problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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Example 11. Isosceles $\triangle A O B$, vertex angle $2 \theta$, altitude on the base $h$, and the altitude is in the positive direction of the $x$-axis. Inside $\triangle A O B$ there is a point $\mathrm{P}$, its distances to $\mathrm{OA}, \mathrm{AB}, \mathrm{OB}$ are $|\mathrm{PD}|,|\mathrm{PE}|,|\mathrm{PF}|$ resp... | As shown in Figure 11, take the positive half-axis of $\mathrm{O} \times$ as the polar axis, and let the coordinates of the moving point $\mathrm{P}$ be $(\rho, \alpha)$.
$\because \mathrm{P}$ is inside $\triangle \mathrm{AOB}$,
\[
\begin{array}{l}
\therefore |\alpha| < \theta, \rho > 0, \\
|P D| = \rho \cdot \sin (\t... | \cos^{2} \theta \left(x^{2} + y^{2}\right) - 2 h x + h^{2} = 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,866 |
In the plane $\boldsymbol{\alpha}$, there is an equilateral triangle $\mathrm{A}^{\prime} \mathrm{BC}$, the line $D E / / B C$, and it intersects $A^{\prime} B$ and $A^{\prime} C$ at $D$ and $E$ respectively. Along $\mathrm{DE}$, fold $\triangle \mathrm{A}^{\prime} \mathrm{DE}$ so that the plane containing $\mathrm{A}^... | Solution 1
Let the equilateral triangle
have a side length of 1,
$F$ be the midpoint of $DE$, and $G$ be the midpoint of $BC$. Connect $AF, AG, FG$, then $\mathrm{AF} \perp \mathrm{DE}, \mathrm{AF} \perp \alpha, \mathrm{AG} \perp \mathrm{BC}, \mathrm{AF} \perp \mathrm{FG}$.
$$
\begin{array}{l}
\text { Let } \mathrm{AF}... | DE \text{ is the midline of } \triangle A^{\prime}BC | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,867 |
Example 3. Given any 10 different 2-digit numbers, it is always possible to select two groups of numbers from them, such that there are no common numbers between the groups, and the sums of the numbers in these groups are equal. (1974 International Mathematical Olympiad Problem). | Let A denote the set formed by these 10 two-digit numbers, which has a total of $2^{10}-1=1023$ different non-empty subsets. Each subset consists of no more than 10 two-digit numbers, so the sum of the elements in each subset does not exceed $10 \times 100=1000$. Therefore, there must be two different subsets whose ele... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 701,868 |
Example 1. Let $\mathrm{P}_{0}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ be a point not on the coordinate axes. Try to prove that there must be two parabolas passing through this point, both having the origin as a common focus, symmetric with respect to the $x$-axis, and these two parabolas are perpendicular to each ... | Because the equations of parabolas symmetric to the $x$-axis with the origin as a common focus are $y^{2}=4 \lambda(x+\lambda)(\lambda \neq 0)$. Using the relationship between the parameter and the desired parabola, and the discriminant of a quadratic equation, the problem is transformed into proving that the quadratic... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,869 |
Example 3. As shown in Figure 2, given the line $\mathrm{y}=2 \mathrm{x}$ and point $\mathrm{M}(6, 2)$, try to find points $A$ and $B$ on the $x$-axis and the line $y=2 x$ respectively, such that the perimeter of $\triangle A \bar{B} M$ is minimized, and find the minimum perimeter. | Solving directly is difficult. If we take the symmetric points of point M with respect to the $x$-axis and the line $\mathrm{y}=2 \mathrm{x}$, denoted as $\mathrm{M}^{\prime}$ and $\mathrm{M}^{\prime \prime}$, respectively, and find their coordinates, the problem then becomes finding the distance between them and the i... | 8 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,871 |
Example 5. Given a trihedral angle S-ABC, if the bisectors SD and SE of $\angle \mathrm{ASB}$ and $\angle \mathrm{BSC}$ are perpendicular to each other, prove that the bisector SF of $\angle \mathrm{ASC}$ is also perpendicular to SD and SE. | Proof: Let the unit vectors on $S A, S B, S C$ be
$$
\left(\overrightarrow{\mathrm{a}_{0}}+\overrightarrow{\mathrm{b}_{0}}\right) \cdot\left(\overrightarrow{\mathrm{b}_{0}}+\overrightarrow{\mathrm{c}_{0}}\right)=0,
$$
i.e.,
$$
\begin{array}{c}
\overrightarrow{a_{0}} \cdot \overrightarrow{b_{0}}+\vec{b}_{0}^{2}+\overri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,873 |
Example 6. Let the point $(x, y)$ vary within the set $\left\{\begin{array}{l}y-x+1 \geqslant 0, \\ 2 x-y-3 \geqslant 0\end{array}\right.$, find the minimum value of $x+2 y$. | This problem originally belongs to the category of quantitative relationship, but the given set $\left\{\begin{array}{l}y-x+1 \geqslant 0, \\ 2 x-y-3 \geqslant 0\end{array}\right.$ can be represented on a coordinate plane, as shown in Figure 4. If we let $x+2 y=k$, then $y=-\frac{1}{2} x+\frac{k}{2}$, which represents ... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,874 |
Example of parameterizing equations
$$
\left\{\begin{array}{l}
x=\frac{-8 k}{4+k^{2}} \\
y=\frac{16-4 k^{2}}{4+k^{2}}
\end{array}\right.
$$
into a common equation, where $\mathrm{k}$ is the parameter. | $$
\begin{array}{l}
\text { Solution One (Single Substitution Method) } \\
\text { (1) Squaring (1) gives } \\
x^{2}=\frac{64 k^{2}}{\left(4+k^{2}\right)^{2}} \text {. } \\
\text { From (2), } k^{2}=\frac{4(4-y)}{y+4} \cdot(y+4 \neq 0) \\
\text { Substituting into (3) and simplifying gives } \\
\frac{x^{2}}{4}+\frac{y^... | \frac{x^{2}}{4}+\frac{y^{2}}{16}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,875 |
Example 1. For the independent variables $\mathrm{x}$, $\mathrm{y}$, under the constraints $x+1 \geqslant 0, y+1 \geqslant 0, x-2 y+2 \geqslant 0, \quad x+y \leqslant 2$, find the maximum and minimum values of the function $z=2 x+y$. | Solve: First, determine the range of values for $x$ and $y$. It is the closed region of a quadrilateral with vertices $\mathrm{A}(-1,-1)$, $\mathrm{B}(3,-1)$, $\mathrm{C}\left(\frac{2}{3}\right.$, $\left.\frac{4}{3}\right)$, and $\mathrm{D}\left(-1, \frac{1}{2}\right)$ (Figure 1).
Let $\mathrm{k}=2 \mathrm{x}+\mathrm{... | z_{\text {max }}=5, z_{\text {min }}=-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,876 |
Example 2. For $x$, $y$ satisfying the inequality $\frac{x^{2}}{4}+y^{2} \leqslant 1$, find the maximum and minimum values of the function $z=x-y$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note at the end is not part of the translation but is provided to clarify the instruct... | Solve: First, determine the domain of $x$ and $y$, which is the closed region $D$ enclosed by an ellipse centered at point $\mathrm{O}$ with a major semi-axis of 2 and a minor semi-axis of 1. (Figure 2).
