考生注意事项 答题前,考生须在试题册指定位置上填写考生编号和考生姓名;在答题卡指定位置上填写报考单位...

Создано: 4 марта 2025 г.

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Вопрос

考生注意事项
答题前,考生须在试题册指定位置上填写考生编号和考生姓名;在答题卡指定位置上填写报考单位,考生姓名和考生编号,并涂写考生编号信息点。
选择题的答案必须涂写在答题卡相应题号的选项上,非选择题的答案必须书写在答题卡指定位置的边框区域内。超出答题区域书写的答案无效;在草稿纸、试题册上答题无效。
填(书)写部分必须使用黑色字迹签字笔书写,字迹工整,笔迹清楚;涂写部分必须使用 2B 铅笔填涂。
考试结束,将答题卡和试题册按规定交回。
一、单选题(1~10小题,每小题5分,共50分)
设函数
f
(
x
)
在区间
(

1
,
1
)
内有定义,且在点
x

0
处连续,则以下结论:
(1)当
lim
x

0
f
(
x
)
3

x

0
时,
f
(
x
)
在点
x

0
处可导;
(2)当
lim
x

0
f
(
x
)
x
2

0
时,
f
(
x
)
在点
x

0
处可导;
(3)当
f
(
x
)
在点
x

0
处可导时,
lim
x

0
f
(
x
)
3

x

0

所有正确结论的序号为

A. (1)
B. (2)
C. (2)(3)
D. (1)(2)
设正项数列
{
a
n
}
,
{
b
n
}
,
{
c
n
}
,则“


n

1
a
n
c
n



n

1
c
n
b
n
均收敛”是“


n

1
a
n
b
n
收敛”的

A. 充分非必要条件
B. 必要非充分条件
C. 充要条件
D. 既非充分又非必要条件

f
(
x
,
y
)
可微,
p
(
x
,
y
)
,
p
0
(
x
0
,
y
0
)
为曲面
z

f
(
x
,
y
)
上的点,则

A.
lim
p

p
0
f
(
p
)
+
[
f
(
p
0
)

grad
f
|
p
0
×
−→
p
0
p
]


−→
p
0
p

0
B.
lim
p

p
0
f
(
p
)

[
f
(
p
0
)

grad
f
|
p
0

−→
p
0
p
]


−→
p
0
p

0
C.
lim
p

p
0
f
(
p
)
+
[
f
(
p
0
)
+
grad
f
|
p
0
×
−→
p
0
p
]


−→
p
0
p

0
D.
lim
p

p
0
f
(
p
)

[
f
(
p
0
)
+
grad
f
|
p
0

−→
p
0
p
]


−→
p
0
p

0
设正项数列
{
a
n
}
单调增加,则以下选项中使得
{
a
n
}
收敛的是

A.
{
(
1
+
a
n
)
1
a
n
}
收敛于 1
B.
{
(
1
+
1
a
n
)
a
n
}
收敛于
e
C.
{
a
n
ln
a
n
}
收敛于 0
D.
{
ln
a
n
a
n
}
收敛于 0

A
为 3 阶矩阵,若
|
E

A
|

1
,
|

E

A
|


1
,
|
2
E

A
|

2
,则
|
3
E

A
|

A. -2
B. -8
C. 8
D. 11

A





1
2
2

3

1
3



,
b




4
1
2



,若对任意 2 维实列向量
x
,均有

b

A
x
0



b

A
x

,则
x
0

A.
(

3

2
)
B.
(
3

2
)
C.
(

3
2
)
D.
(
3
2
)

α
1




1

1
a



,
α
2





1
0
1



,
α
3




1
b
1



是 3 阶实矩阵
A
的 3 个互异特征值的特征向量,则“(a,b)=(1,2)”是“A(a, b)=(1,2) ” 是 “ \boldsymbol{A} 为对称矩阵”的

A. 充要条件
B. 充分非必要条件
C. 必要非充分条件
D. 既非充分又非必要条件
需要识图,忽略,答案A

设总体
X

U
(
a
,
b
)

a
,
b
为未知参数。
X
1
,
X
2
,

,
X
n
(
n

1
)
为来自总体
X
的简单随机样本,
X
(
1
)

min
{
X
1
,
X
2
,

,
X
n
}
,
X
(
n
)

max
{
X
1
,
X
2
,

,
X
n
}
,
¯
X

1
n

n
i

1
X
i
,
S
2
1

1
n

n
i

1
(
X
i

¯
X
)
2
,令
L

b

a
,则
L
的矩估计量与最大似然估计量分别为

A.
¯
X
+

3
S
1
,
X
(
n
)

X
(
1
)
B.
¯
X
+

3
S
1
,
X
(
n
)

X
(
1
)
2
C.
2

3
S
1
,
X
(
n
)

X
(
1
)
D.
2

3
S
1
,
X
(
n
)

X
(
1
)
2
设在条件
X

x
下,
Y
服从正态分布
N
(
x
,
x
2
)

X
的边缘分布为
U
(
0
,
1
)
,则
X
,
Y
的相关系数为

A.

5
12
B.

5
5
C.


5
12
D.


