张宇八套卷错题

所给的二次型不含平方项,故令\(x_1=y_1+y_2,x_2=y_1-y_2,x_3=y_3,\)有

\(f(x_1,x_2,x_3)=2y_1^2-2y_2^2-4y_1y_3+8y_2y_3,\)

再用配方法有

\(f(x_1,x_2,x_3)=2(y_1-y_3)^2-2(y_2-2y_3)^2+6y_3^2\)

令\(z_1=y_1-y_2,z_2=y_2-2y_3,z_3=y_3\),得

\(f(x_1,x_2,x_3)=2z_1^2-2z_2^2+6z_3^2\),又上述所作变化均为可逆线性变换,故其正惯性指数为2.

(1)如图所示,当液面在\(y(-1\le y\le1)\)处时,容器内液体的体积为V=V(t),则

\(\frac{dV}{dt}=\frac{dV}{dt}\cdot\frac{dy}{dt}=2\sqrt{1-y^2}\frac{dy}{dt},\)

令y=0并由条件\(\frac{dV}{dt}=0.2得\frac{dy}{dt}=0.1(m/min),\)故液面下降得速率为0.1m/min.

(2)\(w=\int_{-1}^{1}2\sqrt{1-y^2}(1-y)dy\)

\(=\int_{-1}^{1}2\sqrt{1-y^2}dy-\int_{-1}^{1}2y\sqrt{1-y^2}dy\)

\(=4\int_{0}^{1}\sqrt{1-y^2} \)\(\color{red}{\overrightarrow{\quad y=sint\quad}}\)\(4\int_{0}^{\frac{\pi}{2}}cos^2tdt\)

\(=4\cdot\frac{\pi}{4}=\pi(J)\)

其中,由于\(2y\sqrt{1-y^2}dy\)为奇函数故\(\int_{-1}^{1}2y\sqrt{1-y^2}dy=0\)

注意:本题不是相似:而是合同

\((1)\text{因}A^T=A,\text{故}B^T=(C^TAC)^T=C^TA^TC=C^TAC=B,\text{所以B为对称矩阵,b=3,}\\\)

对于

\( f(x_1,x_2)=(x_1,x_2)\begin{pmatrix}2&2\\2&a\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=2x_{1}^{2}+ax_{2}^{2}+4x_1x_2\\ =2(x_1+x_2)^2+(a-2)x_{2}^{2}\\\)

对于

\(g(y_1,y_2)=(y_1,y_2)\begin{pmatrix}4&3\\3&1\end{pmatrix}\begin{pmatrix}y_1\\y_2\end{pmatrix}=4y_{1}^{2}+y_{2}^{2}+6y_1y_2\\ =4\left(y_1+\frac{3}{4}y_2\right)^2-\frac{5}{4}y_{2}^{2}.\\\)

\( \text{由于A与B合同,故矩阵A与B得正、负指数对应相等,于是a-2<0,即a<2,又a为正整数,故a=1}\\ \text{综上所述:a=1,b=3}\\\)

\((2)\text{由(1)知,}\\ \text{令}\begin{cases}z_1=x_1+x_2\\z_2=x_2\end{cases}\text{,经过可逆线性变换}\begin{pmatrix}z_1\\z_2\\\end{pmatrix}=\begin{pmatrix}1&1\\0&1\\\end{pmatrix}\begin{pmatrix}x_1\\x_2\\\end{pmatrix}=C_1\begin{pmatrix}x_1\\x_2\\\end{pmatrix}\\ \)

\(\text{使}f(x_1,x_2)=2z_{1}^{2}-z_{2}^{2}\\ g(y_1,y_2)=4\left(y_1+\frac{3}{4}\right)^2-\frac{5}{4}y_{2}^{2}, \text{令}\begin{cases}z_1=\sqrt{2}y_1+\frac{3\sqrt{2}}{4}y_2\\z_2=\frac{\sqrt{5}}{2}y_2\\\end{cases}\\ \)

