MikaStars39/nano-eval huggingface dataset:
What was the main reason for the U.S. government's implementation of a neutral policy towards a series of international issues in the 1930s?

Options:
A. To focus on internal political reforms
B. Pacifism and isolationism prevail.
C. To avoid war and ensure self-preservation.
D. Addressing the Economic Crisis
E. Lack of public support for international intervention
F. Uncertainty about the outcome of foreign conflicts
G. Influence of non-interventionist political leaders
H. To reduce military expenditure
I. To strengthen trade relations
J. Military and political strength inferior to Britain and France
Answer: D
In the 1930s, as the "relative stability" of the capitalist world disappeared, Japan's () finally went bankrupt.

Options:
A. Imperialist Agenda
B. Dollar Diplomacy
C. China Policy
D. Coordinated Diplomacy
E. Peaceful Diplomacy
F. Economic Isolationism
G. Expansionist Strategy
H. Trade Partnership
I. Agricultural Reform
J. Military Aggression
Answer: D
Driven by the Second Industrial Revolution, the world market was finally formed, and which country dominated it?

Options:
A. Netherlands
B. Australia
C. Germany
D. Japan
E. Industrial countries of Europe and America
F. Russia
G. Britain
H. France
I. Italy
J. Spain
Answer: E
Who is the thinker who advocates not only the separation of the three kinds of state power to achieve the purpose of mutual checks and balances, but also the better implementation of checks and balances on power through the separation of state and local governments?

Options:
A. Alexis de Tocqueville
B. James Madison
C. John Locke
D. Friedrich Hayek
E. Thomas Jefferson
F. Charles de Secondat, Baron de Montesquieu
G. Immanuel Kant
H. John Stuart Mill
I. Herbert Spencer
J. Jean-Jacques Rousseau
Answer: E
Who was the inventor of the world's first internal combustion engine?

Options:
A. Nikola Tesla
B. Henry Ford
C. Gramme,Zénobe Théophile
D. Marconi, Guglielmo
E. Ernst Werner von Siemens
F. Wilhelm Maybach
G. Carl Friedrich Benz
H. James Watt
I. Gottlieb Daimler
J. Rudolf Diesel
Answer: G
Which year did the whole of India become a British colony?

Options:
A. 1867
B. 1901
C. 1882
D. 1858
E. 1890
F. 1824
G. 1866
H. 1849
I. 1837
J. 1875
Answer: H
What is the main premise of the liberal reform movement in the capitalist countries?

Options:
A. The suppression of socialist ideologies by capitalist states
B. The need of the bourgeoisie for world domination
C. The strength of the industrial bourgeoisie grew
D. The decline of mercantilist practices in Western Europe
E. The creation of a world government to oversee economic disparities
F. The increase in labor unions and worker rights movements
G. The rise of technology and its effect on manual labor industries
H. The impact of Adam Smith's publication of The Wealth of Nations
I. The inevitable requirement for further lifting the feudal bondage
J. Expansion of global trade networks in the 20th century
Answer: C
If the equation $\frac{dy}{dx} = y^2$ is defined in the strip $-2 < x < 5$, then the interval of existence for its solution passing through the point $(1, -1)$ is ().

Options:
A. (0,5)
B. (1,6)
C. (-5,-2)
D. (-1,4)
E. (-2,3)
F. (-3,2)
G. (-3,-5)
H. (2,6)
I. (-2, -4)
Answer: A
Given the second-order differential equation $y^{\prime\prime}+x y^{\prime\prime}+3 y^{\prime}-4 x^{2} y=\mathrm{e}^{2 \prime}$ Which of the following systems of first-order equations is equivalent to the given second-order differential equation?

Options:
A. $$
\left\{\begin{aligned} {{{y_{1}^{\prime}}}} & {{} {{}=3y_{2}}} \\ {{{y_{2}^{\prime}}}} & {{} {{}=y_{3}}} \\ {{{y_{3}^{\prime}}}} & {{} {{}=4 x^{2} y_{1}-3 y_{2}-x y_{4}+e^{2 x}}} \\ \end{aligned} \right.
$$
B. $$
\left\{\begin{aligned} {{{y_{1}^{\prime}}}} & {{} {{}=y_{2}}} \\ {{{y_{2}^{\prime}}}} & {{} {{}=y_{3}}} \\ {{{y_{3}^{\prime}}}} & {{} {{}=4 x^{2} y_{1}-3 y_{2}-x y_{3}+e^{2 x}}} \\ \end{aligned} \right.
$$
C. $$
\left\{\begin{aligned} {{{y_{1}^{\prime}}}} & {{} {{}=y_{2}}} \\ {{{y_{2}^{\prime}}}} & {{} {{}=3y_{3}}} \\ {{{y_{3}^{\prime}}}} & {{} {{}=4 x^{2} y_{1}-3 y_{2}-x y_{3}+e^{2 x}}} \\ \end{aligned} \right.
$$
D. $$
\left\{\begin{aligned} {{{y_{1}^{\prime}}}} & {{} {{}=y_{2}}} \\ {{{y_{2}^{\prime}}}} & {{} {{}=y_{3}}} \\ {{{y_{3}^{\prime}}}} & {{} {{}=4 x^{3} y_{1}-2 y_{2}-x y_{3}+e^{4 x}}} \\ \end{aligned} \right.
$$
E. $$
\left\{\begin{aligned} {{{y_{1}^{\prime}}}} & {{} {{}=y_{2}}} \\ {{{y_{2}^{\prime}}}} & {{} {{}=y_{3}}} \\ {{{y_{3}^{\prime}}}} & {{} {{}=4 x^{3} y_{1}-3 y_{2}-x y_{3}+e^{2 x}}} \\ \end{aligned} \right.
$$
F. $$
\left\{\begin{aligned} {{{y_{1}^{\prime}}}} & {{} {{}=2y_{2}}} \\ {{{y_{2}^{\prime}}}} & {{} {{}=2y_{3}}} \\ {{{y_{3}^{\prime}}}} & {{} {{}=4 x^{2} y_{1}-3 y_{2}-x y_{3}+e^{2 x}}} \\ \end{aligned} \right.
$$
G. $$
\left\{\begin{aligned} {{{y_{1}^{\prime}}}} & {{} {{}=y_{2}}} \\ {{{y_{2}^{\prime}}}} & {{} {{}=y_{3}}} \\ {{{y_{3}^{\prime}}}} & {{} {{}=4 x^{2} y_{1}-2 y_{2}-x y_{3}+e^{3 x}}} \\ \end{aligned} \right.
$$
H. $$
\left\{\begin{aligned} {{{y_{1}^{\prime}}}} & {{} {{}=4y_{2}}} \\ {{{y_{2}^{\prime}}}} & {{} {{}=2y_{3}}} \\ {{{y_{3}^{\prime}}}} & {{} {{}=4 x^{2} y_{1}-3 y_{2}-x y_{3}+e^{2 x}}} \\ \end{aligned} \right.
$$
I. $$
\left\{\begin{aligned} {{{y_{1}^{\prime}}}} & {{} {{}=2y_{2}}} \\ {{{y_{2}^{\prime}}}} & {{} {{}=y_{3}}} \\ {{{y_{3}^{\prime}}}} & {{} {{}=4 x^{3} y_{1}-3 y_{2}-x y_{3}+e^{23x}}} \\ \end{aligned} \right.
$$
J. $$
\left\{\begin{aligned} {{{y_{1}^{\prime}}}} & {{} {{}=2y_{2}}} \\ {{{y_{2}^{\prime}}}} & {{} {{}=y_{3}}} \\ {{{y_{3}^{\prime}}}} & {{} {{}=4 x^{2} y_{1}-3 y_{2}-x y_{3}+e^{2 x}}} \\ \end{aligned} \right.
$$
Answer: B
The equation of the curve that is orthogonal to the family of integral curves defined by the equation $( x \mathrm{l n} x ) \mathrm{d} x-( y-\mathrm{l n} x ) \mathrm{d} y=0$ and passes through the point $( e, 1 )$ is ( ).

Options:
A. $y={\frac{1} {2}} \left( x \ln x+{\frac{x} { \ln x}} \right)$
B. $y={\frac{1} {2}} \left( \ln y+{\frac{1} {x}} \right)$
C. $y={\frac{1} {2}} \left( x \ln x+{\frac{1} { \ln x}} \right)$
D. $y={\frac{1} {2}} \left( \ln x+{\frac{1} {\ln x}} \right)$
E. $y={\frac{1} {2}} \left( \ln x+{\frac{1} {y}} \right)$
F. $y={\frac{1} {2}} \left( x+{\frac{1} {x}} \right)$
G. $y={\frac{1} {2}} \left( \ln y+{\frac{1} {e}} \right)$
H. $y={\frac{1} {2}} \left( \ln x-{\frac{1} {\ln x}} \right)$
I. $y={\frac{1} {2}} \left( \ln x+\ln y \right)$
J. $y={\frac{1} {2}} \left( \ln y+{\frac{1} {\ln y}} \right)$
Answer: D
Let the function$y=y \left( x \right)$ satisfy $\Delta y=\frac{x y} {1+x^{2}} \Delta x+o \left( \Delta x \right)$, and$y \left( 0 \right)=2$ Then $y \left( x \right)=()$ .