5
5
二、填空题(11~16小题,每小题5分,共30分)

f
(
x
,
y
,
z
)

x
2
e
y
z
2
,则
f
(
x
,
y
,
z
)
在点
(

1
,
0
,
1
)
处的方向导数的最小值为 ______。


L
为点
(
0
,
1
)
到点
(
1
,
2
)
的直线段,则

L

x
y

d
s

______。

微分方程
d
y

d
x

x
+
2
y

3
x
+
1
满足
y
(
0
)

2
的解为 ______。


Ω

{
(
x
,
y
,
z
)

x
2

y

x
,
0

z

2

x

y
}
,则

Ω
x
e
y
+
z

d
x

d
y

d
z

______。


R
3
中的两组基:
α
1




1
0
1



,
α
2




2
1
0



,
α
3




1
1
1



,
β
1




1
2

1



,
β
2




2
2

1



,
β
3




2

1

1



,由
α
1
,
α
2
,
α
3

β
1
,
β
2
,
β
3
的过渡矩阵为 ______。

甲、乙两种检验设备均有 90% 的正确检出率且相互独立,现有一批合格率为 10% 的产品,从中任取一件产品,经两种检验设备检验均为合格品,则该产品是合格品的概率为 ______。

Мысль

Evaluating continuity

Interestingly enough, I'm working through the implications of the limit of f(x)/(3√x) as x approaches 0 on f(x)'s differentiability at x=0. This involves assessing f(x)'s continuity and order relative to x.

Analyzing function dynamics

OK, let me see. I’m assessing f(x)=o(x^(1/3)) relative to x^(1/3). It’s clear that f(x)=o(x^(1/3)) doesn’t ensure differentiability at x=0.

Analyzing the implications

I’m working through if f(x)=o(x^2) suggests differentiability at x=0 with a derivative of 0, assuming continuity and f(0)=0. This seems promising but requires more scrutiny.

Confirming differentiability

I’m digging into the idea that if f(x)=o(x^2) and f is continuous at 0 with f(0)=0, then f(x)/x approaches 0 as x approaches 0, suggesting the derivative at 0 might be 0.

Analyzing function dynamics

I'm digging into functions like f(x)=x²sin(1/x²), which meet f(x)=o(x²) but lack a limit as x approaches 0, indicating they might not be differentiable there.

Analyzing derivatives

I’m piecing together that if f(x)=o(x²) and f is continuous at 0, then the derivative at 0 seems to be 0. However, for x in (0,1), this might not hold.

Confirming continuity effects

Interestingly enough, if f(x)=o(x²) and f is continuous at 0 from both sides, it leads to a difference quotient of 0. This confirms differentiability at 0 with a derivative of 0.

Assessing importance

To evaluate the limit, I’m considering if f(0)=0 is a prerequisite. It seems, without it, the limit might not hold, making the statement potentially true.

Piecing it together

Analyzing f(x)/(3√x) as x approaches 0 shows it diverges unless f(0)=0. Thus, differentiability at 0 implies f(0)=0.

Assessing completeness

I'm considering if the statement entirely excludes the need for f(0)=0. If f(0)!=0, the limit that should go to 0 does not.

Pinpointing assumptions

Assuming f(0)=0 to evaluate the limit. This aligns with standard results, suggesting that differentiability at 0 implies f(0)=0.

Considering different functions

I’m digging into various examples like f(x)=x^(2/3) and f(x)=x^(3/2) to evaluate differentiability at 0 and continuity at both sides, ensuring convergence as x approaches 0.

Assessing convergence

I’m examining the series' convergence using tools like the Cauchy product and Schur's test. It’s interesting to see how these relate to the necessary and sufficient conditions.

Assessing convergence

I'm working through whether convergence of series ancn\sum a_n c_n and bncn\sum b_n c_n is a sufficient or necessary condition for anbn\sum a_n b_n to converge.

Analyzing convergence

I’m working through examples like an=1/na_n = 1/n, bn=1/nb_n = 1/n, and cn=1/nc_n = 1/n, to test the convergence of series ancn\sum a_n c_n, bncn\sum b_n c_n, and anbn\sum a_n b_n.

Evaluating convergence implications

OK, let's see. The convergence of ancn\sum a_n c_n and bncn\sum b_n c_n doesn't guarantee anbn\sum a_n b_n converges either, so it's not a sufficient condition.

Assessing necessary condition

I'm digging into whether anbn\sum a_n b_n convergence might necessitate ancn\sum a_n c_n and bncn\sum b_n c_n convergence, given that a counterexample disproves sufficiency.

Analyzing dependencies

OK, let's see. It turns out anbn\sum a_n b_n convergence doesn't obligate ancn\sum a_n c_n or bncn\sum b_n c_n to converge with specific cnc_n values.

Finding the correct choice

OK, I determined that the correct choice is "D" since the convergence of anbn\sum a_n b_n doesn't necessitate ancna_n c_n or bncnb_n c_n convergence.

Identifying alignment

I’m comparing the expressions to establish the limit's criteria by checking for correct signs, properties of dot and cross products, and ensuring the gradient's role is accurately represented.

Revisiting the expression

I rechecked expression (B) and it doesn't fit the standard limit definition. Expression (D) matches perfectly, and I confirmed the correct answer is (D).

Interpreting sequence behavior

I’m noting that statement (A) is tricky since ana_n is positive and monotone increasing. If ana_n \to \infty, (1+an)1/an(1 + a_n)^{1/a_n} might approach something, but it's vague.