\(\text{即经可逆线性变换}\begin{pmatrix}z_1\\z_2\\\end{pmatrix}=\begin{pmatrix}\sqrt{2}&\frac{3\sqrt{2}}{4}\\0&\frac{\sqrt{5}}{2}\\\end{pmatrix}\begin{pmatrix}y_1\\y_2\\\end{pmatrix}=C_2\begin{pmatrix}y_1\\y_2\\\end{pmatrix}\\ \)

\(\text{使}\\ g(y_1,y_1)=2z_{1}^{2}-z_{2}^{2},\\ \)

\(\text{于是有}C_1\begin{pmatrix}x_1\\x_2\\\end{pmatrix}=C_2\begin{pmatrix}y_1\\y_2\\\end{pmatrix}\\ \)

\(\text{故}\begin{pmatrix}x_1\\x_2\\\end{pmatrix}=C_{1}^{-1}C_2\begin{pmatrix}y_1\\y_2\\\end{pmatrix}=C\begin{pmatrix}y_1\\y_2\\\end{pmatrix}\\\)

\( \text{即}C={\begin{pmatrix}1&1\\0&1\\\end{pmatrix}}^{-1}\begin{pmatrix}\sqrt{2}&\frac{3\sqrt{2}}{4}\\0&\frac{\sqrt{5}}{2}\\\end{pmatrix}\\ \)

\(=\begin{pmatrix}1&-1\\0&1\\\end{pmatrix}\begin{pmatrix}\sqrt{2}&\frac{3\sqrt{2}}{4}\\0&\frac{\sqrt{5}}{2}\\\end{pmatrix}\\ =\begin{pmatrix}\sqrt{2}&\frac{3\sqrt{2}-2\sqrt{5}}{4}\\0&\frac{\sqrt{5}}{2}\\\end{pmatrix}\\\)

答案:\(\frac{1}{x}(x>0)\)

解析:由\(\int_{a}^{ab}f(x)dx\)与a无关所以\(\left[\int_{a}^{ab}f(x)dx\right]'\equiv0,\)即

\(f(ab)b-f(a)\equiv0.\)

上式对任意a均成立,所以令a=1亦成立,有\(f(b)b-f(1)=0,f(b)=\frac{1}{b},\)即有\(f(x)=\frac{1}{x}(x>0)\)

可以验算,\(\int_{a}^{ab}\frac{1}{x}dx=ln(ab)-lna=lnb,\)与a无关

解:由\(a=\lim\limits_{x\to+\infty}\frac{y}{x},\)令\(\frac{y}{x}=u,\)则y=ux,代入\(y^3=x(x^2-2y)\),得\(u^3x^3=x(x^2-2ux),\)即

\(u^3=1-\frac{2u}{x},\)

两端取极限,有\(\lim \limits_{x\to+\infty}u^3=\lim \limits_{x\to+\infty}(1-\frac{2u}{x}),\)即\(\lim\limits_{x\to+\infty}(\frac{y}{x})^3=\lim\limits_{x\to+\infty}(1-2\cdot\frac{y}{x}\cdot\frac{1}{x}),\)得

\(a^3=1-2\cdot a\cdot 0, \)即a=1.

又\(b=\lim\limits_{x\to+\infty}(y-ax)=\lim\limits_{x\to+\infty}(y-x)\),令y-x=t,代入\(y^3=x(x^2-2y)\),得

\((x+t)^3=x[x^2-2(x+t)]\)

整理得:\(y^3+3x^2t+3xt^2=-2xt-2x^2\)

两端除以\(x^2\)有,\(\frac{t^3}{x^2}+3t+\frac{3t^2}{x}=-2\frac{t}{x}-2\)

两端再取极限,有\(\lim\limits_{x\to\infty}\left(\frac{t^3}{x^2}+3t+\frac{3t^2}{x}\right)=\lim\limits_{x\to\infty}\left(-2\frac{t}{x}-2\right)\)

即\(\lim\limits_{x\to\infty}\left[\frac{(y-x)^3}{x^2}+3(y-x)+\frac{3(y-x)^2}{x}\right]=\lim\limits_{x\to\infty}\left(-2\frac{y-x}{x}-2\right)\)

得3b=-2,即\(b=-\frac{2}{3}\).