Let $x-y=k$, i.e., $y=x-k$, which represents a family of parallel lines with a slope of 1 and a y-intercept of $-k$... | z_{\max} = \sqrt{5}, z_{\min} = -\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,877 |
Example 3. Let $z=x^{2}+y^{2}-12 x+16 y$, find $z_{\text {max }}$ and $z_{\text {min }}$ when $x^{2}+y^{2} \leqslant 25$. | Solve: First, make the domain of $x$ and $y$, which is a closed circular region with O as the center and a radius of 5 (Figure 3).
$$
\begin{aligned}
& \text { Let } x^{2}+y^{2}-12 x+16 y=k \text {, i.e., }(x-6)^{2} \\
+ & (y+8)^{2}=100+k
\end{aligned}
$$
It represents a family of circles with $M(6,-8)$ as the center ... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,878 |
Example 4. Let $a_{1}, a_{2}, \cdots, a_{n}$ be any $n$ real numbers. Prove: there must exist a $\mathrm{k}: 1 \leqslant \mathrm{k} \leqslant \mathrm{n}$, such that
$$
\sum_{i=1}^{k} a_{i}^{\prime}-\sum_{i=k+1}^{n} a_{i}\left|\leqslant \max _{1 \leqslant i \leqslant n}\right| a_{i} i \text {. }
$$
(Polish Mathematical ... | For convenience, let $a=\max _{1 \leqslant i \leqslant n} a_{i}$, and denote
$$
\begin{array}{l}
b_{0}=a_{1}+a_{2}+\cdots+a_{n-1}+a_{n}, \\
b_{1}=a_{1}+a_{2}+\cdots+a_{n-1}-a_{n}, \\
\cdots \cdots
\end{array}
$$
$$
b_{n-1}=a_{1}-a_{2}-\cdots-a_{n-1}-a_{n}
$$
Let $B=\left\{b_{0}, b_{1}, \cdots, b_{n-1}\right\}, B_{1}=\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,879 |
Example 4. Let $z=\frac{x^{2}}{4}-y^{2}$, when $x^{2}+y^{2} \leqslant 4$, find $z_{max}$ and $z_{min}$. | Solve: First, make the domain of $x, y$ (as shown in Figure 4), which is a closed circular region centered at $O$ with a radius of 2.
$$
\text { Let } \frac{x^{2}}{4}-y^{2}=k \text {. }
$$
(i) When $\mathrm{k}=0$
$$
\begin{array}{l}
\text { then, } \frac{x^{2}}{4}-y^{2}=0 \Leftrightarrow \\
\left(\frac{x}{2}\right)^{2}... | z_{max}=1, z_{min}=-4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,880 |
Example 5. Given $z=y^{2}-2 x$, when $|x+y| \leqslant 1$, find $z_{max}$ and $z_{min}$. | Solve for the domain of $x$ and $y$ (Figure 5), which is a square with vertices at A, B, $A^{\prime}$, and $B^{\prime}$. By the same method, we can derive
$$
\begin{array}{l}
z_{\text {max }}=2, \\
z_{\text {min }}=-2 .
\end{array}
$$
From examples 3, 4, and 5, it can be seen that for a general bivariate function
$z=a... | z_{\text {max }}=2, \quad z_{\text {min }}=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,881 |
Example 1. In isosceles trapezoid $\mathrm{ABCD}$, the lower base $\mathrm{AB}$ is 20, the upper base $\mathrm{CD}$ is 12, and the height is $8 \sqrt{2}$. By folding the trapezoid along the perpendicular bisector MN (Figure 1) of the bases to form a dihedral angle of $120^{\circ}$, find the length of $\mathrm{AC}$ at t... | ```
Figure 2 is the folded shape. Figure 3 is the shape after Figure 2 is supported, where $\mathrm{AE}$ $\mathrm{FM}-\mathrm{PQCN}$ is a right quadrilateral prism, with the two bases being parallel parallelograms (the figure is intuitive, and the construction method is not described again, the same below).
\[
\begin{a... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,882 |
Example 2. Line segment $\mathrm{AB}=6$, through $\mathrm{A}$ construct $\mathrm{PA}$, such that $\mathrm{PA}=4 \sqrt{2}, \angle \mathrm{PAB}=45^{\circ}$. Through $\mathrm{B}$ construct $\mathrm{QB}$, such that $\mathrm{QB}=4, \angle \mathrm{QBA}=120^{\circ}$. Then let $\mathrm{QB}$ rotate around axis $\mathrm{AB}$, so... | Let Figure 4 be a graph that meets the conditions of the problem, and Figure 5 be the supporting graph of Figure 4, where $\mathrm{NAE}-\mathrm{MRQ}$ is a right triangular prism, $\mathrm{MB} / / \mathrm{PA}$, $P L \perp A R$. To meet the conditions given by the formula, it is necessary to find $\mathrm{PL}, \mathrm{QR... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,883 |
Question: As shown in Figure 1, from a point $\mathrm{P}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ outside the parabola $\mathrm{y}^{2}=2 \mathrm{p} \times$, two tangent lines $\mathrm{PA}, \mathrm{PB}$ are drawn to the curve, with $A$ and $B$ being the points of tangency. Try to find: the equation of the chord $A B$... | As shown in Figure 1, let the points of tangency be $\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$,
$$
\begin{array}{c}
B\left(x_{2}, y_{2}\right) \text {, then } \\
\text { the equation of PA is } y_{1} y=p\left(x_{1}+x\right), \\
\text { the equation of PB is } y_{2} y=p\left(x_{2}+x\right) \text {. } \\
\be... | y_{0} y=p\left(x+x_{0}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,885 |
Example 1. Draw a tangent line to the circle $x^{2}+y^{2}=25$ through point C $(6,-8)$, find the distance from C to the line connecting the two points of tangency.
保留了原文的格式和换行。 | Solve: As shown in Figure 2, the equation of the chord of contact A.B is:
$$
6 x-8 y-25=0 \text {. }
$$
The distance from C $(6,-8)$ to the chord of contact $\mathrm{AB}$ is:
$$
\mathrm{d}=\frac{|36+64-25|}{\sqrt{6^{2}+(-8)^{2}}}=7.5 .
$$ | 7.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,886 |
Example 2. The chord AB of the parabola $y=x^{2}$ keeps moving while being tangent to the circle $\odot O$ $x^{2}+y^{2}=1$. Find the locus of the intersection of the tangents to the parabola at points A and B. | Solution 1 As shown in Figure 3, let the chord AB of the parabola be tangent to $\odot \mathrm{O}$ at $\mathrm{P}(\alpha, \beta)$. Then the equation of the tangent line to $\odot \mathrm{O}$ at point $\mathrm{P}$ is
$$
\begin{array}{c}
\alpha \mathrm{x}+\beta \mathrm{y}=1, \\
\text { Let } \mathrm{A}\left(\mathrm{x}_{1... | y^{2}-4 x^{2}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,887 |
Example 3. From any point $\mathrm{P}$ on the directrix of an ellipse, draw two tangents PA and PB to the ellipse. Prove that the chord of contact AB passes through the focus corresponding to the directrix. | Prove that as shown in Figure 4, the equation of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. $P\left(x_{0}, y_{0}\right)$ is a point on the left directrix $x=-\frac{a^{2}}{c}$, and $\mathrm{PA}, \mathrm{PB}$ are two tangent lines to the ellipse.
Then the equation of the chord of contact $A B$ is $\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,888 |
Example 4. From a point $\mathrm{P}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ outside the parabola $\mathrm{y}^{2}=4 \mathrm{x}$, two tangents $P A$ and $P B$ are drawn, with $A$ and $B$ being the points of tangency.
(1) Express the area of $\triangle P A B$ in terms of $x_{0}, y_{0}$;
(2) If $\mathrm{P}\left(x_{0}, ... | (1) As shown in Figure 5, let the points of tangency be \( A\left(x_{1}, y_{1}\right) \) and \( B\left(x_{2}, y_{2}\right) \). Then the equation of the chord of contact \( AB \) is:
$$
y_{0} y=2\left(x+x_{0}\right),
$$
which simplifies to \( 2 x - y_{0} y + 2 x_{0} = 0 \).
Draw \( PC \) perpendicular to the chord \( A... | \frac{29 \sqrt{29}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,889 |
Given the lines $1_{1}: a x-k y+k=0$;
$1_{2}: \mathrm{kx}-\mathrm{y}-1=0$, where $\mathrm{a} \neq 0$, $\mathrm{a}$ is a constant, and $\mathrm{k}$ is a parameter. Coordinates of the point and eccentricity:
(2) When $a>0, y \geqslant 1$, find the minimum distance from point $P(x, y)$ on the trajectory to point $A(0,... | Solve (1) From $\left\{\begin{array}{l}a x-k y+k=0, \\ k x-y-1=0\end{array}\right.$, eliminating the parameter $k$ (using substitution method) yields $-a x^{2}+y^{2}=1$. When $a>0$, the curve is a hyperbola. Converting it to the standard form gives: $-\frac{x^{2}}{\frac{1}{a}}+y^{2}=1, c=\sqrt{1+\frac{1}{a}}$.