Revealing ana_n behavior

OK, let me see. If ana_n heads to infinity, (1+an)1/an(1 + a_n)^{1/a_n} nears 1. But ana_n can't shrink to 0 unless it's 0 forever.

Evaluating convergence trends

I'm thinking through ana_n convergence: statement (B) hints at ana_n \to \infty with ee as the limit. Statement (C) implies an0a_n \to 0.

Analyzing potential outcomes

Considering if an1a_n \to 1 satisfies anln(an)0a_n \ln(a_n) \to 0. Other outcomes are explored, but they lead to contradictions.

Shifting my perspective

I'm noting that if anln(an)0a_n \ln(a_n) \to 0, then ana_n must converge to 1 due to monotone increasing properties and the behavior of xln(x)x \ln(x) as x0x \to 0.

Weighing the options

If anln(an)0a_n \ln(a_n) \to 0, it seems ana_n is likely converging to 1, as other outcomes lead to contradictions. However, ana_n could also approach infinity.

Breaking down polynomial constraints

I'm working through a 3x3 matrix AA with conditions p(1)=1p(1)=1, p(1)=1p(-1)=-1, and p(2)=2p(2)=2, aiming to identify p(3)p(3).

Clarifying polynomial forms

I’m thinking through the polynomial p(t)=t3+αt2+βt+γp(t) = t^3 + \alpha t^2 + \beta t + \gamma for a 3x3 matrix, focusing on coefficients and their signs to match given matrix conditions.

Confirming matrix determinants

I’ve verified the matrix determinant conditions, ensuring consistency with p(1)=1p(-1)=-1 and p(2)=2p(2)=2, confirming polynomial forms and matrix properties.

Isolating coefficients

I’m working through equations from the polynomial conditions to isolate coefficients α\alpha, β\beta, and γ\gamma. After substituting and simplifying, β\beta is found to be 0.

Evaluating polynomial results

I’ve calculated the polynomial coefficients, finding α=2\alpha = -2, β=0\beta = 0, and γ=2\gamma = 2. Substituting these, p(3)=11p(3) = 11.

Reasoned for 2m 54s

Python

python
import sympy as sp x1, x2 = sp.symbols('x1 x2', real=True) A = sp.Matrix([[-1,2],[2,-3],[-1,3]]) b = sp.Matrix([4,1,2]) # We want x0 that satisfies A^T A x0 = A^T b ATA = A.T*A ATb = A.T*b x0_sol = ATA.LUsolve(ATb) x0_sol

Мысль

Investigating eigenvectors

I’m working through a real 3x3 matrix with distinct eigenvalues. The question boasts symmetry requiring mutually orthogonal eigenvectors, which simplifies to a=1,b=2a=1, b=2.

Cracking orthogonality

Working through the dot products, I find a=1a=1 and b=2b=2 for orthogonality. This makes me think symmetry in the matrix is tied to these values.

Crunching orthogonal eigenvectors

Here, I’m thinking about how a matrix with orthogonal eigenvectors is diagonalizable, supporting symmetry. Interestingly, distinct eigenvalues and orthonormal vectors lead to this conclusion.

Confirming symmetry criteria

I’m piecing together that orthogonal eigenvectors (a,b)=(1,2)(a, b) = (1, 2) are both necessary and sufficient for the matrix to be symmetric, fitting both necessary and sufficient criteria.

Pulling together moments

For a uniform distribution U(a,b)U(a,b), L=baL = b - a, and the first sample moment about the mean involves S12=1n(XiXˉ)2S^2_1 = \frac{1}{n} \sum (X_i - \bar{X})^2.

Piecing together estimates

I’m gathering info on the method-of-moment estimate for L=baL = b - a, noting it’s the square root of 12 times the sample standard deviation, specifically using the biased version S1S1.

Calculating covariance

OK, let me see if I can work out the covariance between XX and YY by using their expected values and finding E[XY]E[XY] and E[X]E[Y]E[X]E[Y].

Piecing together covariance

Chicken and egg scenario: I’m gathering info on Cov(X,Y)Cov(X, Y) by finding E[X]E[X], E[Y]E[Y], and E[XY]E[XY]. Progress involves meticulous calculation of these expected values.

Calculating covariance

First, I find E[XY]E[XY] as E[X2]E[X^2], which is 13\frac{1}{3}. Then, Cov(X,Y)Cov(X, Y) comes to 112\frac{1}{12}.

Determining correlation

I just calculated the variances and covariance, leading to the finding that the correlation coefficient is 15\frac{1}{\sqrt{5}}, which is roughly 0.4472.

Taking a closer look

Piecing together the options, option B (55\frac{\sqrt{5}}{5}) aligns with our calculated correlation coefficient of 15\frac{1}{\sqrt{5}}. For the fill-in question, we find the minimal directional derivative using the negative gradient direction.

Calculating gradients

OK, let me see. I’m working out the gradient of f=x2eyz2f = x^2 e^{y z^2} at (1,0,1)(-1,0,1). First, I find the partial derivatives dfdx\frac{df}{dx}, dfdy\frac{df}{dy}, and dfdz\frac{df}{dz}.

Piecing together integral strategies

I’m parametrizing the line segment from (0,1)(0,1) to (1,2)(1,2) by defining r(t)=(t,1+t)\mathbf{r}(t) = (t, 1 + t) for tt in [0,1][0,1].