(1)令\(h(x)=e^x(x^2+2)-2,\)则h(0)=0,且

\(h'(x)=e^x(x^2+2x+2)=e^x[(x+1)^2+1]>0\)

故h(x)再(-∞,+∞)上单调增加,当x<0时,h(x)<h(0)=0,即\(e^x(x^2+2)<2\).

(2)由(1)知,当x≥0时,\(e^x(x^2+2)\ge2.\)因此当x<0时,\(\frac{1}{2}x^2+1<e^{-x}\);当x≥0时,\(\frac{1}{2}x^2+1\ge e^x\),

故

\(f(x)=\begin{cases}e^{-x}, &x<0,\\\frac{1}{2}x^2+1,&x\ge 0, \end{cases}\)且\(g(0)\ge f(0)=1\)

若g(0)=1,则

\(g_{+}^{'}(0)=\lim\limits_{x\to0^+}\frac{g(x)-g(0)}{x-0}\ge \lim\limits_{x\to0^+}\frac{f(x)-f(0)}{x-0} = \lim\limits_{x\to0^+}\frac{\frac{1}{2}x^2+1-1}{x}=0\)

\(g_{-}^{'}(0)=\lim\limits_{x\to0^-}\frac{g(x)-g(0)}{x-0}\ge \lim\limits_{x\to0^-}\frac{f(x)-f(0)}{x-0} = \lim\limits_{x\to0^-}\frac{e^{-x}-1}{x}=-1\)

于是\(g_{+}^{'}(0)\not= g_{-}^{'}(0)\),g(x)在x=0处不可导,与题设矛盾,故g(0)>1.

\(\text{解答} (1)f(x,y,z)=x^T \begin{pmatrix}0&0&1\\0&1&0\\1&0&0\\ \end{pmatrix}x,\text{其二次型矩阵}A=\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\\\end{pmatrix} x=(x,y,z)^T,\)

\(\text{令}\\ |\lambda E-A|=\begin{vmatrix}\lambda&0&-1\\0&\lambda-1&0\\-1&0&1\\\end{vmatrix}=(\lambda-1)^2(\lambda+1)=0 \text{得A的特征值}\lambda_1=\lambda_2=1,\lambda_3=-1.\\ \)

\(\text{当}\lambda_1=\lambda_2=1,\\ (E-A)x=\begin{pmatrix}1&0&-1\\0&0&0\\-1&0&1\\\end{pmatrix}x=0\\\)

\( \text{得线性无关得特征向量}\xi_1=(1,0,1)^T,\xi_2=(0,1,0)^T,\\\)

\( \text{当}\lambda_3=-1, (-E-A)x=\begin{pmatrix}-1&0&-1\\0&-2&0\\-1&0&-1\\\end{pmatrix}x=0\\ \text{得线性无关得特征向量}\)

\(\xi_3=(1,0,-1)^T\\ \xi_1,\xi_2,\xi_3\text{已经正交,只需要单位化}\\ \eta_1=(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})^T, \eta_2=(0,1,0)^T, \eta_3=(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}})^T, \\\)

\(\text{令} Q=(\eta_1,\eta_2,\eta_3)=\begin{pmatrix}\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\0&1&0\\\frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}}\end{pmatrix} \text{即为所求得正交矩阵}.\\\)

\( \text{记}y=(u,v,w)^T,\text{则经过正交变换}x=Qy,\)

\(\text{有} f(x,y,z)=x^TAx=y^TQ^TAQy=y^T\begin{pmatrix}1&&\\&1&\\&&-1\\\end{pmatrix}y=u^2+v^2-w^2.\\\)

\((2)\text{由}(1),f\le u^2+y^2+z^2,\text{又}\\ x^2+y^2+z^2=x^Tx=(Qy)^T(Qy)=y^TQ^TQy=y^Ty=u^2+v^2+w^2,\\ \text{故},g(x,y,z)=\frac{f(x,y,z)}{x^2+y^2+z^2}\le \frac{u^2+v^2+w^2}{u^2+v^2+w^2}=1,\\ \)

\(\text{即函数g(x,y,z)得最大值为1,且当(u,v,w)=(0,1,0)时,g(x,y,z)=1.此时}\\ \)