Options:
A. $$2 \sqrt{1+3x^{2}}$$
B. $$2 \sqrt{x^{2}+3}$$
C. $$2 \sqrt{1+x^{2}}$$
D. $$2 \sqrt{2-x^{2}}$$
E. $$2 \sqrt{1-x^{2}}$$
F. $$\frac{2}{\sqrt{x^{2}+1}}$$
G. $$\frac{2}{\sqrt{1+x^{2}}}$$
H. $$\frac{2}{\sqrt{1-x^{2}}}$$
I. $$2 \sqrt{x^{2}-1}$$
J. $$2 \sqrt{x^{2}+1}$$
Answer: C
Someone's food intake is 2500 calories/day, of which 1200 calories/day is used for basic metabolism, in fitness exercise, he consumes 16 calories/kg/day multiplied by his weight, assuming that the calories stored in the form of fat are 100% effective, and a kilogram of fat contains 10,000 calories, how does the person's weight change over time().

Options:
A. $$w \left( t \right)=\frac{3 2 5} {4}+\left( w_{0}-\frac{3 2 5} {4} \right) \mathrm{e}^{-\frac{t} {5 2 5}}$$
B. $$w \left( t \right)=\frac{3 2 5} {6}+\left( w_{0}-\frac{3 2 5} {6} \right) \mathrm{e}^{-\frac{t} {6 2 5}}$$
C. $$w \left( t \right)=\frac{3 2 5} {4}+\left( 3w_{0}-\frac{3 2 5} {4} \right) \mathrm{e}^{-\frac{t} {6 2 5}}$$
D. $$w \left( t \right)=\frac{3 2 5} {4}+\left( w_{0}-\frac{3 2 5} {4} \right) \mathrm{e}^{-\frac{t} {6 7 5}}$$
E. $$w \left( t \right)=\frac{3 2 5} {4}+\left( 2w_{0}-\frac{3 2 5} {4} \right) \mathrm{e}^{-\frac{t} {6 2 5}}$$
F. $$w \left( t \right)=\frac{3 2 5} {4}+\left( w_{0}-\frac{3 50} {4} \right) \mathrm{e}^{-\frac{t} {6 2 5}}$$
G. $$w \left( t \right)=\frac{3 2 5} {4}+\left( 2w_{0}-\frac{3 5 5} {4} \right) \mathrm{e}^{-\frac{t} {6 2 5}}$$
H. $$w \left( t \right)=\frac{3 2 5} {4}+\left( 6w_{0}-\frac{3 2 5} {4} \right) \mathrm{e}^{-\frac{t} {6 2 5}}$$
I. $$w \left( t \right)=\frac{3 2 5} {4}+\left( w_{0}-\frac{3 2 5} {4} \right) \mathrm{e}^{-\frac{t} {6 2 5}}$$
J. $$w \left( t \right)=\frac{3 2 5} {4}+\left( w_{0}-\frac{3 7 5} {4} \right) \mathrm{e}^{-\frac{t} {6 2 5}}$$
Answer: I
For a known first-order linear homogeneous system of equations $x^{\prime}=A(t)x$, the eigenvalues are 1 and 2, and their corresponding eigenvectors are respectively ${\binom{1} {1}} \cdot{\binom{1} {0}}$,Then the coefficient matrix $\mathrm{A} \ (\mathbf{t})=$______.

Options:
A. $$\left( \begin{matrix} {{1}} & {{-2}} \\ {{0}} & {{2}} \\ \end{matrix} \right)$$
B. $$\left( \begin{matrix} {{1}} & {{1}} \\ {{0}} & {{2}} \\ \end{matrix} \right)$$
C. $$\left( \begin{matrix} {{1}} & {{-1}} \\ {{0}} & {{2}} \\ \end{matrix} \right)$$
D. $$\left( \begin{matrix} {{2}} & {{1}} \\ {{0}} & {{1}} \\ \end{matrix} \right)$$
E. $$\left( \begin{matrix} {{1}} & {{2}} \\ {{2}} & {{1}} \\ \end{matrix} \right)$$
F. $$\left( \begin{matrix} {{1}} & {{0}} \\ {{-1}} & {{2}} \\ \end{matrix} \right)$$
G. $$\left( \begin{matrix} {{2}} & {{-2}} \\ {{0}} & {{1}} \\ \end{matrix} \right)$$
H. $$\left( \begin{matrix} {{2}} & {{-1}} \\ {{0}} & {{1}} \\ \end{matrix} \right)$$
I. $$\left( \begin{matrix} {{2}} & {{0}} \\ {{-1}} & {{1}} \\ \end{matrix} \right)$$
J. $$\left( \begin{matrix} {{1}} & {{-2}} \\ {{1}} & {{2}} \\ \end{matrix} \right)$$
Answer: H
If use substitution$\mathrm{y}=\mathrm{z}^{\mathrm{m}}$differential equation can be $\mathrm{y^{\prime}=a x^{a}+b y^{\beta}}$ $$ ( \alpha\beta\neq0)$$into first-order homogeneous equation $\mathrm{d z} / \mathrm{d x}=\mathrm{f} \ \ ( \mathrm{z} / \mathrm{x} )$, $a$, $\beta$ is () shall meet the conditions.

Options:
A. $$1 / \beta+1 / \alpha=1$$
B. $$2 / \beta-2 / \alpha=1$$
C. $$1 / \beta+1 / \alpha=-1$$
D. $$2 / \beta-1 / \alpha=2$$
E. $$1 / \beta-2 / \alpha=1$$
F. $$2 / \beta-1 / \alpha=1$$
G. $$1 / \beta-1 / \alpha=-1$$
H. $$1 / \beta-1 / \alpha=2$$
I. $$1 / \beta-1 / \alpha=1$$
J. $$1 / \beta-1 / \alpha=0$$
Answer: I
Satisfy the equation $\int_{0}^{1} f \left(t x \right) \! \mathrm{d} t=n f \left(x \right)$(n is a natural number greater than 1) the derivable function f (x) is ().

Options:
A. $$\mathrm{C x}^{(2+x^2) \ / 2x}$$
B. $$\mathrm{C x} ^{( 1-x ) \ / x}$$
C. $$\mathrm{C x}^{(x- 3) \ / 3x}$$
D. $$
\mathrm{C s i n n x}
$$
E. $$\mathrm{C x}^{(x+2) \ / x}$$
F. $$\mathrm{C x}^{(-2+x) \ / 2x}$$
G. $$
\operatorname{Cc o s n x}
$$
H. $$\mathrm{C x} ^{( 1+x ) \ / x}$$
I. $$
\mathrm{C x}
$$
J. $$\mathrm{C x} ^{( x -1 ) \ / 2x}$$
Answer: B
A curve at right angles to each of a family of curves is called the orthogonal trajectory of the given family of curves. If the family of curves is $\mathbf{x}^{2}+\mathbf{y}^{2}=2 \mathbf{c} \mathbf{x}$(c is constant), then the orthogonal trajectory of the family of curves is ______.

Options:
A. $$\mathrm{y} = \mathrm{c}_{1} \ ( \mathrm{y}^{2} + \mathrm{x}^{3} )$$
B. $$\mathrm{y} \!=\! \mathrm{c}_{1} \ \ ( \mathrm{x}^{4} \!+\! \mathrm{y}^{} )$$
C. $$\mathrm{y} \!=\! \mathrm{c}_{1} \ \ ( \mathrm{x}^{4} \!+\! \mathrm{y}^{4} )$$
D. $$\mathrm{y} \!=\! \mathrm{c}_{1} \ \ ( \mathrm{x}^{3} \!+\! \mathrm{y}^{4} )$$
E. $$\mathrm{y} \!=\! \mathrm{c}_{1} \ \ ( \mathrm{x}^{3} \!+\! 2\mathrm{y}^{3} )$$
F. $$\mathrm{y} = \mathrm{c}_{1} \ ( \mathrm{x}^{2} + \mathrm{2y}^{2} )$$
G. $$\mathrm{y} \!=\! \mathrm{c}_{1} \ \ ( \mathrm{3x}^{2} \!+\! \mathrm{y}^{2} +x)$$
H. $$\mathrm{y} \!=\! \mathrm{c}_{1} \ \ ( \mathrm{2x}^{2} \!+\! \mathrm{y}^{2}+2x )$$
I. $$\mathrm{y} \!=\! \mathrm{c}_{1} \ \ ( \mathrm{x}^{2} \!+\! \mathrm{y}^{2} )$$
Answer: I
The train travels at a speed of 20 m/s (equivalent to 72 km/h) on a straight track. When braking, the train achieves an acceleration of -0.4 m²/s. How long does it take for the train to stop after braking and how much distance does the train travel during this time?