Thus, t... | 1_{\text {minimum }}=\left\{\begin{array}{ll}\sqrt{\frac{a b^{2}-a-1}{a(a+1)}}, & b \geqslant \frac{a+1}{a} \text { when. } \\ |1-b|, & b<\frac{a+1}{a} \text { when. }\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,890 |
Theorem 1 If $[a, b, c] \in A$, then $(a, b)$ $=(a, c)=(b, c)=(a, b, c)=1$. Here $(a, b)$ denotes the greatest common divisor of the positive integers $a$ and $b$. | Prove that if $[a, b, c] \in A$ we know $(\mathrm{a}, \mathrm{b}) = 1$, and $a^{2} + b^{2} = c^{2}$, where $a, b, c$ are positive integers. Now we prove $(\mathrm{a}, \mathrm{c}) = 1$. If not, assume $(\mathrm{a}, \mathrm{c}) = \mathrm{k} > 1$, let $\mathrm{a} = \mathrm{ka}^{\prime}, \mathrm{c} = \mathrm{kc}^{\prime}$,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 701,891 |
Theorem 2 If $[a, b, c]$ is a Pythagorean triple, and $(a, b)=d>1$, then $[a, b, c]$ is a derived Pythagorean triple. | Prove that if $(\mathrm{a}, \mathrm{b})=\mathrm{d}>1$, we can set $\mathrm{a}=\mathrm{da}^{\prime}, \mathrm{b}=\mathrm{db}^{\prime},\left(\mathrm{a}^{\prime}, \mathrm{b}^{\prime}\right)=1$, thus $\mathrm{d} \mid \mathrm{c}$, we can set $\mathrm{c}=\mathrm{dc}^{\prime}$. From $\left(\mathrm{da}^{\prime}\right)^{2}+\left... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 701,892 |
Formula representation: $a=2 m n, b=m^{2}-n^{2}, c=m^{2}+n^{2}$. (2) where $m, n$ are positive integers and $m>n,(m, n)=1$, one odd and one even. | Proof: Given that $m, n$ are positive integers and $\mathrm{m}>\mathrm{n}$, we know that $a$, $b, c$ are positive integers. From $(2 m n)^{2}+\left(m^{2}-n^{2}\right)^{2}=\left(11^{2}+n^{2}\right)^{2}$, we know that $[a, b, c]$ is a Pythagorean triplet.
Given that $m, n$ are one odd and one even, we know that $m^{2}-n... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 701,894 |
Theorem 5 The Pythagorean triples that form an arithmetic sequence have only the basic group $[3,4,5]$. | Proof: Let $[a-d, a, a+d] \in A$, then $a-d, a, a+d$ form an arithmetic sequence. From
$$
\begin{array}{c}
(a-d)^{2}+a^{2}=(a+d)^{2} ， \quad \text { we get: } \\
a^{2}-4 a d=0 ， a(a-4 a)=6 ，
\end{array}
$$
Since $\mathrm{a} \neq 0$, we have $\mathrm{a}=4 \mathrm{~d}$. Thus, the numbers are:
$$
a-d=3 \mathrm{~d}, \quad... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 701,895 |
Theorem 6 Any basic group of Pythagorean numbers cannot form a geometric series, nor can it form a harmonic series. | Proof (1) Let $[a, b, c] \in A$, and $a, b, c$ form a geometric series. Since $a, b, c$ have $c$ as the largest, it is clear that $c$ is not the geometric mean. Without loss of generality, let $b = aq, c = aq^2, q$ be a positive integer greater than 1. From
$$
\begin{array}{l}
a^{2}+(a a)^{2}=\left(a q^{2}\right)^{2}, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 701,896 |
Theorem 8 Any odd number greater than 1 must be the smallest positive integer in some primitive Pythagorean triple. | Prove that if $\mathrm{a}$ is any odd number greater than 1, then $\mathrm{a} = a \cdot 1, (a, 1) = 1$. Since $a \geqslant 3$, it follows that:
$$
\text{a} > (1+\sqrt{2}) \cdot 1.
$$
By Theorem 7 (2), we know that (3) determines:
$$
m = \frac{1}{2}(a+1), \quad n = \frac{1}{2}(a-1).
$$
And $\mathrm{m}, \mathrm{n}$ det... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 701,898 |
Theorem 9 If $m>n,(m, n)=1, m n$ is even, then the necessary and sufficient condition for $m, n$ to determine $[a, b, c] \in \bar{A}_{1}$ is $m>(1+\sqrt{2}) n$. | Proof From $\mathrm{m}>\mathrm{n}, \quad(\mathrm{m}, \mathrm{n})=1$ we know that $\mathrm{m}$ and $\mathrm{n}$ cannot both be even. Also, because $m$ and $n$ are coprime, one of $n, n$ is odd and the other is even, so $m, n$ determine $[a, b, c] \in A$. By
$2 n n(1+\sqrt{2}) n$.
Corollary Let $\mathrm{r}$ be a non-nega... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 701,899 |
Theorem 104 Any positive integer multiple greater than 1 of 104 must be the smallest positive integer of some Pythagorean triple. | Prove that for any positive integer multiple of 4 greater than 1, $4t (t>1)$, let $2mn = 4t$, and $m = 2t, n = 1$. Then $m, n$ are one even and one odd, $m > n, (m, n) = 1$. Also, because $\mathrm{m} = 2 \mathrm{t} \geqslant 4 > (1 + \sqrt{2}) \cdot 1 = (1 + \sqrt{2}) \mathrm{n}$, according to Theorem 9, the [a, b, c] ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 701,900 |
刟1. When $0 \leqslant x \leqslant 1$, find the maximum and minimum values of the quadratic function $y=x^{2}-2 a x+4$.
When $0 \leqslant x \leqslant 1$, find the maximum and minimum values of the quadratic function $y=x^{2}-2 a x+4$. | Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-2 \mathrm{ax}+4$ have a maximum value of $\mathrm{M}$ and a minimum value of $\mathrm{m}$.
$$
\begin{aligned}
f(x) & =x^{2}-2 a x+4 \\
& =(x-a)^{2}+\left(4-a^{2}\right) .
\end{aligned}
$$
First, find the maximum value $M$.
(i) When $a>1$ (i.e., the axis of symmetry passes thr... | m=\left\{\begin{array}{ll}
4, & \text{when } a \leq 1; \\
5-2a, & \text{when } a > 1.
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,901 |
Example 3. Solve the equation $\sqrt{4 x-2}+\sqrt{4 x+2}=4$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Solving the given equation:
$$
\frac{1}{\sqrt{4 x+2}+\sqrt{4 x-2}}=\frac{1}{4}.
$$
Rationalizing the denominator on the left side, we get
$$
\frac{\sqrt{4 x+2}-\sqrt{4 x-2}}{4}=\frac{1}{4}.
$$
Thus,
$$
\begin{array}{l}
\sqrt{4 x+2}-\sqrt{4 x-2}=1, \\
-2 \sqrt{4 x-2}=1-4, \\
\therefore \quad x=\frac{17}{16}.