It's all about substitutions

I explore various substitution methods—tanh2(θ)\tanh^2(\theta), sinh2(θ)\sinh^2(\theta), and others. Each approach is detailed, but finding a simpler route continues to be a challenge.

Reasoned for 1m 28s

Python

python
import sympy as sp t=sp.Symbol('t', positive=True) expr = sp.sqrt(t*(1+t)) res = sp.integrate(expr, (t,0,1)) res_simpl = sp.simplify(res) res_simpl

Мысль

Reasoned for a few seconds

Python

python
val = sp.simplify(sp.sqrt(2)*res_simpl) val_simpl = sp.simplify(val) val_approx = val.evalf() val_simpl, val_approx

Мысль

Breaking down integration

I'm converting acosh(sqrt(2)) to ln form for simplification. It leads me to a more elegant expression: 3/2 - (sqrt(2)/4) * ln(sqrt(2) + 1). This approach streamlines the integral.

Recasting the ODE

I'm thinking through the differential equation dy/dx = (x + 2y - 3) / (x + 1), aiming to express it linearly by reworking the terms.

Gathering insights

I’ve identified the ODE and derived the integrating factor μ(x)=1(x+1)2\mu(x) = \frac{1}{(x+1)^2}. This leads me to the next step, progressing through the integration.

Breaking down the integral

I applied partial fractions and term-by-term integration to the integral. This led me to derive the function y=1x+C(x+1)2y = 1 - x + C (x+1)^2 as the solution.

Validating constants

Applying y(0)=2y(0)=2 pinpointed C=1C=1, refining the solution to y=x2+x+2y = x^2 + x + 2. Next, I’ll confirm this solution's accuracy.

Analyzing inequalities

I’m breaking down x2yxx^2 \leq y \leq x and 0z2xy0 \leq z \leq 2 - x - y. It’s clear that xx ranges from 0 to 1.

Integrating with precision

I’ve set up the integral with the order dz,dy,dxdz, dy, dx. Now, I’m carefully integrating to simplify expressions involving exponents and conditions, ensuring accuracy throughout.

Refining the integral

I’m working through the integral, breaking down terms and simplifying expressions. This process will ensure clarity and progress towards a more streamlined solution.

Breaking down the integral

I’m systematically separating and simplifying the integral into distinct terms. This involves integrating each term separately, leading to a clear and organized expression.

Breaking down the integral

I’m separating the integral into three parts: 01xnexdx\int_0^1 x^n e^{-x} dx using integration by parts, recognizing the incomplete gamma function, and calculating with Python.

Reasoned for 1m 1s

Python

python
import sympy as sp x=sp.Symbol('x', positive=True) part1 = sp.exp(2)*sp.integrate((x**2 - x**3)*sp.exp(-x),(x,0,1)) part2 = -sp.integrate(x*sp.exp(x),(x,0,1)) part3 = sp.integrate(x*sp.exp(x**2),(x,0,1)) res = part1+part2+part3 sp.simplify(res)

Мысль

Reasoned for a couple of seconds

Python

python
res_approx = res.evalf() res_simpl, res_approx

Мысль

Reasoned for a second

Python

python
part1_simpl = sp.simplify(part1) part2_simpl = sp.simplify(part2) part3_simpl = sp.simplify(part3) res = part1_simpl + part2_simpl + part3_simpl res_simpl = sp.simplify(res) res_approx = res_simpl.evalf() part1_simpl, part2_simpl, part3_simpl, res_simpl, res_approx

Мысль

Breaking down integrals

I started by simplifying the integrals step by step, arriving at 4e2+232e32-4 e^2 + \frac{23}{2} e - \frac{3}{2}. This involves factorizing and simplifying terms to achieve the final expression.

Reasoned for 10 seconds

Python

python
B1 = sp.Matrix([1,2,-1]) B2 = sp.Matrix([2,2,-1]) B3 = sp.Matrix([2,-1,-1]) A1 = sp.Matrix([1,0,1]) A2 = sp.Matrix([2,1,0]) A3 = sp.Matrix([1,1,1]) B = sp.Matrix.hstack(B1,B2,B3) # 3x3 # We want to solve for c = (c1,c2,c3) such that A1 = B*c C1 = B.LUsolve(A1) C2 = B.LUsolve(A2) C3 = B.LUsolve(A3) P = sp.Matrix.hstack(C1,C2,C3) P_simpl = sp.simplify(P) P_simpl

Python

python
B1 = sp.Matrix([1,2,-1]) B2 = sp.Matrix([2,2,-1]) B3 = sp.Matrix([2,-1,-1]) A1 = sp.Matrix([1,0,1]) A2 = sp.Matrix([2,1,0]) A3 = sp.Matrix([1,1,1]) B = sp.Matrix.hstack(B1,B2,B3) # 3x3 # We want to solve for c = (c1,c2,c3) such that A1 = B*c C1 = B.LUsolve(A1) C2 = B.LUsolve(A2) C3 = B.LUsolve(A3) P = sp.Matrix.hstack(C1,C2,C3) P_simpl = sp.simplify(P) P_simpl

Мысль

Testing product reliability

I’m piecing together the transition matrix and considering Bayes approach to estimate the probability of a product being good based on test outcomes.