Options:
A. The train takes 50 seconds to come to a complete stop after initiating braking, and it travels a distance of 600 meters during this time.
B. The train takes 20 seconds to come to a complete stop after initiating braking, and it travels a distance of 300 meters during this time.
C. The train takes 40 seconds to come to a complete stop after initiating braking, and
D. The train takes 50 seconds to come to a complete stop after initiating braking, and it travels a distance of 500 meters during this time.
E. The train takes 50 seconds to come to a complete stop after initiating braking, and
F. The train takes 60 seconds to come to a complete stop after initiating braking, and it travels a distance of 720 meters during this time.
G. The train takes 70 seconds to come to a complete stop after initiating braking, and
H. The train takes 30 seconds to come to a complete stop after initiating braking, and it travels a distance of 600 meters during this time.
I. The train takes 45 seconds to come to a complete stop after initiating braking, and
J. The train takes 50 seconds to come to a complete stop after initiating braking, and it travels a distance of 550 meters during this time.
Answer: D
Find the integral curve is ____ of $y^{\prime\prime}=x$, which passes through the point $M(0,1)$ and is tangent to the line $y=\frac{x}{2}+1$ at this point.

Options:
A. $$y=\frac{1} {8} x^{3}+\frac{1} {2} x+1$$
B. $$y=\frac{1} {6} x^{3}+x+2$$
C. $$y=\frac{1} {12} x^{3}+\frac{1} {2} x+1$$
D. $$y=x^{3}+\frac{1} {2} x+1$$
E. $$y=\frac{1} {2} x^{4}+\frac{1} {2} x+1$$
F. $$y=\frac{1} {6} x^{3}+x+1$$
G. $$y=\frac{1} {3} x^{3}+\frac{1} {2} x+1$$
H. $$y=\frac{1} {12} x^{-3}+\frac{1} {2} x+1$$
I. $$y=\frac{1} {4} x^{3}+\frac{1} {2} x+1$$
J. $$y=\frac{1} {6} x^{3}+\frac{1} {2} x+1$$
Answer: J
Let the differentiable function $\varphi(x)$ satisfy $\varphi\left(x \right) \operatorname{c o s} x+2 \int_{0}^{x} \varphi\left(t \right) \operatorname{s i n} t \mathrm{d} t=x+1$, find $\varphi(x)$= ().

Options:
A. $$\operatorname3{s i n} 2x+\operatorname{co s}2 x$$
B. $$\operatorname2{s i n} 2x+\operatorname{co s} x$$
C. $$\operatorname{s i n} x+\operatorname{co s} x$$
D. $$\operatorname{s i n} x-\operatorname{co s} x$$
E. $$\operatorname{s i n} x+\operatorname{co s} 2x$$
F. $$\operatorname2{s i n} 2x+\operatorname3{co s} x$$
G. $$\operatorname{s i n} 2x+\operatorname{co s} x$$
H. $$\operatorname{co s} x-2\operatorname{s i n} x$$
I. $$\operatorname{4s i n} 2x+\operatorname{3co s} x$$
J. $$\operatorname{co s2} x+\operatorname{s i n} 3x$$
Answer: C
$\Delta y_ {t} - 3 y_ {t} = 2 t$ for () order difference equation, the $y_ {t+2} - 3 y_ {t} = 2 t$ for () order difference equation.

Options:
A. $$zero;three$$
B. $$one;one$$
C. $$two;zero$$
D. $$one;two$$
E. $$one;three$$
F. $$three;one$$
G. $$three;two$$
H. $$zero;two$$
I. $$two;three$$
J. $$two;one$$
Answer: D
A conical container with a bottom radius of $R$and a height of $H$with an opening facing upward is filled with water. Ask:What is the amount of work required to elevate all the water to the top surface of container h?

Options:
A. $$ \frac{\pi R^{2} H \rho g} {2} \left( h+\frac{2H} {5} \right) $$
B. $$ \frac{\pi R^{2} H \rho g} {3} \left( h+\frac{H} {4} \right) $$
C. $$ \frac{\pi R^{3} H \rho g} {3} \left( h+\frac{H} {5} \right) $$
D. $$ \frac{\pi R^{2} H \rho g} {3} \left( h+\frac{2H} {3} \right) $$
E. $$ \frac{2\pi R^{2} H \rho g} {3} \left( h+\frac{H} {8} \right) $$
F. $$ \frac{\pi R^{2} H \rho g} {4} \left( h+\frac{2H} {3} \right) $$
G. $$ \frac{\pi R^{2} H \rho g} {3} \left( h+\frac{H} {2} \right) $$
H. $$ \frac{\pi R^{3} H \rho g} {5} \left( h+\frac{H} {4} \right) $$
I. $$ \frac{\pi R^{3} H \rho g} {3} \left( h+\frac{H} {7} \right) $$
J. $$ \frac{\pi R^{4} H \rho g} {3} \left( h+\frac{H} {6} \right) $$
Answer: B
Let $f \left(x \right)$be a continuous function and $f \left(x \right)=\operatorname{c o s} x-\int_{0}^{x} \left(x-t \right) f \left(t \right) \mathrm{d} t$, then $f \left(x \right)$=().

Options:
A. $$ \operatorname2{c o s} x+\frac{1} {2} x \operatorname{s i n} x $$
B. $$ \operatorname{c o s} x+\frac{1} {3} x \operatorname{s i n} x $$
C. $$ \operatorname{c o s} x+\frac{3} {4} x \operatorname{s i n} x $$
D. $$ \operatorname{c o s} x+\frac{2} {3} x \operatorname{s i n} x $$
E. $$ \operatorname3{c o s} x+\frac{1} {2} x \operatorname{s i n} x $$
F. $$ \operatorname{c o s} x+\frac{1} {2} x \operatorname{s i n} 2x $$
G. $$ \operatorname{c o s} 2x+\frac{1} {2} x \operatorname{s i n} x $$
H. $$ \operatorname{c o s} x+\frac{1} {2} x \operatorname{s i n} 3x $$
I. $$\operatorname{c o s} x-\frac{1} {2} x \operatorname{s i n} x $$
Answer: I
Let the ground contact points of the front and rear wheels of the bicycle be $P$ and Q, respectively, and set $| P Q |=1$. At the initial time, $P$ is at the origin and Q is at the point $(1, 0)$. If the front wheel moves along the positive direction of the Y-axis, the movement trajectory of point Q is ().

Options:
A. $$y=\operatorname{l n} \frac{1+\sqrt{1-3x^{2}}} {x}-\sqrt{1-x^{2}}$$
B. $$y=\operatorname{l n} \frac{1+\sqrt{1-x^{2}}} {x}-\sqrt{1-x^{2}}$$
C. $$y=\operatorname{l n} \frac{1+\sqrt{1-x^{2}}} {x}-\sqrt{3-x^{2}}$$
D. $$y=\operatorname{l n} \frac{2+\sqrt{1-x^{2}}} {x}-\sqrt{1-x^{2}}$$
E. $$y=\operatorname{l n} \frac{3+\sqrt{1-x^{2}}} {x}-\sqrt{1-x^{2}}$$
F. $$y=\operatorname{l n} \frac{1+\sqrt{1-x^{2}}} {x}+2\sqrt{1-x^{2}}$$
G. $$y=\operatorname{l n} \frac{1+\sqrt{1-x^{2}}} {x}+3\sqrt{1-x^{2}}$$
H. $$y=\operatorname{l n} \frac{1+\sqrt{1-2x^{2}}} {x}-\sqrt{1-x^{2}}$$
I. $$y=\operatorname{l n} \frac{1+\sqrt{1-x^{2}}} {x}+\sqrt{1-x^{2}}$$
J. $$y=\operatorname{l n} \frac{1+\sqrt{1-x^{2}}} {x}-\sqrt{2-x^{2}}$$
Answer: B
Let $f \left(x \right)$be a continuous positive value function, $x \in[0,+\infty)$, If the plane region $R_{t}=\left\{\left(x, y \right) \left| 0 \leqslant x \leqslant t, 0 \leqslant y \leqslant f \left(x \right) \right. \right\} \quad(t > 0)$The ordinate of the centroid is equal to the curve $y=f \left(x \right)$ on the sum of the corresponding curved trapezoidal area on $[0, t]$and $\frac{1} {2}$,to find ().

Options:
A. $$ f \left( x \right)=\mathrm{e}^{2 x} $$
B. $$ f \left( x \right)=3\mathrm{e}^{4 x} $$
C. $$ f \left( x \right)=5\mathrm{e}^{ x} $$
D. $$ f \left( x \right)=2\mathrm{e}^{3 x} $$
E. $$ f \left( x \right)=2\mathrm{e}^{2 x} $$
F. $$ f \left( x \right)=\mathrm{e}^{ x} $$
G. $$ f \left( x \right)=\mathrm{e}^{4 x} $$
H. $$ f \left( x \right)=\mathrm{e}^{3 x} $$
I. $$ f \left( x \right)=2\mathrm{e}^{4 x} $$
Answer: G
The general solution to the difference equation $y_{x+1}-y_x = x 2^x$ is ____.

Options:
A. $$y = C + (x-2) 2^{x+1} \, (C \text{ is any constant})$$
B. $$y = C + (x-2) 3^x \, (C \text{ is any constant})$$
C. $$y = C + (x-1) 3^x \, (C \text{ is any constant})$$
D. $$y = C + (x-1) 2^x \, (C \text{ is any constant})$$
E. $$y = C + (x+1) 3^x \, (C \text{ is any constant})$$
F. $$y = C + (x+2) 2^x \, (C \text{ is any constant})$$
G. $$y = C + (x-2) 2^x \, (C \text{ is any constant})$$
H. $$y = C + (x+2) 3^x \, (C \text{ is any constant})$$
I. $$y = C + (x+1) 2^x \, (C \text{ is any constant})$$
J. $$y = C + (x-2) 2^{x+2} \, (C \text{ is any constant})$$
Answer: G
We know that the function $y=y \left(x \right)$ satisfies $x \left( \operatorname{l n} x-1 \right) y^{\prime} \left( x \right)+\left( 3-\operatorname{l n} x^{2} \right) y \left( x \right)=0, \quad x > e, \, \mathrm{ and} \, y \left( \mathrm{e}^{2} \right)=\frac{e^4}{2},$ $y=y \left( x \right)$ the minimum value of ().