\(\begin{pmatrix}x\\y\\z\\\end{pmatrix}=Q\begin{pmatrix}0\\1\\0\\\end{pmatrix}=\begin{pmatrix}\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\0&1&0\\\frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}}\end{pmatrix}\begin{pmatrix}0\\1\\0\\\end{pmatrix}=\begin{pmatrix}0\\1\\0\\\end{pmatrix},\\ \)

\(\text{故g(x,y,z)得一个最大值点时(0,1,0)}.\)

(1)若\(f'(x_1)=f'(x_2),\)则取\(c=x_1\)或\(x_2\),即得\(f'(c)=\sqrt{f'(x_1)f'(x_2)}\)

若\(f'(x_1)\not=f'(x_2)\),不妨设\(f'(x_1)<f'(x_2)\),则\(f'(x_1)<\sqrt{f'(x_1)f'(x_2)}<f'(x_2).\)根据连续函数介质定理存在\(c\in(a,b)\),使得\(f'(c)=\sqrt{f'(x_1)f'(x_2)}\)

(2)因为f'(x)>0,所以\(a\le f(a)<f(b)\le b.\)对f(x)在区间[a,b]与[f(a),f(b)]上用拉格朗日中值定理,存在\(\eta_1 \in(a,b),\eta_2\in(f(a), f(b))\),使得

\(f(b)-f(a)=f'(\eta_1)(b-a),\)

\(f[f(a)]-f[f(b)]=f'(\eta_2)[f(b)-f(a)]\)

由此得

\(f[f(a)]-f[f(b)]=f'(\eta_1)f'(\eta_2)(a-b)\)

由(1)知,存在\(\xi\in(a,b)\)使得\(f'(\xi)=\sqrt{f'(\eta_1)f'(\eta_2)}\),代入公式,即得

\(f[f(a)]-f[f(b)]=[f'(\xi)]^2(a-b)\)

分析:记\(\begin{pmatrix}1&-1&1\\-1&2&-3\\1&-3&6\\\end{pmatrix}=A\),则所对应的二次型为

\(f(x_1,x_2,x_3)=x^TAx=x_{1}^{2}+2x_{x}^{2}+6x{3}^{2}-2x_1x_2+2x_1x_3-6x_2x_3\)

\(=(x_1-x_2+x_3)^2+(x_2-2x_3)^2+x_{3}^{2}\)

\(=(x_1-x_2+x_3,x_2-2x_3,x_3)\begin{pmatrix}x_1-x_2+x_3\\ x_2-2x_3\\ x_3 \end{pmatrix}\)

\(=(x_1,x_2,x_3)\)\(\begin{pmatrix}1&0&0\\-1&1&0\\1&-2&1\\\end{pmatrix}\)\(\begin{pmatrix}1&-1&1\\0&1&-2\\0&0&1\\\end{pmatrix}\)

\(=(x_1,x_2,x_3)D^TD\begin{pmatrix}x_1\\x_2\\x_3\\\end{pmatrix}\)

其中\(A=D^TD,D=\begin{pmatrix}1&-1&1\\0&1&-2\\0&0&1\\\end{pmatrix}\)

记\(P=\begin{pmatrix}x_1&x_2\\x_3&x_4\end{pmatrix},\)由题知\(a+d=\lambda+\lambda,\)即\(a-\lambda=\lambda-d,\)

由\(AP=P\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}\),有

\(\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x_1&x_2\\x_3&x_4\end{pmatrix}\)\(=\begin{pmatrix}x_1&x_2\\x_3&x_4\end{pmatrix}\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}\),

即\(\begin{cases}ax_1+bx_3=\lambda x_1,\\ ax_2+bx_4=x_1+\lambda x_2,\\ cx_1+dx_3=\lambda x_3,\\ cx_2+dx_4=x_2+\lambda x_4,\\ \end{cases}\)

化简得\(\begin{cases}(a-\lambda)x_1+bx_3= 0,\\ -x_1+(a-\lambda)x_2+bx_4=0,\\ cx_1+(d-\lambda)x_3= 0,\\ cx_2-x_3+(d-\lambda)x_4=0,\\ \end{cases} (*)\)

