8.设A为3阶矩阵,非齐次线性方程组\(AX=\begin{pmatrix} 2\\ -2\\ 4 \end{pmatrix}\)    有通解\(X=k_1\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}+k_2 \begin{pmatrix} 2\\ 0\\ 1 \end{pmatrix}+ \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}\)则与\((A-E)^*\)相似对角矩阵为()
\(A.\begin{pmatrix} -1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}\)
\(B.\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}\)
\(C.\begin{pmatrix} -1&0&0\\ 0&-1&0\\ 0&0&1 \end{pmatrix}\)
\(D.\begin{pmatrix} -1&0&0\\ 0&-1&0\\ 0&0&-1 \end{pmatrix}\)
C   

答案C,解:令\(A=(a_1,a_2,a_3),\)由\(A\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}=0, A\begin{pmatrix} 2\\ 0\\ 1\\ \end{pmatrix}=0\)

得\(\begin{cases} a_1+a_2=0,\\ 2a_1+a_3=0 \end{cases}\)解得\(\begin{cases} a_2=-a_1,\\ a_3=-2a_1, \end{cases}\)

再由\(A\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}=a_1+a_2+a_3=-2a_1 =\begin{pmatrix} 2\\ -2\\ 4 \end{pmatrix}\)

得\(a_1= \begin{pmatrix} -1\\ 1\\ -2 \end{pmatrix}\)

从而\(​​​​a_2= \begin{pmatrix} 1\\ -1\\ 2 \end{pmatrix} ​​​​a_3= \begin{pmatrix} 2\\ -2\\ 4 \end{pmatrix}\)

即\(A=\begin{pmatrix} -1&1&2\\ 1&-1&-2\\ -2&2&4 \end{pmatrix}\)

显然A得特征值为\(\lambda_1=\lambda_2=0, \lambda_3=2,\)

由\(r(0E-A)=1\)得A可对角化

A-E的特征值为\(\lambda_1=\lambda_2=-1, \lambda_3=1,\)

由|A-E|=1,得\((A-E)^*\)得特征值为-1,-1,1,

故与\((A-E)^*\)相似对角矩阵为\(\begin{pmatrix} -1&0&0\\ 0&-1&0\\ 0&0&1 \end{pmatrix}\)应选C