Options:
A. $$3e^2$$
B. $$3e^4$$
C. $$3e^3$$
D. $$5e^3$$
E. $$2e^3$$
F. $$e^4$$
G. $$e^3$$
H. $$e^2$$
I. $$2e^2$$
J. $$2e^4$$
Answer: G
Equation $( x^{2} y+x \mathrm{s i n} x ) \mathrm{d} x+x N ( x, y ) \mathrm{d} y=0$ have integral factor $\mu( x, y )={\frac{1} {x^{2}}}$. Then ().

Options:
A. $N (x, y)=x [2x+\varphi(y)]$, where $\varphi(x)$ is any continuously differentiable function
B. $N (x, y)= 2x+\varphi(y)$, where $\varphi(x)$ is any continuously differentiable function
C. $N (x, y)= 2x+3\varphi(y)$, where $\varphi(x)$ is any continuously differentiable function
D. $N (x, y)=x [x+\varphi(y)]$, where $\varphi(x)$ is any continuously differentiable function
E. $N (x, y)=2x [x+\varphi(y)]$, where $\varphi(x)$ is any continuously differentiable function
F. $N (x, y)=3x [x+2\varphi(y)]$, where $\varphi(x)$ is any continuously differentiable function
G. $N (x, y)=3x [2x+\varphi(y)]$, where $\varphi(x)$ is any continuously differentiable function
H. $N (x, y)= x+\varphi(y)$, where $\varphi(x)$ is any continuously differentiable function
I. $N (x, y)= x+2\varphi(y)$, where $\varphi(x)$ is any continuously differentiable function
J. $N (x, y)=x [x+2\varphi(y)]$, where $\varphi(x)$ is any continuously differentiable function
Answer: D
It starts to snow in the morning, all day long, at noon a snowplow starts to sweep the snow, the volume of snow is constant every hour, to $2$ $PM$ snow $2km$, to $4$ $PM$ snow $1km$, ask when the snow is started?

Options:
A. The snow began at $12:36$
B. The snow began at $11:40$
C. The snow began at $10:46$
D. The snow began at $10:00$
E. The snow began at $11:46$
F. The snow began at $11:36$
G. The snow began at $12:46$
H. The snow began at $10:40$
I. The snow began at $10:36$
J. The snow began at $11:00$
Answer: C
After $5 k g$ soap is dissolved in $3 0 0 L$ water, clean water is poured in at a rate of $1 0 L$ per minute, and the evenly mixed soapy water is pumped out. What's the time when there's only $1 kg$ of soap left in the soapy water?

Options:
A. $$ 2 8 \operatorname{l n} 4 \operatorname{m i n} $$
B. $$ (2 0 \operatorname{l n} 5 +2)\operatorname{m i n} $$
C. $$ 3 0 \operatorname{l n} 5 \operatorname{m i n} $$
D. $$ 3 2 \operatorname{l n} 5 \operatorname{m i n} $$
E. $$ 3 5 \operatorname{l n} 5 \operatorname{m i n} $$
F. $$ 3 0 \operatorname{l n} 6 \operatorname{m i n} $$
G. $$ 3 0 \operatorname{l n} 4 \operatorname{m i n} $$
H. $$ 2 9 \operatorname{l n} 5 \operatorname{m i n} $$
I. $$ 2 5 \operatorname{l n} 6 \operatorname{m i n} $$
J. $$ (3 0 \operatorname{l n} 5 -2)\operatorname{m i n} $$
Answer: C
The mirror surface of the condenser mirror of the searchlight is a surface of rotation, whose shape is formed by a curve $L$ on the $x O y$ coordinate surface rotated about the $x$ axis. According to the performance requirements of the condenser mirror, all the light emitted at a point $O$ on its axis of rotation ($x$ axis) is parallel to the axis of rotation after it is reflected. The equation for the curve $L$ is ().

Options:
A. $$y^{2}=2 C_{1} \left( 3x+\frac{C_{1}} {2} \right)$$
B. $$y^{2}=2 C_{1} \left( x^2+\frac{C_{1}} {2} \right)$$
C. $$y^{2}=2 C_{1} \left( 2x+\frac{C_{1}} {2} \right)$$
D. $$y^{2}=2 C_{1} \left( x^3+\frac{C_{1}} {2} \right)$$
E. $$y^{2}=2 C_{1} \left( x+\frac{C_{1}} {2} \right)$$
F. $$y^{2}=2 C_{1} \left( x+\frac{C_{1}} {4} \right)$$
G. $$y^{2}=3C_{1} \left( x+\frac{C_{1}} {2} \right)$$
H. $$y^{2}=2 C_{1} \left( x+\frac{C_{1}} {3} \right)$$
I. $$y^{2}=2 C_{1} \left( x-\frac{C_{1}} {2} \right)$$
J. $$y^{2}= C_{1} \left( x+\frac{C_{1}} {2} \right)$$
Answer: E
Suppose a winery has a batch of good new wine. If ($t=0$) is sold now, the total revenue is $R_{0}$. If the cellar is stored to be sold at the old wine price in the future, $t$ at the end of the gross income of $R=R_{0} \mathrm{e}^{\frac{2} {5} \sqrt{t}}$. Assume that the bank's annual interest rate is $r$, and interest is calculated by continuous compounding, try to find the number of years in which the present value of the total income can be maximized by selling the hoard is(), and find the value of $t$ when $r=0. 0 6$ is ().

Options:
A. $$\frac{1} {3 0 r^{2}}, t=\frac{1 5 5} {9}$$
B. $$\frac{1} {2 5 r^{3}}, t=\frac{1 1 0} {9}$$
C. $$\frac{1} {2 0 r^{2}}, t=\frac{9 5} {9}$$
D. $$\frac{7} {2 5 r^{2}}, t=\frac{1 3 0} {9}$$
E. $$\frac{1} {2 5 r^{2}}, t=\frac{105} {9}$$
F. $$\frac{9} {2 5 r^{2}}, t=\frac{1 1 0} {9}$$
G. $$\frac{1} {2 5 r^{2}}, t=\frac{1 0 0} {9}$$
H. $$\frac{1} {2 5 r^{3}}, t=\frac{1 0 0} {9}$$
I. $$\frac{3} {2 5 r^{2}}, t=\frac{1 4 0} {9}$$
J. $$\frac{1} {2 5 r^{2}}, t=\frac{1 2 5} {9}$$
Answer: G
Let $\mathrm{F}_{1} \ \left( \mathbf{x} \right)$ and $\mathrm{F}_{2} \ \left( \mathbf{x} \right)$ be two distribution functions, with corresponding probability densities $\mathrm{f}_{1} \ \ ( \mathrm{x} )$ and $\mathrm{f}_{2} \ \ ( \mathrm{x} )$ being continuous functions. Then, the option that must be a probability density is ().

Options:
A. $\mathrm{f_{1} ( x ) \ \ f_{2} \ \ ( x )}$
B. $\mathrm{f}_{1} \; \; ( \mathrm{x} ) \; \; \mathrm{f}_{2} \; \; ( \mathrm{x} ) \; \; \mathrm{F}_{1} \; \; ( \mathrm{x} )$
C. $\mathrm{f}_{1} \; \; ( \mathrm{x} ) \; \; + \; \; \mathrm{f}_{2} \; \; ( \mathrm{x} )$
D. $\mathrm{F}_{1} \; \; ( \mathrm{x} ) \; \; \mathrm{f}_{2} \; \; ( \mathrm{x} )$
E. $\mathrm{f}_{1} \; \; ( \mathrm{x} ) \; \; \mathrm{F}_{2} \; \; ( \mathrm{x} ) \; \;+\mathrm{f}_{2} \; \; ( \mathrm{x} ) \; \; \mathrm{F}_{1} \; \; ( \mathrm{x} )$
F. $\mathrm{F}_{1} \; \; ( \mathrm{x} ) \; \; \mathrm{F}_{2} \; \; ( \mathrm{x} )$
G. $\mathrm{F}_{1} \; \; ( \mathrm{x} ) \; \; - \; \; \mathrm{F}_{2} \; \; ( \mathrm{x} )$
H. $\mathrm{f}_{1} \; \; ( \mathrm{x} ) \; \; \mathrm{f}_{2} \; \; ( \mathrm{x} )$
I. $2 \mathrm{f}_{2} \ \ ( \mathrm{x} ) \ \mathrm{F}_{1} \ \ ( \mathrm{x} )$
J. $\mathrm{f_{1} ( x ) \ \ F_{2} \ \ ( x )}$
Answer: E
In a certain region, the probabilities of adults having diseases A and B are 0.015 and 0.08, respectively. Assuming that the occurrences of these two diseases are independent, the probability of any adult in this region having both diseases simultaneously is____.