\end{arra... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,904 |
3. Prove: $1984^{3}$ can be expressed as the difference of squares of two natural numbers.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Prove the general case first:
$$
\begin{aligned}
n^{3} & =\frac{n^{2} \cdot 4 n}{4} \\
& =\frac{n^{2} (n+1)^{2}-(n-1)^{2}}{2} \\
& =\left[\frac{n(n+1)}{2}\right]^{2}-\left[\frac{n(n-1)}{2}\right]^{2} \\
& (n \in N)
\end{aligned}
$$
When \( n=1984 \), substituting into the identity (1) yields
$$
1984^{3}=\left[\frac{19... | proof | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 701,906 |
4 . Find the sum of $1^{3}+2^{3}+3^{3}+\cdots+1984^{3}$. | Solving: Using the proof from the previous question that the cube of any natural number can be expressed as the difference of the squares of two natural numbers, we can find the sum of the cubes of the first $n$ natural numbers:
$$
1^{3}+2^{3}+3^{3}+\cdots+n^{3} \text {. }
$$
Substituting $1,2,3 \cdots, n$ into the id... | \left[\frac{1984 \cdot 1985}{2}\right]^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,907 |
5 . Find the number of divisors of 1984, as well as the sum of all its divisors. | Let the number of divisors of 1984 be $\mathrm{n}$, and the sum of all divisors be $S$. Since $1984=2^{8} \cdot 31$, the following product
$$
\left(1+2+2^{2}+\cdots+2^{8}\right)(1+31)
$$
the terms of the expanded form of this product are all divisors of 1984, and 1984 has no other divisors. Therefore, the number of di... | n=14, S=4064 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 701,908 |
Lemma 1 Suppose $\mathrm{f}(\mathrm{x}), \mathrm{B}$, left point $x_{0}$ is attainable, and $f\left(x_{0}\right)=g\left(x_{0}\right)$, then
$\lim _{x \rightarrow x_{0}} \frac{f(x)-g(x)}{x-x_{0}}$ exists, and
$$
\lim _{x \rightarrow x_{0}} \frac{f(x)-g(x)}{x-x_{0}}=f^{\prime}\left(x_{0}\right)-g^{\prime}\left(x_{0}\righ... | Prove $\because \frac{f(x)-g(x)}{x-x_{0}}$
$$
\begin{array}{l}
=\frac{f(x)-f\left(x_{0}\right)+g\left(x_{0}\right)-g(x)}{x-x_{0}} \\
=\frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}-\frac{g(x)-g\left(x_{0}\right)}{x-x_{0}},
\end{array}
$$
By the given information,
$$
\begin{array}{l}
\lim _{x \rightarrow x_{0}} \frac{f(x)-f\... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,910 |
Theorem 2 If the functions $\mathrm{f}(\mathrm{x}), \mathrm{g}(\mathrm{x})$ are differentiable at $\mathrm{x}_{0}$, and $f\left(x_{0}\right)=g\left(x_{0}\right)$, then max $\{f(x), g(x)\}$ is not differentiable at the point $x_{0}$ if and only if
$$
\begin{array}{l}
\lim _{x \rightarrow x_{0}} \frac{f(x)-g(x)}{x-x_{0}}... | Proof: Sufficiency. $\because \max f(x), g(x)$, at the point $x_{0}$, the left and right derivatives are respectively $\frac{1}{2}\left\{f^{\prime}+\left(x_{0}+g^{\prime}+\left(x_{0}\right)\right.\right.$
$\left.-\left|f^{\prime}+\left(x_{0}\right)-g^{\prime},\left(x_{0}\right)\right|\right\}$,
$\frac{1}{2}\left\{f^{\p... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,913 |
Question 21. Let the sequence $\left\{a_{n}\right\}$ be defined as
$$
a_{n+1}=2 a_{n}+2 a_{n-1} \quad(n=2,3,4, \cdots)
$$
(1) If the sequence $\left\{a_{n+1}-\alpha a_{n}\right\}$ is a geometric sequence with common ratio $\beta$ $(\alpha, \beta$ are real numbers), then $\alpha$, $\beta$ are the roots of the equation $... | (1) From $\left\{a_{n}+1-\alpha a_{n}\right\}$ being a geometric sequence with common ratio $\beta$, we get
$$
a_{n+1}-\alpha a_{n}=\beta\left(a_{u}-\alpha: 1_{n-1}\right) .
$$
That is, $2 a_{n}+2 a_{n-1}-\alpha a_{n}=\beta a_{n}-\alpha \beta a_{n-1}$.
By comparing the coefficients on both sides of the equation, we ha... | a_{n}=-\frac{1}{2 \sqrt{3}}\left[(1-\sqrt{3})^{n}-(1+\sqrt{3})^{n}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,914 |
Question 23. Find the minimum value of $2 \lg (x-3)-\lg (x-4)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Since the argument of a logarithm is positive,
\[
\therefore x-3>0, x-4>0 \text{. }
\]
Therefore, in the range $x>4$,
\[
\begin{array}{l}
2 \lg (x-3)-\lg (x-4)=\lg \frac{(x-3)^{2}}{x-4} . \\
\text{Let } y=\frac{(x-3)^{2}}{x-4}, \text{ then} \\
y=\frac{x^{2}-6 x+9}{x-4}=-\frac{(x-2)(x-4)+1}{x-4} \\
\quad=x-2+\frac{1}{x... | 2 \lg 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,916 |
Question 24. Let the equation $x^{2}+m x+m=0$ have the complex root $a+b i(b>0)$, as the real number $m$ varies, what kind of figure does the point $(\mathrm{a}, \mathrm{b})$ trace out? | Solve $x^{2}+m x+m=0$.
Equation (1) has a root $a+b i$, and thus also has the conjugate root, $a-bi$, so
$$
\begin{array}{l}
(a+b i)+(a-b i)=-m, \\
(a+b i)(a-b i)=m .
\end{array}
$$
From (2) we get $n_{1}=-2 a$
From (3) we get $a^{2}+b^{2}=m$.
Substituting (3), we get $a^{2}+2 a+b^{2}=0$,
which is $(a+1)^{2}+b^{2}=1$.... | (a+1)^2 + b^2 = 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,917 |
Question 25. The sum of several consecutive natural numbers equals 1000, find these natural numbers. | Let the natural numbers be
$$
m+1, m+2, m+3, \cdots, m+n
$$
( $m, n$ are natural numbers). Then
$$
\frac{(m+1)+(m+n)}{2} \cdot n = 1000,
$$
which simplifies to $2 m n + n^{2} + n = 2000$.
Thus, $m = \frac{1000}{n} - \frac{n}{2} - \frac{1}{2}$.
Clearly, $m > 0$. Therefore, $\frac{1000}{n} > \frac{n}{2}$,
which implies... | \begin{array}{l}
198, 199, 200, 201, 202; \\
55, 56, 57, \cdots, 70; \\
28, 29, 30, \cdots, 52.
\end{array} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 701,918 |
2. The diagonal of a square is $2 \sqrt{2}$, and the two diagonals coincide with the coordinate axes. Then the distance from the origin to one side of the square is:
(A) 1;
(B) $\backslash 2$;
(C) 2 ;
(D) $2 \sqrt{2}$. | Answer: (A).
As shown in the figure, $\triangle O \mathrm{AD}$ is an isosceles right triangle, with the legs being $\sqrt{2}$, and the hypotenuse being $2.0$. $\mathrm{O}$ is the distance from $\mathrm{O}$ to $\mathrm{AD}$, then $\mathrm{OM}$ $=1$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,920 |
3. $a>1, b=$
rational number, $1 . \mathrm{a}^{b}+\mathrm{a}^{-4}=2 \sqrt{ } 2$, then $\mathrm{a}^{b}-\mathrm{a}^{-b}$ equals:
(A) $\sqrt{6}$;
(B) 2 or -2 ;
(C) -2 ;
(D) 2 . | Answer: (D).
From $a^{b}+a^{-b}=2 \sqrt{2}$, we get $a^{2 b}+2+a^{-2 b}=8$.
$\therefore\left(a^{b}-a^{-b}\right)^{2}=4, a^{b}-a^{-b}= \pm 2$.
Given $a>1, b>0$, we have $a^{a}>1, a^{-4}a^{-b}, \therefore a^{b}-a^{-b}=2$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,921 |
4. One side of a triangle is 2, the median of this side is 1, and the sum of the other two sides is $1+\sqrt{3}$, then the area of this triangle is:
(A) 1;
(B) $\frac{1}{2} \sqrt{3}$;
(C) $\sqrt{3}$;
(D)
Cannot be determined. | Answer: (B).
From the given side of the triangle being 2, and the median to this side being 1, we can determine that this triangle is a right triangle, with the hypotenuse being 2. Since the sum of the two legs is $1+\sqrt{3}$, we can find that one leg is 1 and the other is $\sqrt{3}$. Therefore, the area of this trian... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,923 |
5. The number of real roots of the equation $x|=i-3| x \mid+2=0$ is:
(A) 1; (B) 2; (C) 3; (D) 4. | Answer: (C).
When $x>0$, the equation $x^{2}-3 x+2=0$ yields $x=1$ or 2;
When $x<0$, the equation is $-x^{2}+3 x+2=0$, yielding $x=\frac{3 \pm \sqrt{17}}{2}$. Since $x<0$, we should discard $x=\frac{3+\sqrt{17}}{2}$.
Therefore, there are three real roots. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,924 |
6 . For the real-coefficient quadratic equation $2 \mathrm{kx}^{2}+(8 \mathrm{k}+1) \mathrm{x}+8 \mathrm{k}=0$ to have two distinct real roots, the range of $\mathrm{k}$ is:
(A) $\mathrm{k}-\frac{1}{16}$ s
(C) $k \geqslant-\frac{1}{16}$,
(D) None of the above results are correct. | Answer: (D).