Reasoned for 9 seconds

Ответ

答案汇总:


一、单选题

  1. 设函数 f(x)f(x) 在区间 (1,1)(-1,1) 内有定义,且在点 x=0x=0 处连续,则以下结论:
  • (1) 当 limx0f(x)x3=0\lim\limits_{x\to 0}\frac{f(x)}{\sqrt[3]{x}}=0 时,f(x)f(x) 在点 x=0x=0 处可导;
  • (2) 当 limx0f(x)x2=0\lim\limits_{x\to 0}\frac{f(x)}{x^2}=0 时,f(x)f(x) 在点 x=0x=0 处可导;
  • (3) 当 f(x)f(x) 在点 x=0x=0 处可导时,limx0f(x)x3=0\lim\limits_{x\to 0}\frac{f(x)}{\sqrt[3]{x}}=0

分析要点:

  • (1) 不一定为真,可举反例 f(x)=x2/3f(x)=x^{2/3} 等来说明。
  • (2) 若 limx0f(x)x2=0\lim_{x\to 0}\frac{f(x)}{x^2}=0,可推出 limx0f(x)x=0\lim_{x\to 0}\frac{f(x)}{x}=0,从而可导且导数为零。
  • (3) 若 ff 在 0 处可导(且通常隐含 f(0)=0f(0)=0),则 f(x)xf(0)\frac{f(x)}{x} \to f'(0),而 x3\sqrt[3]{x}xx“大”得多,因此 f(x)x30\frac{f(x)}{\sqrt[3]{x}}\to 0。故为真。

正确结论是(2)和(3)。

答案:C (2)(3)(2)(3)


  1. 设正项数列 {an},{bn},{cn}\{a_n\},\{b_n\},\{c_n\},则“ancn\sum a_n c_ncnbn\sum c_n b_n 均收敛”是“anbn\sum a_n b_n 收敛”的什么条件?

分析要点:

  • 可以通过构造反例来检验“充分”或“必要”:
    • ancn\sum a_n c_nbncn\sum b_n c_n 同时收敛,不一定能推出 anbn\sum a_n b_n 收敛(可构造 an=bn=1na_n=b_n=\frac{1}{\sqrt{n}}, cn=1nc_n=\frac{1}{n} 反例)。
    • 也不必然是“必要条件”。
  • 故既非充分又非必要。

答案:D (既非充分又非必要条件)


  1. f(x,y)f(x,y) 可微,p(x,y),p0(x0,y0)p(x,y),p_0(x_0,y_0) 为曲面 z=f(x,y)z=f(x,y) 上的点,则下列极限描述了可微的定义:
limpp0f(p)[f(p0)  +  fp0(p0p)]p0p  =  0.\lim_{p\to p_0}\frac{f(p) - \bigl[f(p_0) \;+\;\nabla f|_{p_0}\cdot(\overrightarrow{p_0p})\bigr]}{\|\overrightarrow{p_0p}\|}\;=\;0.

对照选项可知,唯一符合

f(p)    [f(p0)  +  fp0(p0p)]f(p)\;-\;\bigl[f(p_0)\;+\;\nabla f|_{p_0}\cdot(\overrightarrow{p_0p})\bigr]

的形式是

答案:D


  1. 设正项数列 {an}\{a_n\} 单调增加,则以下选项中使得 {an}\{a_n\} 收敛的是
  1. {(1+an)1an}\bigl\{(1+a_n)^{\tfrac{1}{a_n}}\bigr\} 收敛于 1
  2. {(1+1an)an}\bigl\{(1+\tfrac{1}{a_n})^{a_n}\bigr\} 收敛于 ee
  3. {anlnan}\{a_n\ln a_n\} 收敛于 0
  4. {lnanan}\{\tfrac{\ln a_n}{a_n}\} 收敛于 0

分析要点:

  • 若 (1) 1\to 1,往往暗示 ana_n\to\infty;不保证 {an}\{a_n\} 有有限极限。
  • 若 (2) e\to e,同样多半暗示 ana_n\to\infty
  • 若 (3) anlnan0a_n\ln a_n\to 0,只能在 an>0a_n>0 且单调情况下推知 an1a_n\to 1(因为 an0+a_n \to 0^+ 与单调递增矛盾,anc>1a_n\to c>1limanlnan0\lim a_n\ln a_n \neq 0)。
    • 事实上这强迫 an1a_n\to 1,从而收敛于有限值。
  • 若 (4) lnanan0\frac{\ln a_n}{a_n}\to 0 可能对应 ana_n\to\infty(因为 lnx/x0\ln x / x \to 0xx\to\infty)。

因此只有 (3) 才能保证 {an}\{a_n\} 收敛到某个有限值。

答案:C


  1. AA 为 3 阶矩阵,若 EA=1,EA=1,2EA=2\lvert E - A\rvert = 1,\quad \lvert -E - A\rvert = -1,\quad \lvert 2E - A\rvert = 2, 则 3EA= ?\lvert 3E - A\rvert = \ ?

分析要点:

  • p(t)=det(tEA)p(t)=\det(tE - A)p(t)p(t) 是三次多项式,系数由 trace(A)\mathrm{trace}(A)det(A)\det(A) 等确定。
  • 给定 p(1)=1,  p(1)=1,  p(2)=2p(1)=1,\;p(-1)=-1,\;p(2)=2,解出 p(3)p(3)
  • 具体求解得 p(t)=t32t2+2p(t)=t^3 -2t^2 +2,代入 t=3t=3,结果为 11。

答案:D (11)


A=(122313),b=(412).A= \begin{pmatrix} -1 & 2\\ 2 & -3\\ -1 & 3 \end{pmatrix},\quad b= \begin{pmatrix} 4\\ 1\\ 2 \end{pmatrix}.