Options:
A. 0.0012
B. 0.0120
C. 0.00012
D. 0.0102
E. 0.0021
F. 0.1120
G. 0.1200
H. 0.0003
I. 0.00021
J. 0.0210
Answer: A
In the pocket, there are 7 white balls and 3 black balls. If you randomly draw two balls.(1) How many white and black balls are there in the pocket?(2) What is the probability of both balls drawn being of the same color?(3) What is the probability of the two balls being of different colors?

Options:
A. 7/12,1/18,7/36
B. 7/12,7/18,3/36
C. 7/12,13/18,3/36
D. 7/12,3/18,5/36
E. 7/12,9/18,11/36
F. 7/12,7/18,1/36
G. 1/12,7/18,7/36
H. 3/12,5/18,8/36
I. 7/12,1/18,9/36
J. 7/12,11/18,3/36
Answer: F
A gambler believes that the chance of rolling at least one 6 in four throws of a single die is equal to the chance of rolling at least one double 6 in 24 throws of two dice. what the possibilities of the two evens?

Options:
A. P(A)=0.5077,P(B)=0.4915
B. P(A)=0.5277,P(B)=0.4914
C. P(A)=0.5077,P(B)=0.4814
D. P(A)=0.5176,P(B)=0.4914
E. P(A)=0.5177,P(B)=0.4914
F. P(A)=0.5178,P(B)=0.4914
G. P(A)=0.5176,P(B)=0.4915
H. P(A)=0.5177,P(B)=0.4915
I. P(A)=0.5277,P(B)=0.4915
J. P(A)=0.5276,P(B)=0.4915
Answer: E
Set random variables X and Y have the same distribution, the density function of x is that:
$$
p(x)= \begin{cases} {{{{ \frac{3} {8}} \, x^{2},}}} & {{{0 < x < 2}}} \\ {{{0,}}} & {{{others}}} \\ \end{cases}
$$
Given that events A={X>a} and B={Y>a} are independent events, and $P(A \cup B)=3/4$, calculate the constant variable a = ______.

Options:
A. 1.2337
B. 1.5038
C. 1.2599
D. 1.5874
E. 1.8992
F. 1.6735
G. 1.8171
H. 1.4422
I. 1.7284
J. 1.3159
Answer: D
Suppose the number of colds a person catches in a year follows a Poisson distribution with parameter $\lambda=5$. There is a certain cold-preventing medication that is effective for 75% of people (reducing the Poisson distribution parameter to $\lambda=3$), but ineffective for the remaining 25%. If a person takes this medication and catches a cold twice in a year, what is the probability that the medication is effective for them?

Options:
A. $P ( B \mid A )=0.600$
B. $P ( B \mid A )=0.475$
C. $P ( B \mid A )=0.640$
D. $P ( B \mid A )=0.889$
E. $P ( B \mid A )=0.750$
F. $P ( B \mid A )=0.820$
G. $P ( B \mid A )=0.555$
H. $P ( B \mid A )=0.910$
I. $P ( B \mid A )=0.925$
J. $P ( B \mid A )=0.870$
Answer: D
Given that the equation $1 / \ln\ \left( 1+x \right) \ -1 / x=k$ has real roots in the interval $( 0, \ 1 )$ what is the range of values of the constant k?

Options:
A. $$ -2 / \mathrm{l n} 2-1 < \mathrm{k} < 3 / 2 $$
B. $$ -1 / \mathrm{l n} 2-1 < \mathrm{k} < 3 / 2 $$
C. $$ -2 / \mathrm{l n} 2-1 < \mathrm{k} < 1 / 2 $$
D. $$ 2 / \mathrm{l n} 2-1 < \mathrm{k} < 3 / 2 $$
E. $$ 2 / \mathrm{l n} 2-1 < \mathrm{k} < 1 / 2 $$
F. $$ 2 / \mathrm{l n} 2-1 < \mathrm{k} < 1 $$
G. $$ 1 / \mathrm{l n} 2-1 < \mathrm{k} < 1 / 2 $$
H. $$ 1 / \mathrm{l n} 2-1 < \mathrm{k} < 3 / 2 $$
I. $$ 1 / \mathrm{l n} 2-1 < \mathrm{k} < 1 $$
J. $$ -1 / \mathrm{l n} 2-1 < \mathrm{k} < 1 / 2 $$
Answer: G
Given that the general solution of the differential equation $\mathrm{y^{\prime\prime}+a y^{\prime}+b y=c e^{x}}$ is $\mathrm{y=~ ( \, C_{1}+C_{2} x \, ) ~ \ e^{-x}+e^{x}}$ , what are the values of $\mathrm{a, ~ b, ~ c}$ in order?

Options:
A. $$
1 \,, \; \; 0 \,, \; \; 2
$$
B. $$
1 \,, \; \; 0 \,, \; \; 1
$$
C. $$
2 \,, \; \; 0 \,, \; \; 3
$$
D. $$
2 \,, \; \; 1 \,, \; \; 3
$$
E. $$
1 \,, \; \; 0 \,, \; \; 3
$$
F. $$
2 \,, \; \; 0 \,, \; \; 2
$$
G. $$
2 \,, \; \; 1 \,, \; \; 4
$$
H. $$
2 \,, \; \; 1 \,, \; \; 2
$$
I. $$
1 \,, \; \; 1 \,, \; \; 1
$$
J. $$
2 \,, \; \; 0 \,, \; \; 1
$$
Answer: G
We know that the function $\mathbf{u} \mathbf{( x, \ y )}$ satisfies:
$$
2 \frac{\partial^{2} u} {\partial x^{2}}-2 \frac{\partial^{2} u} {\partial y^{2}}+3 \frac{\partial u} {\partial y}=0
$$
What is the value of a, b, such that under the transformation $\mathrm{u}( \mathrm{x} \,, \mathrm{y} ) \mathrm{~}=\mathrm{v} \mathrm{~} ( \mathrm{x} \,, \mathrm{~ y} ) \mathrm{~} e^{\mathrm{a x} \,+\, \mathrm{b y}}$, the above equation can be reduced to the function $\mathrm{v}( \mathbf{x},$ $y )$ without the first partial derivative?

Options:
A. $$
\mathrm{ a=1 \,, ~ b=1/ 2}
$$
B. $$
\mathrm{ a=1 / 2 \,, ~ b=1}
$$
C. $$
\mathrm{ a=1 \,, ~ b=1}
$$
D. $$
\mathrm{ a=3 / 4 \,, ~ b=1}
$$
E. $$
\mathrm{ a=0 \,, ~ b=3 / 4}
$$
F. $$
\mathrm{ a=0 \,, ~ b=1 / 2}
$$
G. $$
\mathrm{ a=1 \,, ~ b=1 / 4}
$$
H. $$
\mathrm{ a=1 \,, ~ b=3 / 4}
$$
I. $$
\mathrm{ a=0 \,, ~ b=1}
$$
J. $$
\mathrm{ a=1 / 4 \,, ~ b=1}
$$
Answer: E
Let the vectors $\alpha_{1}=( k,-2,-\frac{1} {2} ) \;, \; \; \; \alpha_{2}=( 1, 2,-3 ) \; \; \;, \alpha_{3}=( 4, 1, 0 )$ be linearly dependent. What is the value of $k$ ?

Options:
A. $$
\mathrm{k=-\frac{5 5} {9}}
$$
B. $$
\mathrm{k=-\frac{5 5} {3}}
$$
C. $$
\mathrm{k=\frac{5 5} {3}}
$$
D. $$
\mathrm{k=\frac{4 5} {6}}
$$
E. $$
\mathrm{k=-\frac{5 5} {12}}
$$
F. $$
\mathrm{k=-\frac{4 5} {6}}
$$
G. $$
\mathrm{k=\frac{5 5} {12}}
$$
H. $$
\mathrm{k=-\frac{5 5} {6}}
$$
I. $$
\mathrm{k=\frac{5 5} {9}}
$$
J. $$
\mathrm{k=\frac{5 5} {6}}
$$
Answer: H
What are the values of the following matrices?
$$
\left| \begin{matrix} {{{-a b}}} & {{{a c}}} & {{{a e}}} \\ {{{b d}}} & {{{-c d}}} & {{{d e}}} \\ {{{b f}}} & {{{c f}}} & {{{-e f}}} \\ \end{matrix} \right|
$$

Options:
A. $$
5a b c d e f
$$
B. $$
2 a b c d e f
$$
C. $$
-2 a b c d e f
$$
D. $$
-3 a b c d e f
$$
E. $$
-5 a b c d e f
$$
F. $$
-4 a b c d e f
$$
G. $$
3 a b c d e f
$$
H. $$
- a b c d e f
$$
I. $$
4 a b c d e f
$$
J. $$
a b c d e f
$$
Answer: I
Let $V$ be the region enclosed by the surface $x^{2}+y^{2}+z^{2}-2 a ( x+y+z )+2 a^{2}=0 \left( a > 0 \right)$ , then what is the value of $\iiint_{V} ( x^{3}+y^{3}+z^{3} ) d x d y d z$ ?

