problem stringlengths 29 1.36k | solution stringlengths 3 6.02k | answer stringlengths 1 100 | subject stringclasses 5
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Given that point $P$ is on the ellipse $\frac{y^{2}}{8} + \frac{x^{2}}{4} = 1$, $F\_1$, $F\_2$ are its two foci, and $\angle F\_1PF\_2 = 60^{\circ}$, find the area of $\triangle F\_1PF\_2$. | Since point $P$ is on the ellipse $\frac{y^{2}}{8} + \frac{x^{2}}{4} = 1$, and $F\_1$, $F\_2$ are its two foci with $\angle F\_1PF\_2 = 60^{\circ}$,
We have $|PF\_1| + |PF\_2| = 4\sqrt{2}$,
$\Rightarrow |PF\_1|^2 + |PF\_2|^2 + 2|PF\_1| \cdot |PF\_2| = 32$, $(1)$
And $(|PF\_1|^2 + |PF\_2|^2 - 2|PF\_1| \cdot |PF\_2| \... | \frac{4\sqrt{3}}{3} | Geometry | b800b2fd-c5b6-5cbd-8960-07c1ad3532b5 |
Zadam Heng bets Saniel Dun that he can win in a free throw contest. Zadam shoots until he has made $5$ shots. He wins if this takes him $9$ or fewer attempts. The probability that Zadam makes any given attempt is $\frac{1}{2}$. What is the probability that Zadam Heng wins the bet?
[i]2018 CCA Math Bonanza Individual R... | To find the probability that Zadam Heng wins the bet, we need to calculate the probability that he makes exactly 5 successful shots in 9 or fewer attempts. The probability of making any given shot is \(\frac{1}{2}\).
1. **Probability of making exactly 5 shots in 5 attempts:**
\[
P(\text{5 shots in 5 attempts}) =... | \frac{1}{2} | Combinatorics | ca7ce028-bc46-53b5-82c9-2f88ed44d236 |
10 Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=\frac{1}{4}, a_{2}=\frac{3}{4}, a_{n+1}=2 a_{n}-a_{n-1}(n \geqslant$
2), the sequence $\left\{b_{n}\right\}$ satisfies: $b_{1} \neq \frac{1}{4}, 3 b_{n}-b_{n-1}=n(n \geqslant 2)$, and the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is ... | 10 (1) Since $a_{n+1}=2 a_{n}-a_{n-1}(n \geqslant 2)$, we have
$$
a_{n+1}-a_{n}=a_{n}-a_{n-1}(n \geqslant 2).
$$
Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence.
Given $a_{1}=\frac{1}{4}, a_{2}=\frac{3}{4}$, we have $a_{n}=\frac{2 n-1}{4}$.
From $3 b_{n}-b_{n-1}=n$, we get $b_{n}=\frac{b_{n-1}+n}{3}$. Thus... | S_{n}=\frac{1}{4}n^{2}-(\frac{1}{3})^{n}+1 | Algebra | c7e04f9a-be86-5592-975f-01042b9d6ac1 |
8. In triangle $A B C$, the lengths of the sides are known: $|A B|=12,|B C|=13,|C A|=15$. On side $A C$, a point $M$ is taken such that the radii of the circles inscribed in triangles $A B M$ and $B C M$ are equal. Find the ratio $|A M|:|M C|$. | 8. Let $|A M|:|M C|=k$. The condition of equality of the radii of the circles inscribed in triangles $A B M$ and $B C M$ means that their areas are in the same ratio as their perimeters. From this, since the ratio of the areas is $k$, we get ; $B M \left\lvert\,=\frac{13 k-12}{1-k}\right.$. From this equality, in parti... | \frac{22}{23} | Geometry | 8df0764c-c802-55f3-b29a-954f74dd8ebe |
In $\Delta ABC$, $a+c=6$, and $(3-\cos A)\tan \frac{B}{2}=\sin A$, then the maximum area of $\Delta ABC$ is. | **Analysis**
This problem examines the application of the sine theorem and the formula for the area of a triangle in solving triangles. First, based on the given conditions and trigonometric formulas, combined with the sine theorem, we obtain $ac \leqslant 9$. Then, using the area formula, we can derive the result. Th... | 2\sqrt{2} | Geometry | e94a6201-25ad-5418-8e21-bca758de0124 |
Given an ellipse $C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\left(a \gt b \gt 0\right)$, with the left vertex $A$, the top vertex $B$, and $|AB|=\sqrt{7}$. A line $l$ is drawn passing through the right focus $F$, and when the line $l$ passes through point $B$, the slope is $-\sqrt{3}$.