若对任意 2 维实列向量 xx,都有 bAx0    bAx\|\,b - A\,x_0\|\;\le\;\|\,b - A\,x\|, 则求 x0=?x_0=?

分析要点:

  • 这是最小二乘问题:minxbAx\min_x \|b - A x\|
  • 其解满足正规方程 ATAx0=ATbA^\mathsf{T} A\,x_0 = A^\mathsf{T}b
  • 直接解即可。

计算得

x0=(32).x_0 = \begin{pmatrix} 3\\[6pt] 2 \end{pmatrix}.

答案:D   (3,2)\;(3,\,2)


  1. α1=(1,1,a),  α2=(1,0,1),  α3=(1,b,1)\alpha_{1}=(1,-1,a),\;\alpha_{2}=(-1,0,1),\;\alpha_{3}=(1,b,1) 是 3 阶实矩阵 AA 的 3 个互异特征值的特征向量,则“(a,b)=(1,2)(a,b)=(1,2)”是“AA 为对称矩阵”的?

分析要点:

  • 对称实矩阵在不同特征值处对应的特征向量必两两正交。
  • α1α2=0\alpha_1\cdot \alpha_2 = 0α2α3=0\alpha_2\cdot \alpha_3 = 0α1α3=0\alpha_1\cdot \alpha_3 = 0 即可解出 (a,b)=(1,2)(a,b)=(1,2)
  • 反之若 (a,b)=(1,2)(a,b)=(1,2) 确保这三向量两两正交,也可构成正交特征向量组,说明可以是对称矩阵。

故该条件既是充分又是必要。

答案:A (充要条件)


  1. 设总体 XU(a,b)X\sim U(a,b)a,ba,b未知,令 L=baL=b-a。样本 X1,,XnX_1,\dots,X_n 来自该总体,记
X(1)=min{Xi},X(n)=max{Xi},X=1nXi,S12=1n(XiX)2.X_{(1)}=\min\{X_i\},\quad X_{(n)}=\max\{X_i\},\quad \overline{X}=\frac{1}{n}\sum X_i,\quad S^2_1=\frac{1}{n}\sum\bigl(X_i-\overline{X}\bigr)^2.

LL 的矩估计量与最大似然估计量分别是?

分析要点:

  • 矩估计:基于 Var(X)=(ba)212\mathrm{Var}(X) = \frac{(b-a)^2}{12};题中 S12S^2_1 采取 1n\frac{1}{n} 而非 1n1\frac{1}{n-1},因此 L矩估计=12S12=23S1L_\text{矩估计}=\sqrt{12}\,\sqrt{S^2_1}=2\sqrt{3}\,S_1
  • 最大似然估计:对均匀分布 [a,b][a,b] 的极大似然给出 a^=X(1),  b^=X(n)\hat{a}=X_{(1)},\;\hat{b}=X_{(n)},故 L^=X(n)X(1)\hat{L}=X_{(n)}-X_{(1)}

答案:C (23S1,  X(n)X(1))\bigl(2\sqrt{3}\,S_1,\;X_{(n)} - X_{(1)}\bigr)


  1. 设在条件 X=xX=x 下,YN(x,x2)Y\sim N(x,x^2),且 XU(0,1)X\sim U(0,1)。则 X,YX,Y 的相关系数为?

分析要点: 用公式

ρX,Y  =  Cov(X,Y)Var(X)Var(Y).\rho_{X,Y} \;=\;\frac{\mathrm{Cov}(X,Y)}{\sqrt{\mathrm{Var}(X)}\,\sqrt{\mathrm{Var}(Y)}}.
  • E[X]=12,Var(X)=112\mathrm{E}[X]=\tfrac12,\quad \mathrm{Var}(X)=\tfrac{1}{12}
  • E[Y]=E[E[YX]]=E[X]=12\mathrm{E}[Y]=\mathrm{E}[\mathrm{E}[Y\mid X]]= \mathrm{E}[X]=\tfrac12
  • E[XY]=E[XE[YX]]=E[X2]=13\mathrm{E}[XY] = \mathrm{E}[\,X\,\mathrm{E}[Y\mid X]\,]=\mathrm{E}[X^2]=\tfrac13.
    • 因此 Cov(X,Y)=1314=112\mathrm{Cov}(X,Y)=\tfrac13-\tfrac14=\tfrac{1}{12}
  • Var(Y)=E[Var(YX)]+Var(E[YX])=E[X2]+Var(X)=13+112=512.\mathrm{Var}(Y)= \mathrm{E}[\mathrm{Var}(Y\mid X)] + \mathrm{Var}(\mathrm{E}[Y\mid X]) = \mathrm{E}[X^2] + \mathrm{Var}(X) = \tfrac13 + \tfrac{1}{12}=\tfrac{5}{12}.

ρX,Y=112112512=1121125=15=55.\rho_{X,Y} = \frac{\tfrac{1}{12}}{\sqrt{\tfrac{1}{12}}\,\sqrt{\tfrac{5}{12}}} = \frac{\tfrac{1}{12}}{\tfrac{1}{12}\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}.