Options:
A. $$3 \pi a^{5}$$
B. $${\frac{1 2} {5}} \pi a^{5}$$
C. $${\frac{1 9} {5}} \pi a^{5}$$
D. $$2 \pi a^{5}$$
E. $${\frac{1 3} {5}} \pi a^{5}$$
F. $${\frac{1 7} {5}} \pi a^{5}$$
G. $${\frac{1 4} {5}} \pi a^{5}$$
H. $${\frac{1 6} {5}} \pi a^{5}$$
I. $${\frac{1 8} {5}} \pi a^{5}$$
J. $${\frac{1 1} {5}} \pi a^{5}$$
Answer: H
What is a fundamental solution system of the following homogeneous linear equations?
$$ \left\{\begin{matrix} {{{x_{1}-3 x_{2}+x_{3}-2 x_{4}=0}}} \\ {{{-5 x_{1}+x_{2}-2 x_{3}+3 x_{4}=0}}} \\ {{{-x_{1}-1 1 x_{2}+2 x_{3}-5 x_{4}=0}}} \\ {{{3 x_{1}+5 x_{2}+ x_{4}=0}}} \\ \end{matrix} \right. $$

Options:
A. $$
\left(-\frac{5} {1 4}, \frac{3} {1 4}, 1, 0 \right)^{T}, \left( \frac{1} {2},-\frac{1} {14}, 1, 1 \right)^{T}
$$
B. $$
\left(-\frac{5} {1 4}, \frac{3} {1 4}, 1, 1 \right)^{T}, \left( \frac{1} {2},-\frac{1} {2}, 0, 1 \right)^{T}
$$
C. $$
\left(-\frac{5} {1 4}, \frac{3} {1 4}, 1, 0 \right)^{T}, \left( \frac{1} {2},-\frac{1} {2}, 1, 1 \right)^{T}
$$
D. $$
\left(-\frac{5} {1 4}, \frac{3} {1 4}, 1, 0 \right)^{T}, \left( \frac{1} {2},-\frac{1} {2}, 0, 0 \right)^{T}
$$
E. $$
\left(-\frac{5} {1 4}, -\frac{3} {1 4}, 1, 0 \right)^{T}, \left( \frac{1} {2},-\frac{1} {2}, 0, 1 \right)^{T}
$$
F. $$
\left(\frac{5} {1 4}, \frac{3} {1 4}, 1, 0 \right)^{T}, \left( \frac{1} {2},-\frac{1} {2}, 0, 1 \right)^{T}
$$
G. $$
\left(-\frac{5} {1 4}, \frac{3} {1 4}, 1, 0 \right)^{T}, \left( \frac{1} {2},-\frac{1} {2}, 0, 1 \right)^{T}
$$
H. $$
\left(-\frac{5} {1 4}, \frac{3} {1 4}, -1, 0 \right)^{T}, \left( \frac{1} {2},-\frac{1} {2}, 0, 1 \right)^{T}
$$
I. $$
\left(-\frac{5} {1 4}, \frac{3} {1 4}, 1, 0 \right)^{T}, \left( \frac{1} {2},\frac{1} {2}, 0, 1 \right)^{T}
$$
J. $$
\left(-\frac{5} {1 4}, \frac{3} {1 4}, 1, 0 \right)^{T}, \left( -\frac{1} {2},-\frac{1} {2}, 0, 1 \right)^{T}
$$
Answer: G
In euclidean space $R^{4}$ , there is a unit vector that is orthogonal to $\alpha_{1} \!=\! ( 1, 1, 0, 0 ), \ \ \alpha_{2} \!=\! ( 1, 1, \ -1, \ -1 ), \ \ \alpha_{3} \!=\! ( 1, \ -1, 1, \ -1 )$ , so what is this vector?

Options:
A. $$
\varepsilon=\left( \frac{1} {2},\frac{1} {2},-\frac{1} {2}, \frac{1} {2} \right)
$$
B. $$
\varepsilon=\left(- \frac{1} {2},\frac{1} {2},-\frac{1} {2}, \frac{1} {2} \right)
$$
C. $$
\varepsilon=\left( -\frac{1} {2},\frac{1} {2},\frac{1} {2}, -\frac{1} {2} \right)
$$
D. $$
\varepsilon=\left( \frac{1} {2},\frac{1} {2},\frac{1} {2}, -\frac{1} {2} \right)
$$
E. $$
\varepsilon=\left( -\frac{1} {2},\frac{1} {2},\frac{1} {2}, \frac{1} {4} \right)
$$
F. $$
\varepsilon=\left( \frac{1} {2},\frac{1} {2},\frac{1} {2}, -\frac{1} {4} \right)
$$
G. $$
\varepsilon=\left( \frac{1} {2},-\frac{1} {2},-\frac{1} {2}, \frac{1} {2} \right)
$$
H. $$
\varepsilon=\left( \frac{1} {2},\frac{1} {2},\frac{1} {2}, \frac{1} {2} \right)
$$
I. $$
\varepsilon=\left(- \frac{1} {2},-\frac{1} {2},-\frac{1} {2}, \frac{1} {2} \right)
$$
J. $$
\varepsilon=\left( -\frac{1} {2},\frac{1} {2},\frac{1} {2}, \frac{1} {2} \right)
$$
Answer: G
Given the n-order determinant $D \!=\! \left| \! \begin{array} {c c c c c} {{{1}}} & {{{2}}} & {{{3}}} & {{{\cdots}}} & {{{n}}} \\ {{{1}}} & {{{2}}} & {{{0}}} & {{{\cdots}}} & {{{0}}} \\ {{{1}}} & {{{0}}} & {{{3}}} & {{{\cdots}}} & {{{0}}} \\ {{{\cdots}}} & {{{}}} & {{{\cdots}}} & {{{}}} & {{{\cdots}}} \\ {{{1}}} & {{{0}}} & {{{0}}} & {{{\cdots}}} & {{{n}}} \\ \end{array} \! \right|$ , what is the sum of the cofactors $A_{1 1}+A_{1 2}+\ \cdots\ +A_{1 n}$ of the elements in row 1?

Options:
A. $$
n! ( 1-\sum_{i=2}^{n} {\frac{1} {i-1}} )
$$
B. $$
(n + 1)! ( 1-\sum_{i=2}^{n} {\frac{1} {i}} )
$$
C. $$
n! ( 1-\sum_{i=2}^{n} {\frac{1} {i+1}} )
$$
D. $$
n! ( 1+\sum_{i=2}^{n} {\frac{1} {1-i}} )
$$
E. $$
n! ( 1-\sum_{i=2}^{n} {\frac{1} {i}} )
$$
F. $$
n! ( 1+\sum_{i=2}^{n} {\frac{1} {1+i}} )
$$
G. $$
n! ( 1+\sum_{i=2}^{n} {\frac{1} {i}} )
$$
H. $$
(n - 1)! ( 1-\sum_{i=2}^{n} {\frac{1} {i}} )
$$
I. $$
(n + 2)! ( 1-\sum_{i=2}^{n} {\frac{1} {i}} )
$$
J. $$
(n - 2)! ( 1-\sum_{i=2}^{n} {\frac{1} {i}} )
$$
Answer: E
Let the square matrix of order $n$ $A,B$ satisfy the condition $A+B=A B$ . Known $B {=} \left( \begin{matrix} {{{1}}} & {{{-3}}} & {{{0}}} \\ {{{2}}} & {{{1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{2}}} \\ \end{matrix} \right)$ , then what is the value of $A$?

Options:
A. $$
A=\left( \begin{matrix} {{{1}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ {{{\frac{1} {3}}}} & {{{1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{2}}} \\ \end{matrix} \right)
$$
B. $$
A=\left( \begin{matrix} {{{1}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ {{{-\frac{1} {3}}}} & {{{1}}} & {{{0}}} \\ {{{1}}} & {{{0}}} & {{{2}}} \\ \end{matrix} \right)
$$
C. $$
A=\left( \begin{matrix} {{{-1}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ {{{-\frac{1} {3}}}} & {{{1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{2}}} \\ \end{matrix} \right)
$$
D. $$
A=\left( \begin{matrix} {{{1}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ {{{-\frac{1} {3}}}} & {{{1}}} & {{{1}}} \\ {{{0}}} & {{{0}}} & {{{2}}} \\ \end{matrix} \right)
$$
E. $$
A=\left( \begin{matrix} {{{1}}} & {{{-\frac{1} {2}}}} & {{{0}}} \\ {{{-\frac{1} {3}}}} & {{{1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{2}}} \\ \end{matrix} \right)
$$
F. $$
A=\left( \begin{matrix} {{{1}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ {{{-\frac{1} {3}}}} & {{{1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{-2}}} \\ \end{matrix} \right)
$$
G. $$
A=\left( \begin{matrix} {{{1}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ {{{-\frac{1} {3}}}} & {{{-1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{2}}} \\ \end{matrix} \right)
$$
H. $$
A=\left( \begin{matrix} {{{1}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ {{{-\frac{1} {3}}}} & {{{1}}} & {{{0}}} \\ {{{0}}} & {{{1}}} & {{{2}}} \\ \end{matrix} \right)
$$
I. $$
A=\left( \begin{matrix} {{{1}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ {{{-\frac{1} {3}}}} & {{{1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{2}}} \\ \end{matrix} \right)
$$
J. $$
A=\left( \begin{matrix} {{{1}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ {{{-\frac{1} {2}}}} & {{{1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{2}}} \\ \end{matrix} \right)
$$
Answer: I
If all real symmetric matrices of order $n$ are classified by contract, that is, two real symmetric matrices of order $n$ belong to the same class if and only if they contract, how many classes are there?