Solution 1: The condition for the equation to have two unequal real roots is $\Delta>0 \quad$ and $k \neq 0$.
That is, $\left\{\begin{array}{l}(8 k+1)^{2}-4 \cdot 2 k \cdot 8 k>0, \\ k \neq 0 .\end{array}\right.$
Solving this, we get $k>-\frac{1}{16}$ and $k \neq 0$.
Therefore, options A, B, and C are inco... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,925 |
7. For $\triangle \mathrm{ABC}$ with sides $\mathrm{a}, \mathrm{b}, \mathrm{c}$, the heights to these sides are $\mathrm{h}_{\mathrm{s}}=6$, $\mathrm{h}_{\mathrm{b}}=4, \mathrm{~h}_{0}=3$. Then the ratio of the sides $\mathrm{a}: \mathrm{b}: \mathrm{c}$ is:
(A) $1: 2: 3$;
(B) $2: 3: 4$,
(C) $3: 4: 6$;
(D) Uncertain. | 笭: $(\mathrm{B})$.
From $a h_{2}=b h_{b}=c b_{0}$, we can find that $a: b: c=2: 3: 4$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,926 |
8. If $x>0, y>0$, and $2 \lg (x-2 y)$ $=\lg x+\lg y$, then $\mathrm{x}: \mathrm{y}$ is:
(A) 43
(B) Is
(C) 1 or 43
(D) $\frac{1}{4}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Answer: (A).
Solution 1 From the given equation, we have $(x-2 y)^{2}=x y$.
$$
x^{2}-5 x y+4 y^{2}=0 . \quad x=4 y \text { or } x=y .
$$
When $x=y$, $\lg (x-2 y)=\lg (-y)$ is impossible, so we discard $x=y$. $\therefore x=4 y$, i.e., $x: y=4$.
Solution 2 When $x$ : $y$ is 1 or $\frac{1}{4}$, $\lg (x-2 y)$ is undefine... | A | Other | math-word-problem | Yes | Yes | cn_contest | false | 701,927 |
When $\frac{1}{p}-\frac{1}{q}-\frac{1}{p+q}=0$, find the value of $\frac{q}{p}+\frac{p}{q}$.
| Solution 1 From the given, we have
$$
(p+q) q-(p+q) p-p q=0 \text {. }
$$
which simplifies to $q^{2}-p q-p^{2}=0$.
Dividing by $p^{2}$, we get $\left(\frac{q}{p}\right)^{2}-\left(\frac{q}{p}\right)-1=0$.
$$
\therefore \quad \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2} \text {. }
$$
When $\frac{q}{p}=\frac{1+\sqrt{5}}{2}$,
$$... | \pm \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,931 |
Given real numbers $x, y, z$ satisfy $x=6-y, z^{2}=x y-9$. Prove: $x=y$.
The text above is translated into English, keeping the original text's line breaks and format. | Proof 1: Substitute $x=6-y$ into $z^{2}=x y-9$, we get $z^{2}=(6-y) y-9=6 y-y^{2}-9=-(y-3)^{2}$. Since $y, z$ are both real numbers, $\therefore z^{2} \geqslant 0, (y-3)^{2} \geqslant 0$. Also, since $z^{2}=-(y-3)^{2}$, $\therefore z^{2}=0, (y-3)^{2}=0$. Therefore, $y=3, x=6-3=3$.
$\therefore \quad x=y$.
Proof 2: Since... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,932 |
Example 1. Let $\mathrm{P}$ be a fixed point inside the circle $\odot \mathrm{O}$, and draw two perpendicular lines through $\mathrm{P}$ intersecting the circle at points $\mathrm{A}$ and $\mathrm{B}$. Find the equation of the locus of the midpoint of chord $\mathrm{AB}$. | As shown in Figure 1, take the line PO as the x-axis, with O as the origin to establish a Cartesian coordinate system. (Set up the coordinate system)
Let the radius of the circle be R, |PO| = a, and the midpoint of chord AB be M(x, y). (Set points)
Connect PM and OM, then |PM| = |MB|.
Also, |OB|^2 = |OM|^2 + |MB|^2,
T... | x^2 + y^2 + ax - \frac{R^2 - a^2}{2} = 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,933 |
Example 2. In $\triangle \mathrm{ABC}$, the length of $\mathrm{AB}$ is $8, \angle C=45^{\circ}$. If the position of $\mathrm{AB}$ remains unchanged, find the equation of the locus of vertex C. | Solve As shown in Figure 3, take the line AB as the x-axis, and the perpendicular bisector of AB as the y-axis, so A(-4,0), B(4, 0). Let the coordinates of the moving point C be (x, y).
i) If point C is above the x-axis, $\angle \mathrm{CAB}+45^{\circ}=\angle \mathrm{CBX}$.
$$
\begin{array}{l}
\because \operatorname{tg... | x^{2}+(y-4)^{2}=32 \text{ and } x^{2}+(y+4)^{2}=32 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,935 |
Example 1. Let $f(x)=\left\{\begin{array}{ll}0, & -1 \leqslant x<0, \\ 1, & 0<x \leqslant 1 .\end{array}\right.$ Then $f(x)$ is integrable on $[-1,1]$, but has no antiderivative. | Given that $f(x)$ is bounded on $[-1,1]$, and is continuous everywhere except at $x=0$, it is thus integrable. However, $f(x)$ does not have an antiderivative on $[-1,1]$. If it did, by the following Theorem 3,
$$
F(x)=\int_{-1}^{x} f(t) d t
$$
would be an antiderivative of $f(x)$. Through simple calculations, we can... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,936 |
Theorem 1 If $\mathrm{f}(\mathrm{x})$ is continuous on $[\mathrm{a}, \mathrm{b}]$, then it always has an antiderivative $\mathrm{G}(\mathrm{x})$, and the upper limit function is one of the antiderivatives, i.e.,
$$
G(x)=\int_{a}^{x} f(t) d t+C .
$$
(C is a constant) | To prove [1]: $\forall x \in[a, b]$, it is always true that $G^{\prime}(x)=f(x)$. For this, we find $G^{\prime}(x)$:
$$
\begin{array}{l}
\because \triangle G(x)=G(x+\Delta x)-G(x) \\
=\int_{a}^{x+\Delta x} G(t) d t-\int_{a}^{x} G(t) d t \\
=\int_{x}^{x+\Delta x} G(t) d t, \\
\therefore \frac{\Delta G(x)}{\Delta x}=\fra... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,937 |
Example 1. Take any two points $\mathrm{P}, \mathrm{Q}$ on an ellipse centered at the origin $\mathrm{O}$, and let $\angle \mathrm{POQ}=\frac{\pi}{2}$. Prove that $\frac{1}{\mathrm{OP}^{2}}+\frac{1}{\mathrm{OQ}^{2}}$ is a constant. | Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,
$\angle XOP=\theta$, then the parametric equations of the lines $\mathrm{OP}$ and $\mathrm{OQ}$ are respectively $\left\{\begin{array}{l}x=t \cos \theta, \\ y=t \sin \theta .\end{array}\right.$
and $\left\{\begin{array}{l}x=t \cos \left(\fr... | \frac{a^{2}+b^{2}}{a^{2} b^{2}} | Geometry | proof | Yes | Yes | cn_contest | false | 701,938 |
Example 2. A straight line intersects a hyperbola and its asymptotes, prove that the segments intercepted between the asymptotes and the curve are equal.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let the hyperbola be
$$
b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2},
$$
then its asymptotes are $b^{2} x^{2}-a^{2} y^{2}=0$.
Suppose the line is given by $\left\{\begin{array}{l}x=h+t \cos \alpha, \\ y=k+t \sin \alpha .\end{array}\right.$
$(h, k, t, \alpha$
are variables)
The line (3) intersects the hyperbola
(1) at points $... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,939 |
Example 3. Find the equation of the line passing through $\mathrm{A}$ $(-6,3)$ and tangent to the parabola $2 y^{2}=9 x$.
untranslated text remains the same as requested. | Let the required tangent line equation be
$$
\left\{\begin{array}{l}
x=-6+t \cos \alpha, \\
y=3+t \sin \alpha .