选项中对应 5/5\sqrt{5}/5

答案:B


二、填空题

下面按照题目序号(11~16)依次给出答案(或简要计算过程)。


(11)f(x,y,z)=x2eyz2f(x,y,z)=x^2 e^{yz^2},则它在点 (1,0,1)(-1,0,1) 处的方向导数的最小值为多少?

  • 梯度 f=(fx,fy,fz),\nabla f = \Bigl(\,\frac{\partial f}{\partial x},\,\frac{\partial f}{\partial y},\,\frac{\partial f}{\partial z}\Bigr), 其中 fx=2xeyz2,fy=x2eyz2z2,fz=x2eyz2(2yz).\frac{\partial f}{\partial x}=2x\,e^{yz^2},\quad \frac{\partial f}{\partial y}=x^2 e^{yz^2}\,z^2,\quad \frac{\partial f}{\partial z}=x^2 e^{yz^2}\,(2yz).
  • 在点 (1,0,1)(-1,0,1) 处,e0=1e^{0}=1,且 y=0y=0 导致第 3 项为 0。代入得 f(1,0,1)=(2,  1,  0).\nabla f(-1,0,1) = (-2,\;1,\;0).
  • 方向导数在单位向量 u\mathbf{u} 方向的值为 fu\nabla f\cdot \mathbf{u},其最小值即 f-\|\nabla f\| 对应与 -梯度方向。
  • f=(2)2+12+02=5.\|\nabla f\|=\sqrt{(-2)^2 + 1^2 + 0^2}=\sqrt{5}.

答案:5\boxed{-\sqrt{5}}


(12)LL 为点 (0,1)(0,1) 到点 (1,2)(1,2) 的直线段,求弧线积分 Lxyds\displaystyle\int_{L}\sqrt{x\,y}\,ds.

  • 参数化:从 (0,1)(0,1)(1,2)(1,2) 可令 r(t)=(x(t),y(t))=(t,1+t),t[0,1].\mathbf{r}(t) = (x(t),\,y(t)) = (t,\,1+t),\quad t\in [0,1].
  • ds=(dxdt)2+(dydt)2dt=12+12dt=2dt.ds=\sqrt{\bigl(\tfrac{dx}{dt}\bigr)^2+\bigl(\tfrac{dy}{dt}\bigr)^2}\,dt=\sqrt{1^2 + 1^2}\,dt=\sqrt2\,dt.
  • 被积函数 xy=t(1+t)\sqrt{x\,y}=\sqrt{\,t\,(1+t)}.
  • 故积分为 01t(1+t)2dt=201t+t2dt.\int_0^1 \sqrt{t(1+t)}\,\sqrt2\,dt = \sqrt2 \,\int_0^1 \sqrt{t+t^2}\,dt.
  • 该定积分的结果可用符号计算或合适的替换。最终可化得闭形式 201t(1+t)dt  =  32    24ln(2+1).\sqrt2\int_0^1 \sqrt{t(1+t)}\,dt \;=\;\frac{3}{2}\;-\;\frac{\sqrt2}{4}\,\ln\bigl(\sqrt2+1\bigr)\,. 数值约为 1.18841.1884\ldots

答案(任意等价形式均可):

32    24ln(2+1).\boxed{\frac{3}{2}\;-\;\frac{\sqrt{2}}{4}\ln(\sqrt{2}+1)\,.}

(13) 微分方程 dydx=x+2y3x+1\frac{dy}{dx}=\frac{x+2y-3}{x+1} 满足 y(0)=2y(0)=2 的解。

  • 改写为 (x+1)dydx=x+2y3(x+1)\,\frac{dy}{dx}=x+2y-3,即 dydx2x+1y  =  x3x+1.\frac{dy}{dx}-\frac{2}{\,x+1\,}\,y\;=\;\frac{x-3}{\,x+1\,}.
  • 这是线性方程,积分因子 μ(x)=exp ⁣(2x+1dx)=1(x+1)2\mu(x)=\exp\!\Bigl(\int -\tfrac{2}{x+1}dx\Bigr)=\tfrac{1}{(x+1)^2}
  • 解得 y=1x+C(x+1)2.\displaystyle y=1 - x + C\,(x+1)^2.
  • 由初值 y(0)=2y(0)=2 算得 C=1C=1,故 y=1x+(x+1)2  =  x2+x+2.y = 1 - x + (x+1)^2 \;=\; x^2 + x + 2.

答案:y=x2+x+2\boxed{y=x^2 + x +2}


(14)

Ω={(x,y,z)  x2yx,  0z2xy},\Omega=\Bigl\{(x,y,z)\,\Bigm|\;x^2 \le y\le x,\;0\le z\le 2 - x - y\Bigr\},

则三重积分 Ωxey+zdxdydz=?\displaystyle\iiint_\Omega x\,e^{\,y+z}\,dxdydz = ?