Options:
A. $$
{\frac{1} {2}}n ( n+2 )
$$
B. $$
{\frac{1} {4}} n( n+1 )
$$
C. $$
n^2
$$
D. $$
{\frac{1} {4}}n ( n+2 )
$$
E. $$
{\frac{1} {2}} ( n+1 ) ( n+2 )
$$
F. $$
n( n+1 )
$$
G. $$
{\frac{1} {4}} ( n+1 ) ( n+2 )
$$
H. $$
n ( n+2 )
$$
I. $$
( n+1 ) ( n+2 )
$$
J. $$
{\frac{1} {2}} n( n+1 )
$$
Answer: E
Set $x^{4} \!-\! 2 x^{2} \!+\! 3 \!=\! c_{0} \!+\! c_{1} ( x \!+\! 2 ) \!+\! c_{2} ( x \!+\! 2 )^{2} \!+\! c_{3} ( x \!+\! 2 )^{3} \!+\! c_{4} ( x \!+\! 2 )^{2}$ , try to use synthetic division to find the $c_{0}, \, c_{1}, \, c_{2}, \, c_{3}, \, c_{4}$ . What are their solutions?

Options:
A. $$
c_{0}=1 1, \; \; c_{1}=-2 4, \; \; c_{2}=2 2, \; \; c_{3}=-8, \; \; c_{4}=1 1
$$
B. $$
c_{0}=1 1, \; \; c_{1}=-2 4, \; \; c_{2}=2 0, \; \; c_{3}=-8, \; \; c_{4}=1
$$
C. $$
c_{0}=1 1, \; \; c_{1}=-2 4, \; \; c_{2}=2 2, \; \; c_{3}=-1 8, \; \; c_{4}=1
$$
D. $$
c_{0}=1 1, \; \; c_{1}=-2 , \; \; c_{2}=2 2, \; \; c_{3}=-8, \; \; c_{4}=1
$$
E. $$
c_{0}=1 1, \; \; c_{1}=-2 0, \; \; c_{2}=2 2, \; \; c_{3}=-8, \; \; c_{4}=1
$$
F. $$
c_{0}=1 1, \; \; c_{1}=-2 4, \; \; c_{2}=2 2, \; \; c_{3}=-8, \; \; c_{4}=1 0
$$
G. $$
c_{0}=1 , \; \; c_{1}=-2 4, \; \; c_{2}=2 2, \; \; c_{3}=-8, \; \; c_{4}=1
$$
H. $$
c_{0}=1 1, \; \; c_{1}=-2 4, \; \; c_{2}=2 2, \; \; c_{3}=-8, \; \; c_{4}=1
$$
I. $$
c_{0}=1 1, \; \; c_{1}=-2 4, \; \; c_{2}=2 , \; \; c_{3}=-8, \; \; c_{4}=1
$$
J. $$
c_{0}=1 1, \; \; c_{1}=-2 4, \; \; c_{2}=2 2, \; \; c_{3}=-8, \; \; c_{4}=1 2
$$
Answer: H
Let $\{\alpha_{1}, \, \alpha_{2}, \cdots, \, \alpha_{n} \}$ be a basis of vector space $V$ , what is the transition matrix from this basis to $\{\alpha_{3}, \, \alpha_{4}, \, \cdots, \, \alpha_{n}, \, \, \, \alpha_{1}, \, \alpha_{2} \}$ ?

Options:
A. $$
\left( \begin{matrix} {{-2}} & {{I_{2}}} \\ {{I_{n-2}}} & {{0}} \\ \end{matrix} \right)
$$
B. $$
\left( \begin{matrix} {{0}} & {{I_{2}}} \\ {{I_{n-2}}} & {{-2}} \\ \end{matrix} \right)
$$
C. $$
\left( \begin{matrix} {{1}} & {{I_{2}}} \\ {{I_{n-2}}} & {{0}} \\ \end{matrix} \right)
$$
D. $$
\left( \begin{matrix} {{2}} & {{I_{2}}} \\ {{I_{n-2}}} & {{0}} \\ \end{matrix} \right)
$$
E. $$
\left( \begin{matrix} {{-1}} & {{I_{2}}} \\ {{I_{n-2}}} & {{0}} \\ \end{matrix} \right)
$$
F. $$
\left( \begin{matrix} {{0}} & {{I_{2}}} \\ {{I_{n-2}}} & {{0}} \\ \end{matrix} \right)
$$
G. $$
\left( \begin{matrix} {{0}} & {{I_{2}}} \\ {{I_{n-2}}} & {{2}} \\ \end{matrix} \right)
$$
H. $$
\left( \begin{matrix} {{0}} & {{I_{2}}} \\ {{I_{n-2}}} & {{-1}} \\ \end{matrix} \right)
$$
I. $$
\left( \begin{matrix} {{1}} & {{I_{2}}} \\ {{I_{n-2}}} & {{1}} \\ \end{matrix} \right)
$$
J. $$
\left( \begin{matrix} {{0}} & {{I_{2}}} \\ {{I_{n-2}}} & {{1}} \\ \end{matrix} \right)
$$
Answer: F
When is the range of values for $a$ such that the quadratic form $f ( x_{1}, x_{2}, x_{3} )=x_{1}^{2}+x_{2}+5 x_{3}+2 a x_{1} x_{2}-2 x_{1} x_{3}+4 x_{2} x_{3}$ ?

Options:
A. $$
- \frac{1} {5} < a < \frac{1} {5}
$$
B. $$
- 2 < a < 2
$$
C. $$
- 1< a < 1
$$
D. $$
- \frac{4} {5} < a < \frac{4} {5}
$$
E. $$
- \frac{3} {5} < a < \frac{3} {5}
$$
F. $$
- \frac{2} {5} < a < \frac{2} {5}
$$
G. $$
- \frac{4} {9} < a < 0
$$
H. $$
- \frac{4} {3} < a < 0
$$
I. $$
- \frac{4} {7} < a < 0
$$
J. $$
- \frac{4} {5} < a < 0
$$
Answer: J
It is known that $\xi=\left( \begin{matrix} {{{1}}} \\ {{{1}}} \\ {{{-1}}} \\ \end{matrix} \right)$ is an eigenvector of $A=\left( \begin{matrix} {{{2}}} & {{{-1}}} & {{{2}}} \\ {{{5}}} & {{{a}}} & {{{3}}} \\ {{{-1}}} & {{{b}}} & {{{-2}}} \\ \end{matrix} \right)$ , can the values of the constants $a$ and $b$ be determined?

Options:
A. $$
a=-1, b=0
$$
B. $$
a=0, b=0
$$
C. $$
a=-3, b=2
$$
D. $$
a=2, b=0
$$
E. $$
a=-3, b=0
$$
F. $$
a=-2, b=0
$$
G. $$
a=-3, b=1
$$
H. $$
a=1, b=0
$$
I. $$
a=3, b=0
$$
J. $$
a=-3, b=3
$$
Answer: E
Let vectors $\alpha=\left( a_{1}, a_{2}, \Lambda\,, a_{n} \right) r, \, \, \, \beta=\left( b_{1}, b_{2}, \Lambda\,, b_{n} \right) r$ be both nonzero vectors. And condition $\alpha_{T} \beta=0$ is satisfied. What is the value of $A_{2}$in the n-order matrix $A=\alpha \beta \tau$ ?

Options:
A. $$
A_{2}=E^2
$$
B. $$
A_{2}=\beta\tau
$$
C. $$
A_{2}=1-E^3
$$
D. $$
A_{2}=E
$$
E. $$
A_{2}=O
$$
F. $$
A_{2}=0
$$
G. $$
A_{2}=E^2-E
$$
H. $$
A_{2}=E^3
$$
I. $$
A_{2}=E^2-1
$$
J. $$
A_{2}=E^n
$$
Answer: E
Let $A$ be a 4th order real symmetric matrix, and $A^2 + A = 0$. If $R(A) = 3$, then to which matrix is $A$ similar?