\end{array}\right.
$$
Substitute (1) into the parabola equation, eliminate $x$ and $y$ to get $2 t^{2} \sin ^{2} \alpha+(12 \sin \alpha-9 \cos \alpha) t+72=0$.
By the definition of the tangent line and the ... | 3 x+4 y+6=0 \text{ and } x-4 y+18=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,940 |
Example 4. Find the length of the tangent from a point $P\left(x_{1}, y_{1}\right)$ outside a circle $(x-a)^{2}$ $+(y-b)^{2}=r^{2}$. | Let the parametric equation of the line passing through point $P\left(x_{1}, y_{1}\right)$ be
$$
\left\{\begin{array}{l}
x=x_{1}+t \cos \theta \\
y=y_{1}+t \sin \theta
\end{array}\right.
$$
Substituting (1) into the equation of the circle and simplifying, we get
$$
\begin{array}{l}
t^{2}+2\left[\left(x_{1}-a\right) \... | l^{2}=\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}-r^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,941 |
Example 5. Through the origin, any line intersects the parabola $y^{2}=x^{2}-x+3$ at points $A$ and $B$. Find the locus of the midpoint of $AB$.
Keep the original text's line breaks and format, and output the translation result directly. | Let the equation of the line passing through the origin be
$$
\mathrm{y}=\mathrm{kx} . \quad(\mathrm{k} \text { is a parameter) }
$$
Then the coordinates of points A and B are the solutions to the system of equations
$$
\left\{\begin{array}{l}
y=k x, \\
y=x^{2}-x+3
\end{array}\right.
$$
Eliminating \( y \) from (1) a... | y=2x^2-x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,942 |
Example 6. Prove that the midpoints of a set of parallel chords in a hyperbola lie on
a straight line. | Let the equation of a hyperbola be
$$
b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2} \text {. }
$$
A set of parallel chords is given by $\left\{\begin{array}{l}x=h+t \cos \alpha, \\ y=k+t \sin \alpha \text {. }\end{array}\right.$, where $\alpha$ is a constant.
Let $\mathrm{P}_{1}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right), \mat... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,943 |
Example 7. One of a set of similar right hyperbolas centered at a fixed point $\mathrm{C}$ intersects with one of a set of circles of fixed radius $\mathrm{r}$ at four points $P, Q, R, S$. Prove that $\mathrm{PC}^{2}+\mathrm{QC}^{2}+\mathrm{RC}^{2}+\mathrm{SC}^{2}$ is a constant. | To prove that with a fixed point $\mathrm{C}$ as the origin, and the two asymptotes of a rectangular hyperbola as the coordinate axes, the equation of this set of similar rectangular hyperbolas can be written as
$$
x y=k,
$$
where $k$ is a parameter.
The equation of a set of circles with radius $r$ is
$$
(x-\alpha)^{2... | 4r^2 | Geometry | proof | Yes | Yes | cn_contest | false | 701,944 |
Example 1. Solve the inequality
$$
\frac{1-\sqrt{1-4 x^{2}}}{x}<3 .
$$ | The set of permissible values for the unknown $x$ is $-\frac{1}{2} \leqslant x < 0, 0 < x \leqslant \frac{1}{2}$. At this time, the arithmetic root $\sqrt{1-4 x^{2}}$ satisfies the inequality
$$
0 \leqslant \sqrt{1-4 x^{2}} < 1 \text{. }
$$
(1) When $-\frac{1}{2} \leqslant x < 0$, noting (2), the left end of inequality... | not found | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,945 |
Example 3. Solve the inequality $x+\sqrt{x+1}>\frac{11}{4}$. | Solution 1 The set of allowed values for the unknown $x$ is $x \geqslant-1$. The original inequality is equivalent to the following inequality:
$$
\sqrt{\mathrm{x}+1}>\frac{11}{4}-\mathrm{x} \text {. }
$$
This inequality is also equivalent to the following two systems of inequalities:
$$
\begin{array}{l}
\text { (1) }... | x>\frac{5}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,946 |
Theorem 2 If $f(x)$ is continuous on $[a, b]$, then $\mathrm{f}(\mathrm{x})$ is integrable on $[\mathrm{a}, \mathrm{b}]$, and the “Fundamental Theorem of Calculus” holds. That is, if $F(x)$ is any antiderivative of $f(x)$, then $\int_{a}^{b} f(x) d x=F(b)-F(a)$. | Proof [1] By Theorem 1, the upper limit function $G(x) = \int_{a}^{x} f(t) dt$ is an antiderivative of $f(x)$, so $G'(x) = f(x)$. Also, $F(x)$ is an antiderivative of $f(x)$, hence $F'(x) = f(x)$, thus $G'(x) = F'(x)$.
$\therefore F(x) = G(x) + C. \quad(C$ is a constant)
In the above equation, let $x = a$,
$\because G(... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,948 |
Example 4. Solve the inequality
$$
\frac{2 x^{2}}{1+x^{2}}-\frac{5 x}{\sqrt{1+x^{2}}}+2>0 .
$$ | Let $x=\operatorname{tg} \theta\left(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\right)$,
i.e., $2 \sin ^{2} \theta-5 \sin \theta+2>0$,
or $(\sin \theta-2)(2 \sin \theta-1)>0$.
Since $\sin \theta-2<0$, it follows that $2 \sin \theta-1<0$, $\sin \theta<\frac{1}{2},-\frac{\pi}{2}<\theta<\frac{\pi}{6}$. Also, $\operatorname{tg} ... | x<\frac{\sqrt{3}}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,949 |
Example 5. Solve the inequality $2 \sqrt{x} \geq 3-\frac{1}{x}$. | The set of permissible values for the unknown $x$ is $x>0$.
Let $f(x)=2 \sqrt{x}-\left(3-\frac{1}{x}\right)$, then the inequality $f(x) \geqslant 0$ is equivalent to the original inequality.
$$
f^{\prime}(x)=x^{-\frac{1}{2}}-x^{-2}=x^{-2}\left(x^{-\frac{3}{2}}-1\right) \text {. }
$$
Solve the equation $f^{\prime}(x)=0... | x>0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,950 |
Example 7. Solve the inequality
$$
\sqrt{\frac{1}{2} x+1}-1>\frac{1}{5}(2 x-7) .
$$ | The set of permissible values for the unknown $x$ is $x \geqslant-2$. (1)
$$
\text { Let } y_{1}=\sqrt{\frac{1}{2} x+1}-1, y_{2}=\frac{1}{5}(2 x-7) \text {, }
$$
Then the graphs of $\mathrm{y}_{1}$ and $\mathrm{y}_{2}$ are shown in Figure 1.
Now find the x-coordinate of the intersection point $\mathrm{A}$ of the parab... | -2 \leqslant x < 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,952 |
Example 8. Solve the inequality $\sqrt{1-x}-\sqrt{x}>\frac{1}{\sqrt{3}}$. | Solve for the set of permissible values of the unknown $\mathrm{x}$ as $0 \leqslant \mathrm{x} \leqslant 1$.
Let $\mathrm{y}_{1}=\sqrt{1-\mathrm{x}}, \mathrm{y}_{2}=\sqrt{\mathrm{x}}+\frac{1}{\sqrt{3}}$, then the graphs of $\mathrm{y}_{1}, \mathrm{y}_{2}$ are shown in Figure 2.
Now find the x-coordinate of the interse... | 0 \leqslant \mathrm{x}<\frac{3-\sqrt{5}}{6} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,953 |
1恻9. 筈不等式
$$
3 \sqrt{1-\frac{x^{2}}{4}}<\frac{1}{2} x+2
$$
1 side 9. Inequality
$$
3 \sqrt{1-\frac{x^{2}}{4}}<\frac{1}{2} x+2
$$ | We use the mixed system
$$
y<\frac{1}{2} x+2, y^{2}=9\left(1-\frac{x^{2}}{4}\right).
$$
$y \geqslant 0$ to replace the original inequality. The equation
$y^{2}=9\left(1-\frac{x^{2}}{4}\right)$ represents an ellipse (Figure 4).