  • 先确定 xx 范围:由 x2xx^2 \le x(且 x2yxx^2\le y\le x),可知 x[0,1]x\in[0,1]
  • 对固定 xxy[x2,x]y\in[x^2,\,x];再对固定 (x,y)(x,y)z[0,2(x+y)]z\in[\,0,\,2-(x+y)\,]
  • 积分写作 01 ⁣ ⁣x2x ⁣ ⁣02xyxey+zdzdydx.\int_{0}^{1} \!\!\int_{\,x^2}^{x} \!\!\int_{\,0}^{2-x-y} x\,e^{\,y+z}\,dz\,dy\,dx\,.
  • 先对 zz 积分:02xyeyezdz=ey[ez]02xy=ey(e2xy1)\int_0^{2-x-y} e^{\,y} e^{\,z}\,dz = e^y \bigl[e^z\bigr]_{0}^{\,2-x-y} = e^y\bigl(e^{\,2-x-y}-1\bigr). 乘以 xxxey(e2xy1)x\,e^y\bigl(e^{\,2-x-y}-1\bigr).
  • 展开并对 yy 积分,再对 xx 积分,最后可用分部或软件求得精确值。经计算结果为(用符号化简): 4e2  +  232e    32    0.2040 -4\,e^2 \;+\; \frac{23}{2}\,e \;-\;\frac{3}{2} \;\approx\;0.2040\ldots 或可写成 12[8e2+23e3].\tfrac12\,\bigl[-8\,e^2 \,+\,23\,e \,-\,3\bigr].

答案:4e2  +  232e    32\displaystyle\boxed{-\,4\,e^2 \;+\;\tfrac{23}{2}\,e \;-\;\tfrac{3}{2}}


(15)R3R^3 中两组基:

α1=(101),  α2=(210),  α3=(111),β1=(121),  β2=(221),  β3=(211).\alpha_{1}= \begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix},\; \alpha_{2}= \begin{pmatrix} 2\\ 1\\ 0 \end{pmatrix},\; \alpha_{3}= \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}, \quad \beta_{1}= \begin{pmatrix} 1\\ 2\\ -1 \end{pmatrix},\; \beta_{2}= \begin{pmatrix} 2\\ 2\\ -1 \end{pmatrix},\; \beta_{3}= \begin{pmatrix} 2\\ -1\\ -1 \end{pmatrix}.

{αi}\{\alpha_i\}{βi}\{\beta_i\} 的过渡矩阵是?

  • 要求把每个 αi\alpha_i 表示成 β\beta-基的坐标向量 c\mathbf{c},即 αi=Bc\alpha_i = B\,\mathbf{c}
  • 将这三个 αi\alpha_i 分别解出坐标 ci\mathbf{c_i},堆为矩阵的三列,即得过渡矩阵。
  • 计算(或用软件)可得: P=(3238373323131).P= \begin{pmatrix} -3 & -2 & -3 \\ \frac{8}{3} & \frac{7}{3} & 3 \\ -\tfrac{2}{3} & -\tfrac{1}{3} & -1 \end{pmatrix}.

答案:(\displaystyle \begin{pmatrix} -3 & -2 & -3\ \frac{8}{3} & \frac{7}{3} & 3\ -\frac{2}{3} & -\frac{1}{3} & -1 \end{pmatrix}).


(16) 甲、乙两种检验设备均有 90% 的正确检出率且相互独立,现有一批合格率为 10% 的产品。任取一件,经两种设备检验均判为“合格品”,则其真正合格的概率是多少?

  • 设事件 GG 表示“产品真合格”,则 P(G)=0.1P(G)=0.1。事件 BB 表示“产品不合格”,P(B)=0.9P(B)=0.9
  • 两种设备若产品合格,则判“合格”概率均 0.9;若产品不合格,则判“合格”概率是 0.1(因为其正确率 0.9 判不合格)。
  • 事件 TT =“两台设备都测为合格”。 P(TG)=0.9×0.9=0.81,P(TB)=0.1×0.1=0.01.P(T\mid G)=0.9\times 0.9 =0.81,\quad P(T\mid B)=0.1\times 0.1 =0.01.
  • 由 Bayes 公式: P(GT)=P(TG)P(G)P(TG)P(G)  +  P(TB)P(B)=0.81×0.10.81×0.1  +  0.01×0.9=0.0810.081+0.009=0.9.P(G \mid T) = \frac{P(T\mid G)\,P(G)}{\,P(T\mid G)\,P(G)\;+\;P(T\mid B)\,P(B)\,} = \frac{0.81\times 0.1}{\,0.81\times 0.1 \;+\;0.01\times 0.9\,} = \frac{0.081}{0.081+0.009} = 0.9.

答案:0.9\boxed{0.9}


总结

  • 单选 (1)-(9) 的答案依次是:

    1. C
    2. D
    3. D
    4. C
    5. D
    6. D
    7. A
    8. C
    9. B
  • 填空 (11)-(16) 的最终结果分别为:

    1. 5-\sqrt{5}
    2. 32    24ln(2+1)\displaystyle \frac{3}{2}\;-\;\frac{\sqrt{2}}{4}\ln(\sqrt{2}+1)
    3. y=x2+x+2y=x^2 + x +2
    4. 4e2  +  232e    32\displaystyle -\,4\,e^2 \;+\;\frac{23}{2}\,e \;-\;\frac{3}{2}
    5. (\displaystyle \begin{pmatrix} -3 & -2 & -3 \ \tfrac{8}{3} & \tfrac{7}{3} & 3 \ -\tfrac{2}{3} & -\tfrac{1}{3} & -1 \end{pmatrix})
    6. 0.90.9

祝考试顺利。

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