又因\(\lambda\)是A得二重特征值,故\(\begin{pmatrix}\lambda-a&-b\\-c&\lambda-d\end{pmatrix}=0\)

对于方程组(*)的系数做初等行变换,得

\(\begin{pmatrix}a-\lambda&0&b&0\\-1&a-\lambda&0&b\\c&0&d-\lambda&0\\0&c&-1&d-\lambda\end{pmatrix}\)

\(\begin{matrix} r3+\frac{\lambda-d}{b}\\ \overrightarrow{\quad r_3\leftrightarrow r_4 \quad}\\ \end{matrix} \begin{pmatrix}a-\lambda&0&b&0\\-1&a-\lambda&0&b\\0&c&-1&d-\lambda\\0&0&0&0\end{pmatrix}\)

\( \overrightarrow{\quad r_3+\frac{\lambda-d}{b}r_2 \quad} \begin{pmatrix} a-\lambda&0&b&0\\ -1&a-\lambda&0&b\\ \frac{d-\lambda}{b}&0&-1&0\\ 0&0&0&0 \end{pmatrix}\)

\( \overrightarrow{\quad r_3+\frac{1}{b}r_1 \quad} \begin{pmatrix} a-\lambda&0&b&0\\ -1&a-\lambda&0&b\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}\)

则基础解系为\(\xi_1=\)\(\begin{pmatrix}b\\ 0\\ \lambda-a\\ 1 \end{pmatrix}\),\(\xi_2=\begin{pmatrix} 0\\ b\\ 0\\ a-\lambda\\ \end{pmatrix}\)

故通解为:

\(\begin{pmatrix}x_1\\ x_2\\ x_3\\ x_4 \end{pmatrix}=k_1\xi_1+k_2\xi_2=\begin{pmatrix} k_1b\\ k_2b\\ k_1(\lambda-a)\\ k_1+k_2(\lambda-q) \end{pmatrix},\)

又\(\lambda-a=\frac{1}{2}(d-a),\)故所求矩阵为

\(p=\begin{pmatrix}k_1b&k_2b\\ \frac{1}{2}k_1(d-a)&k_1+k_2(d-a) \end{pmatrix},\)|\(|p|=k_{1}^{2}b\not=0,\)

其中\(k_1\not=0,k_2\)为任意常数

17.解:由极值条件,得方程组

\(\begin{cases}\frac{\partial f}{\partial x}=3-2ax-2by=0\\ \frac{\partial f}{\partial y}=4-4ay-2bx=0\\ \end{cases}\)

整理得

\(\begin{cases}2ax+2by=3\\ bx+2ay=2\\ \end{cases}\)

当\(\begin{vmatrix}2a&2b\\ b&2a\\ \end{vmatrix}=4a^2-2b^2\not=0,\)即\(2a^2-b^2\not=0,\)方程由唯一解

\(x_0=\frac{3a-2b}{2a^2-b^2}, y_0=\frac{4a-3b}{4a^2-2b^2},\)

又由于\(A=\frac{\partial^2f}{\partial x^2}|_{(x_0,y_0)}=-2a\),\(B =\frac{\partial^2f}{\partial x \partial y}|_{(x_0,y_0)}=-2b\),\(C =\frac{\partial^2f}{\partial y^2}|_{(x_0,y_0)}=-4a\)

故当\(AC-B^2=8a^2-4b^2>0,\)且\(A=-2a<0,\)即\(2a^2-b^2>0\)时f(x,y)有唯一的极大值.