The solution sought is the part of the ellipse that lies below the line $y=\frac{1}{2} x+2$,... | -2 \leqslant x < \frac{-2-3 \sqrt{6}}{5}, \frac{-2+3 \sqrt{6}}{5} < x \leqslant 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,954 |
$\triangle \mathrm{ABC}$ has vertices $\mathrm{A}(-\mathrm{a}, 0)$ and $\mathrm{B}(\mathrm{b}, 0) \quad(\mathrm{a}>0, \mathrm{b}>0)$ fixed, the third vertex is $\mathrm{C}(\mathrm{x}, \mathrm{y})$, if $\mathrm{CO}(\mathrm{O}$ is the origin) is the bisector of $\angle \mathrm{ACB}$, find the equation of the locus of poi... | Given $\angle \mathrm{ACO}=\angle \mathrm{BOC}$, we have $|\mathrm{CA}|:|\mathrm{CB}| = |A O|:|O B|$, i.e., $\frac{\sqrt{(x+a)^{2}+y^{2}}}{\sqrt{(x-b)^{2}+y^{2}}}=\frac{a}{b}$. Squaring both sides and simplifying, we get $(a-b) x^{2}+(a-b) y^{2}-2 a b x=0$.
(i) When $a=b$, the equation of the locus of point $C$ is $x=0... | x^{2}+y^{2}-\frac{2 a b}{a-b} x=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,955 |
Example 1. Given $A(-a, 0), B(a, 0)$, a moving point $M$ is on the circle $x^{2}+y^{2}=r^{2}$, $a>r>0$, and equilateral triangles $\mathrm{AMC}$ and BMD are constructed above the broken line AMB with $AM$ and $BM$ as sides, respectively. Find the equation of the trajectory of the midpoint $\mathrm{P}$ of $C D$. | Consider the x-axis as the real axis in the complex plane, and the y-axis as the imaginary axis. Let $Z_{\boldsymbol{\mu}}=r(\cos \theta + i \sin \theta) \cdot (0 \leqslant \theta < 2 \pi)$
$\overrightarrow{M_{1}}$ corresponds to the complex number: $(r \cos \theta - a) + ir \sin \theta$,
$\vec{A} M$ corresponds to t... | x^{2} + \left( y - \frac{\sqrt{3}}{2} a \right)^{2} = \frac{r^{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,956 |
Example 6. A geometric sequence with the first term and common ratio both being positive numbers, and an arithmetic sequence have the same first term, last term, and number of terms, then the sum of this geometric sequence is not greater than the sum of this arithmetic sequence. | Proof: Let the first term of a geometric sequence be $\mathrm{a}$, the common ratio be $q$, and the number of terms be $\mathrm{n}(\mathrm{a}, \mathrm{q}>0)$, then the sum of the first $\mathrm{n}$ terms is
$$
S_{n}=a\left(1+q+q^{2}+\cdots+q^{n-1}\right) .
$$
The sum of the first $\mathrm{n}$ terms of an arithmetic se... | S_{n} \leqslant Q_{n} | Algebra | proof | Yes | Yes | cn_contest | false | 701,958 |
Theorem 3 If $\mathrm{f}(\mathrm{x})$ is integrable on $[\mathrm{a}, \mathrm{b}]$ and has an antiderivative $F(x)$, then we still have:
$$
F(x)=\int_{a}^{x} f(t) d t+C .
$$ | Proof [2] Let $F^{\prime}(x)=f(x), \forall x \in[a, b]$.
Insert partition points $a=x_{0}<x_{1}<\cdots$
$$
\begin{array}{l}
<x_{n}=b, \text { then } F(b)-F(a)=\sum_{i=1}^{n}\left[F\left(x_{i}\right)\right. \\
\left.-F\left(x_{i-1}\right)\right]=\sum_{i=1}^{n} F^{\prime}\left(\xi_{i}\right)\left(x_{i}-x_{i-1}\right)
\en... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,959 |
$\begin{array}{l}1 \cdot \frac{1000^{2}}{252^{2}-248^{2}} \text { equals } \\ \text { (A) } 62500 \text {; (B) } ; 1000 \text {; (C) } 500 \text {; (D) } 250 \text {; } \\ \text { (E) } \frac{1}{2} \text {. }\end{array}$ | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,960 |
3. Let $\mathrm{n}$ be the smallest integer satisfying the following conditions: greater than 1, has no prime factors less than 10, and is not a prime number. Then
(A) $100<\mathrm{n} \leqslant 110$;
(B) $110<n \leqslant 120$;
(C) $120<\mathrm{n} \leqslant 130$;
(D) $130<\mathrm{n} \leqslant 140$;
(E) $140<\mathrm{n} \... | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 701,962 |
4. A rectangle intersects with a circle as shown in the figure, $\mathrm{AB}=4$, $\mathrm{BC}=5, \mathrm{DE}=3$, then $\mathrm{EF}$ equals
(A) 6 ; (B) 7 ;
(C) $\frac{20}{3}$; (D) 8 ;
(E) 9 . | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,963 |
5. The largest integer $n$ that satisfies $\mathrm{n}^{200}<5^{300}$ is
(A) 8 ;
(B) 9 ;
(C) 10 ;
(D) 11;
(E) 12 . | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,964 |
6. The number of boys in a school is three times the number of girls, and the number of girls is nine times the number of teachers. The numbers of boys, girls, and teachers are represented by the letters b, g, and t, respectively. The total number of boys, girls, and teachers can be expressed as
(A) $31 \mathrm{~b}$;
(... | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,965 |
7. Dave (a person's name) walks to school at an average of 90 steps per minute, with each step being 75 cm long, and it takes him 16 minutes to get to school. His brother Jack (a person's name) walks the same route to school, at an average of 100 steps per minute, with each step being 60 cm long. How long does it take ... | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,966 |
$9.4^{10} 5^{25}$ 的位数（用十进位制）是 (A) 31: (B) 30; (C) 29; (D) 28; (E) 27.
The number of digits (in decimal notation) of $9.4^{10} 5^{25}$ is (A) 31: (B) 30; (C) 29; (D) 28; (E) 27. | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,968 |
10. In the complex plane, the four vertices of a square represent four complex numbers, three of which are $1+2 \mathrm{i}, -2+\mathrm{i}$, and $-1-2 \mathrm{i}$. The fourth is
(A) $2+\mathrm{i}$;
(B) $2-\mathrm{i}$;
(C) $1-2 \mathrm{i}$;
(D) $-1+2 \mathrm{i}$;
(E) $-2-\mathrm{i}$. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,969 |
Theorem 4 Suppose $f(x)$ is integrable on $[a, b]$, and except for a finite number of points $\left\{x_{k}\right\}(k=0,1,2, \cdots, m+1)$ (let $a=x_{0}<x_{1}<\cdots<x_{n}=b$), there is $F^{\prime}(x)=f(x)$. At each point $x_{k}$ $(k=0,1, \cdots, m+1)$, $F(x)$ has a discontinuity of the first kind, then the formula is:
... | To prove that on each interval $\left[\mathbf{x}_{\mathbf{k}}, \mathbf{x}_{\mathbf{k}+1}\right]$, the function
$$
\Phi(\mathrm{x})=\left(\begin{array}{l}
F\left(\mathrm{x}_{\mathrm{k}}+0\right), \mathrm{x}=\mathrm{x}_{k}, \\
F(\mathrm{x}), \mathrm{x} \in\left(\mathrm{x}_{\mathbf{k}}, \mathrm{x}_{\mathrm{k}+1}\right), \... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,970 |
11. A computer has one key for performing a square operation, and another key for performing a reciprocal operation. Initially, a number $x \neq 0$ is input, and then the square operation and the reciprocal operation are alternately performed $n$ times each, with the final result denoted as $y$. Assuming the computer i... | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,971 |
12. If the sequence $\left\{a_{n}\right\}$ is defined by $a_{1}=2, a_{n+1}$ $=a_{n}+2 n(n \geqslant 1)$, then $a_{100}$ equals
(A) 9900;
(B) 9902;
(C) 9904;
(D) 10100;
(E) 10102. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,972 |
13. $\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$ equals