Options:
A. $$
\left( \begin{matrix} {{{-1}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{-1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{-1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ \end{matrix} \right)
$$
B. $$
\left( \begin{matrix} {{{1}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{-1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{0}}} & {{{1}}} \\ \end{matrix} \right)
$$
C. $$
\left( \begin{matrix} {{{2}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{-1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ \end{matrix} \right)
$$
D. $$
\left( \begin{matrix} {{{1}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{-1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{0}}} & {{{2}}} \\ \end{matrix} \right)
$$
E. $$
\left( \begin{matrix} {{{1}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{-1}}} & {{{0}}} \\ {{{0}}} & {{{1}}} & {{{0}}} & {{{0}}} \\ \end{matrix} \right)
$$
F. $$
\left( \begin{matrix} {{{1}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ \end{matrix} \right)
$$
G. $$
\left( \begin{matrix} {{{1}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{-1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ \end{matrix} \right)
$$
H. $$
\left( \begin{matrix} {{{1}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{-2}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ \end{matrix} \right)
$$
I. $$
\left( \begin{matrix} {{{1}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{-1}}} & {{{1}}} \\ {{{0}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ \end{matrix} \right)
$$
J. $$
\left( \begin{matrix} {{{1}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{-1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{-1}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{0}}} & {{{0}}} \\ \end{matrix} \right)
$$
Answer: A
If the eigenvalues of an $n$-order matrix $A$ are all zero, how many of the following conclusions are correct?
(1) $| A |=0$
(2) $t r ( A )=0$
(3) $R ( A )=0$
(4) $\mid\lambda E-A \mid=\lambda n$
(5) $| A |>0$
(6) $t r ( A )>0$
(7) $R ( A )>0$
(8) $\mid\lambda E-A \mid>\lambda n$

Options:
A. $$5
$$
B. $$8
$$
C. $$0
$$
D. $$4
$$
E. $$9
$$
F. $$1
$$
G. $$7
$$
H. $$2
$$
I. $$3
$$
J. $$6
$$
Answer: I
Given $A=\left( \begin{matrix} {{{{4}}}} & {{{{-5}}}} & {{{{2}}}} \\ {{{{5}}}} & {{{{-7}}}} & {{{{3}}}} \\ {{{{6}}}} & {{{{-9}}}} & {{{{4}}}} \\ \end{matrix} \right)$ , what are the eigenvectors of $\mathsf{A}$?

Options:
A. $$
( 1, 0, 0 )^{T}
$$
B. $$
( 3, 1, 0 )^{T}
$$
C. $$
( 1, 2, 3 )^{T}
$$
D. $$
( 1, 0, -3 )^{T}
$$
E. $$
( 1, 1, 1 )^{T}
$$
F. $$
( 1, 1, -3 )^{T}
$$
G. $$
( 1, 1, 0 )^{T}
$$
H. $$
( 1, 1, 2 )^{T}
$$
I. $$
( 1, 4, 0 )^{T}
$$
J. $$
( 1, 1, 3 )^{T}
$$
Answer: E
Let the matrix $A=\left( \begin{matrix} {{{-1}}} & {{{0}}} & {{{0}}} \\ {{{-2}}} & {{{1}}} & {{{0}}} \\ {{{-2}}} & {{{a}}} & {{{1}}} \\ \end{matrix} \right)$, so what is the value of $a$ when the matrix $A$ can be diagonalized similarly?

Options:
A. $$
a=-3
$$
B. $$
a=-4
$$
C. $$
a=0
$$
D. $$
a=2
$$
E. $$
a=1
$$
F. $$
a=-2
$$
G. $$
a=-1
$$
H. $$
a=3
$$
I. $$
a=5
$$
J. $$
a=4
$$
Answer: C
The known characteristic value of matrix $A=\left( \begin{matrix} {{7}} & {{4}} & {{z}} \\ {{4}} & {{7}} & {{-1}} \\ {{-4}} & {{-4}} & {{4}} \\ \end{matrix} \right)$ is $\lambda_ {{} _ {1, 2}} = 3, \lambda_ {{} _ {3}} = 1, 2$. So what is the value of $z$ ?

Options:
A. $$
z=4
$$
B. $$
z=-4
$$
C. $$
z=-3
$$
D. $$
z=-1
$$
E. $$
z=1
$$
F. $$
z=-2
$$
G. $$
z=2
$$
H. $$
z=-5
$$
I. $$
z=0
$$
J. $$
z=3
$$
Answer: D
Take two bases in $R_{3}$ : $\alpha_{1}=( 1, 2, 1 ) r, \alpha_{2}=( 2, 3, 3 ) r, \alpha_{3}=( 3, 7, 1 ) r;$
$\beta_{1}=( 3, 1, 4 ) \tau\,, \beta_{2}=( 5, 2, 1 ) \tau\,, \beta_{3}=( 1, 1,-6 ) \tau\,$.
What is the transition matrix from base $\alpha_{1}, \alpha_{2}, \alpha_{3}$ to base $\beta_{1}, \beta_{2}, \beta_{3}$ ?

Options:
A. $$
\left( \begin{matrix} {{-2 7}} & {{-7 1}} & {{-4 1}} \\ {{9}} & {{2 }} & {{9}} \\ {{4}} & {{1 2}} & {{8}} \\ \end{matrix} \right)
$$
B. $$
\left( \begin{matrix} {{ 7}} & {{-7 1}} & {{-4 1}} \\ {{9}} & {{2 0}} & {{9}} \\ {{4}} & {{1 2}} & {{8}} \\ \end{matrix} \right)
$$
C. $$
\left( \begin{matrix} {{-2 7}} & {{-7 1}} & {{-4 1}} \\ {{9}} & {{2 0}} & {{9}} \\ {{4}} & {{1 2}} & {{8}} \\ \end{matrix} \right)
$$
D. $$
\left( \begin{matrix} {{- 7}} & {{-7 1}} & {{-4 1}} \\ {{9}} & {{2 0}} & {{9}} \\ {{4}} & {{1 2}} & {{8}} \\ \end{matrix} \right)
$$
E. $$
\left( \begin{matrix} {{-2 }} & {{-7 1}} & {{-4 1}} \\ {{9}} & {{2 0}} & {{9}} \\ {{4}} & {{1 2}} & {{8}} \\ \end{matrix} \right)
$$
F. $$
\left( \begin{matrix} {{-2 7}} & {{- 1}} & {{-4 1}} \\ {{9}} & {{2 0}} & {{9}} \\ {{4}} & {{1 2}} & {{8}} \\ \end{matrix} \right)
$$
G. $$
\left( \begin{matrix} {{-2 7}} & {{-7 }} & {{-4 1}} \\ {{9}} & {{2 0}} & {{9}} \\ {{4}} & {{1 2}} & {{8}} \\ \end{matrix} \right)
$$
H. $$
\left( \begin{matrix} {{-2 7}} & {{-7 1}} & {{-4 }} \\ {{9}} & {{2 0}} & {{9}} \\ {{4}} & {{1 2}} & {{8}} \\ \end{matrix} \right)
$$
I. $$
\left( \begin{matrix} {{2 }} & {{-7 1}} & {{-4 1}} \\ {{9}} & {{2 0}} & {{9}} \\ {{4}} & {{1 2}} & {{8}} \\ \end{matrix} \right)
$$
J. $$
\left( \begin{matrix} {{-2 7}} & {{-7 1}} & {{-4 1}} \\ {{-9}} & {{2 0}} & {{9}} \\ {{4}} & {{1 2}} & {{8}} \\ \end{matrix} \right)
$$
Answer: C
Let the matrix $A=\left( \begin{matrix} {{{1}}} & {{{0}}} & {{{1}}} \\ {{{0}}} & {{{2}}} & {{{0}}} \\ {{{3}}} & {{{0}}} & {{{1}}} \\ \end{matrix} \right)$ be $A_{2} B-A-B=E$ , then what is the calculated value of $| B |$ ?

Options:
A. $${\frac{1} {4}}
$$
B. $$-{\frac{1} {4}}
$$
C. $$-{\frac{1} {6}}
$$
D. $$-{\frac{1} {5}}
$$
E. $${\frac{1} {3}}
$$
F. $${\frac{1} {6}}
$$
G. $${\frac{1} {2}}
$$
H. $$-{\frac{1} {3}}
$$
I. $$-{\frac{1} {2}}
$$
J. $${\frac{1} {5}}
$$
Answer: H
Consider matrix $A=\left( \begin{matrix} {{{2}}} & {{{5}}} & {{{0}}} & {{{0}}} \\ {{{1}}} & {{{3}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{0}}} & {{{2}}} & {{{1}}} \\ {{{0}}} & {{{0}}} & {{{1}}} & {{{2}}} \\ \end{matrix} \right)$ , using block matrix operations calculate the value of $\left| A^{2012} \right|$, what is the result?

Options:
A. $$
8^{2 0 1 1}
$$
B. $$
8^{2 0 1 2}
$$
C. $$
2^{2 0 1 2}
$$
D. $$
5^{2 0 1 2}
$$
E. $$
8^{2 0 1 4}
$$
F. $$
8^{2 0 1 0}
$$
G. $$
3^{2 0 1 2}
$$
H. $$
6^{2 0 1 2}
$$
I. $$
7^{2 0 1 2}
$$
J. $$
4^{2 0 1 2}
$$
Answer: B
Given $A=\left( \begin{matrix} {{{1}}} & {{{2}}} & {{{3}}} \\ {{{2}}} & {{{3}}} & {{{1}}} \\ {{{3}}} & {{{1}}} & {{{2}}} \\ \end{matrix} \right)$ , what is the value of ${d i g} \binom{{}} {A^{*}}$ ?

Options:
A. $$\frac{A} {1 8}
$$
B. $$-\frac{A} {1 4}
$$
C. $$-\frac{A} {3}
$$
D. $$-\frac{A} {1 0}
$$
E. $$-\frac{A} {1 2}
$$
F. $$-\frac{A} {4}
$$
G. $$-\frac{A} {7}
$$
H. $$-\frac{A} {1 8}
$$
I. $$-\frac{A} {8}
$$
J. $$-\frac{A} {2}
$$
Answer: H