综上所述,当\(2a^2-b^2>0\)且a>0时,f(x,y)有唯一极大值

\(\int_{-1}^{2}(arctan\:nx)^3dx=\int_{1}^{2}(arctan\:nx)^3dx\)

\(=\frac{1}{n}\int_{n}^{2n}(arctan\:nx)^3dx=\frac{1}{n}(2n-n)(arctan\:\xi_n)^3\)

=\((arctan\:\xi_n)^3\)

其中\(n\le\xi_n\le2n\),因为\(\lim\limits_{n\to\infty}arctan\:n=\frac{\pi}{2},\)所以有

\(\lim\limits_{x\to\infty}\int_{-1}^{2}(arctan\:nx)^3dx=\lim\limits_{x\to\infty}(arctan\:\xi_n)^3=(\frac{\pi}{2})^3=\frac{\pi^3}{8}\)

记t时刻容器哦水深为z(t),则注入水的体积为

\(V(t)=\int_{0}^{z(t)}\pi f^2(u)du,\)

水面的面积\(S(t)=\pi f^2(z),\)

由题意,

\(\frac{dV(t)}{dt}=\pi f^2[z(t)]\cdot \frac{dz}{dt}=3,\)

\(\frac{dS(t)}{dt}=\pi \cdot2f[z(t)]\cdot \frac{dz}{dt}=\pi,\)

故

\(\frac{df(z)}{dz}\cdot \frac{2}{f(z)}=\frac{\pi}{3},\)

分离变量得

\(\frac{df(z)}{f(z)}=\frac{\pi}{6}dz\)

于是\(\int\frac{df(z)}{f(z)}=\frac{\pi}{6}\int dz\)

得

\(ln|f(z)|=\frac{\pi}{6}z+lnc_1, \)\(f(z)=\pm c_1e^{\frac{\pi}{6}z}=ce^{\frac{\pi}{6}z}(z\le0),\)

又底面积为\(16\pi cm^2\),即f(0)=4,得c=4,即\(f(z)=4e^{\frac{\pi}{6}z}.\)

解:曲线\(x^2+y^2=a(x+\sqrt{x^2+y^2})\)是条什么曲线?采用极坐标.令

\(r=\sqrt{x^2+y^2},x=rcos\theta\),

于是该方程化为\(r=a(1+cos\theta),a>0,\)他是一条心形线,如图所示.

由于函数\(\varphi(x,y)=x^2ln(y+\sqrt{1+x^2})\)是关于y的奇函数,积分区域关于x轴对称,所以

\(\iint_{D}\varphi(x,y)d\sigma=0.\)

因此

\(I=\iint_D[x^2ln(y+\sqrt{1+y^2})+x\sqrt{x^2+y^2}]d\sigma\)

\(=\iint_Dx\sqrt{x^2+y^2}d\sigma\)

\(=\int_{0}^{2\pi}d\theta\int_{0}^{a(1+cos\theta)}r^3cos\theta dr\)

\(=\frac{\pi^4}{4}\int_{0}^{2\pi}(1+cos\theta)^4cos\theta d\theta\)

\(=\frac{\pi^4}{4}\int_{0}^{2\pi}(1+4cos\theta+6cos^2\theta+4cos^3\theta+cos^4\theta)cos\theta d\theta\)

\(=a^4\int_{0}^{\frac{\pi}{2}}(4cos^2\theta +4cos^4\theta)d\theta\)

\(=a^4(4×\frac{1}{2}×\frac{\pi}{2}+4×\frac{3}{4}×\frac{1}{2}×\frac{\pi}{2})=\frac{7}{4}\pi a^4\)

(1)根据题设条件,A*的特征值为\(\lambda=2,-2,-1,\)所以A+2E的特征值为4,0,1,故存在正交矩阵P,使得

\(P^{-1}(A^*+2E)P=\begin{pmatrix}4&0&0\\0&0&0\\0&0&1\\ \end{pmatrix}\)

因此r(A*+2E)=2.

(2)利用公式\(A^*A=|A|E\)两边取行列式,可得\(|A|^2=|A^*|=4\),所以\(A=\pm2.\)

若|A|=2,则\(A=|A|(A^*)^{-1}\)的特征值为\(\frac{|A|}{\lambda}:1,-1,-2;\)

若|A|=-2,则\(A=|A|(A^*)^{-1}\)的特征值为\(\frac{|A|}{\lambda}:-1,1,2.\)

因为二次型\(g(x_1,x_2,x_3)=x^TAx\)的真惯性指数为2,所以应该排除第一种情形,于是A+2E的特征值为1,3,4,行列式|A+2E|=1×3×4=12.