(A) $\sqrt{2}+\sqrt{3}-\sqrt{5}$;
(B) $4-\sqrt{2}-\sqrt{3}$;
(C) $\sqrt{2}+\sqrt{3}+\sqrt{6}-5$,
(D) $\frac{1}{2}(\sqrt{2}+\sqrt{5}-\sqrt{3})$;
(E) $\frac{1}{3}(\sqrt{3}+\sqrt{5}-\sqrt{2}$). | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,973 |
14. The product of all real roots of the equation $\mathbf{x}^{\log _{10} \mathbf{x}}=10$ is (A) 1 ; (B) -1 ; (C) 10 ; (D) $10^{-1}$; $(\mathrm{E})$ a different answer from $(\mathrm{A}) \sim(\mathrm{D})$. | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,974 |
15. If $\sin 2 x \sin 3 x=\cos 2 x \cos 3 x$, then one value of $\mathrm{x}$ is
(A) $18^{\circ}$;
(B) $30^{\circ}$; (C) $36^{\circ}$;
(D) $45^{\circ}$;
(E) $60^{\circ}$. | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,975 |
```
17. In the right triangle }\textrm{ABC}\mathrm{, }\textrm{AB}\mathrm{ is the hypotenuse, AC
=15, the altitude CH divides
AB into segments AH
and HB HB=16,
(A)120; (B)
144; (C) 150; (D) 216; (E) 144
``` | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,977 |
18. Choose a point $(\mathrm{x}, \mathrm{y})$ on the coordinate plane such that its distances to the $x$-axis, $y$-axis, and the line $x+y=2$ are all equal. Then $\mathbf{x}$ is
(A) $\sqrt{2}-1$; (B) $\frac{1}{2}$;
(C) $2-\sqrt{2}$;
(D) 1 ;
(E) not uniquely determined. | $\mathrm{E}$
18. Notice that the point $(x, y)$ sought is precisely the incenter and excenter of the triangle formed by the line $x+y=2$, the $x$-axis, and the $y$-axis, hence there are four points sought, i.e., (E). | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,978 |
19. A box contains 11 balls, each marked with a number $1,2, \cdots, 11$. If 6 balls are randomly drawn, the probability that the sum of their numbers is odd is
(A) $\frac{100}{231}$;
(B) $\frac{115}{231}$
(C) $\frac{1}{2}$,
(D) $\frac{118}{231}$;
(E) $\frac{6}{11}$. | D
19. The number of combinations when 6 balls are randomly taken from 11 balls is $\mathrm{C}_{11}^{\circ}$; and the number of combinations where the sum of the numbers on these 6 balls is odd is $\mathrm{C}_{0}^{1} \mathrm{C}_{5}^{5}+\mathrm{C}_{8}^{3} \mathrm{C}_{5}^{3}+\mathrm{C}_{0}^{5} \mathrm{C}_{5}^{1}$, the req... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 701,979 |
Example 3. Let $f(x)=\left\{\begin{array}{ll}0, & -1 \leqslant x<0, \\ 1, & 0<x \leqslant 1 .\end{array}\right.$ Find $\int_{-1}^{1} f(x) d x$.
untranslated text is kept in its original format and alignment. | Solve $\because F(x)=\left\{\begin{array}{ll}0, & -1 \leqslant x \leqslant 0, \\ 1, & 0<x \leqslant 1 .\end{array}\right.$
By formula (*), we get
$$
\begin{array}{c}
\int_{-1}^{1} f(x) d x=\int_{-1}^{0} 0 \cdot d x+\int_{0}^{1} 1 \cdot d x \\
=F(0)-F(0+0)+F(0-0)-F(-1)=1 .
\end{array}
$$ | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,981 |
22. Let $\mathrm{a}$ and $\mathrm{c}$ be fixed positive numbers, and for any real number $\mathrm{t}$, let $\left(x_{t}, y_{t}\right)$ be the vertex of the parabola $y=a x^{2}+t x+c$. For all real numbers $t$, the set of points $\left(x_{t}, y_{t}\right)$ drawn on the plane forms the graph of
(A) a straight line;
(B) a... | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,983 |
23. $\frac{\sin 10^{\circ}+\sin 20^{\circ}}{\cos 10^{\circ}+\cos 20^{\circ}}$ equals
(A) $\operatorname{tg} 10^{\circ}+\operatorname{tg} 20^{\circ}$;
(B) $\operatorname{tg} 30^{\circ}$
(C) $\frac{1}{2}\left(\operatorname{tg} 10^{\circ}+\operatorname{tg} 20^{\circ}\right)$,
(D) $\operatorname{tg} 15^{\circ}$;
(E) $\frac... | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,984 |
24. If $a$ and $b$ are positive real numbers, and the equations $x^{2}+a x+2 b=0$ and $x^{2}+2 b x+a=0$ both have real roots, then the minimum value of $a+b$ is
(A) 2;
(B) 3;
(C) 4;
(D) 5;
(E) 6 | $\mathrm{E}$
24. Since the two equations have real roots, their corresponding discriminants are not less than 0, i.e., $\mathrm{a}^{2}-8 \mathrm{~b} \geqslant 0$ and $\mathrm{b}^{2}-\mathrm{a} \geqslant 0$. Using the fact that $\mathrm{a}, \mathrm{b}$ are positive real numbers and the transitivity of inequalities, we g... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,985 |
25. The sum of the areas of all faces of a rectangular prism is $22 \mathrm{~cm}^{2}$, and the total length of all its edges is 24 cm. Then the length of each diagonal (in $\mathrm{cm}$) is
(A) $\sqrt{11}$;
(B) $\sqrt{12}$; (C) $\sqrt{13}$,
(D) $\sqrt{14}$;
(E) not uniquely determined. | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,986 |
26. In obtuse triangle $\mathrm{ABC}$, $\mathrm{AM}=\mathrm{MB}$, $\mathrm{MD} \perp \mathrm{BC}$, $\mathrm{EC} \perp \mathrm{BC}$. If the area of $\triangle \mathrm{ABC}$ is 24, then the area of $\triangle B E D$ is
(A) 9, (B) 12;
(C) 15;
(D) 18;
(E) Not uniquely determined. | B
26. Connect $CM$, using the equal areas of $\triangle AMC$ and $\triangle BMC$ we get the area of $\triangle BMC$ is 12. Then, using the equal areas of $\triangle DME$ and $\triangle DMC$ we get the area of $\triangle SED$ is 12, i.e., (B). | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,987 |
27. In $\triangle \mathrm{ABC}$, $\mathrm{D}$ and $\mathrm{F}$ are on $\mathrm{AC}$ and $\mathrm{BC}$ respectively, and $A B \perp A C, A F \perp B C, B D=$ $\mathrm{DC}=\mathrm{FC}=1$, find $\mathrm{AC}$.
(A) $\sqrt{2}$;
(B) $\sqrt{3}$;
(C) $\sqrt[3]{2}$;
(D) $\sqrt[3]{3}$,
(E) $\sqrt[4]{3}$. | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,988 |
28. The number of different integer pairs $(x, y)$ that satisfy $0<\mathrm{x}<\mathrm{y}$ and $\sqrt{1984}=\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}$ is
(A) 0,
(B) 1;
(C) 3;
(D) 4;
(E) 7. | C
28. From $\sqrt{1984}=8 \sqrt{ } 31$ and $0<x<y$, we can get $\sqrt{1984}=\sqrt{31}+7 \sqrt{31}=2 \sqrt{31}+6 \sqrt{31}$ $=3 \sqrt{31}+5 \sqrt{31}$, thus there are three pairs of positive integers $(\mathrm{x}, \mathrm{y})$, i.e., (C). | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,989 |
29. Let the real number pair $(x, y)$ satisfy $(x-3)^{2}+(y-$ $3)^{2}=6$, find the maximum value of $\frac{y}{x}$.
(A) $3+2 \sqrt{2}$;
(B) $2+\sqrt{3}$;
(C) $3 \sqrt{3}$;
(D) 6 ;
(E) $6+2 \sqrt{3}$. | A
29. Let $\frac{y}{x}=k$, substitute into the given equation, and use the discriminant to find the maximum value of $k$, which is (A). | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,990 |
30. For any complex number $\mathrm{W}=\mathrm{a}+\mathrm{bi}$, its modulus $| \mathrm{W} |$ is defined as the real number $\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$. If $\mathrm{W}=\cos 40^{\circ} + \mathrm{i} \sin 40^{\circ}$, then $\mid \mathrm{W}+2 \mathrm{~W}^{2}+3 \mathrm{~W}^{3}+\cdots + 9 \mathrm{~W}^{9} \mid^{-1}$... | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,991 |
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