$(1)$ Find the equation of $C$;
$(2)... | ### Solution:
#### Part (1): Finding the Equation of $C$
Given that the left vertex is $A(-a,0)$ and the top vertex is $B(0,b)$, with the right focus being $F(c,0)$, we have:
1. The distance between $A$ and $B$ is $|AB| = \sqrt{a^2 + b^2} = \sqrt{7}$.
2. The slope of the line $FB$ is given by $\frac{b-0}{0-c} = -\sq... | 2 | Geometry | d0e86bdb-ddb1-5819-b499-e73b447f7fc6 |
(1) Simplify: $\dfrac{\tan(3\pi{-}\alpha)\cos(2\pi{-}\alpha)\sin({-}\alpha{+}\dfrac{3\pi}{2})}{\cos({-}\alpha{-}\pi)\sin({-}\pi{+}\alpha)\cos(\alpha{+}\dfrac{5\pi}{2})}$;
(2) Simplify $\dfrac{\cos (\dfrac{\pi}{2}+\alpha)\sin (\dfrac{3\pi}{2}{-}\alpha)}{\cos (\pi{-}\alpha)\tan (\pi{-}\alpha)}$ | (1) The original expression:
$$\dfrac{\tan(3\pi{-}\alpha)\cos(2\pi{-}\alpha)\sin({-}\alpha{+}\dfrac{3\pi}{2})}{\cos({-}\alpha{-}\pi)\sin({-}\pi{+}\alpha)\cos(\alpha{+}\dfrac{5\pi}{2})}$$
Using the trigonometric identities:
- $\tan(a-b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}$
- $\cos(a-b) = \cos a \cos b + \sin ... | \cos \alpha | Algebra | 1ec0c1fb-c4cc-5169-bdb3-89d345e8e449 |
Problem 9.5. On the board, the graph of the function $y=x^{2}+a x+b$ is drawn. Yulia drew two lines parallel to the $O x$ axis on the same drawing. The first line intersects the graph at points $A$ and $B$, and the second line intersects the graph at points $C$ and $D$. Find the distance between the lines if it is know... | Answer: 24.
Solution. Let the first line have the equation $y=s$, and the second line have the equation $y=t$. Then the distance between the lines is $(t-s)$.
The length of segment $A B$ is equal to the absolute value of the difference of the roots of the equation $x^{2}+a x+b=s$. We can express the difference of the... | 24 | Algebra | 72e78fea-af9b-5c7d-b960-1cf102da2f44 |
Consider the ellipse $C$: $\frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}}=1 (a > b > 0)$ with an eccentricity of $\frac{1}{2}$. The distance from its left focus to the point $P(2,1)$ is $\sqrt{10}$.
(I) Find the standard equation of the ellipse $C$;
(II) If the line $l$: $y=kx+m$ intersects the ellipse $C$ at two points $A$ ... | (I) Since the distance from the left focus $(-c,0)$ to the point $P(2,1)$ is $\sqrt{10}$, we have $\sqrt{(2+c)^{2}+1}=\sqrt{10}$, which gives $c=1$.
Also, $e=\frac{c}{a}=\frac{1}{2}$, thus $a=2$ and $b^2=a^2-c^2=3$.
Therefore, the standard equation of the ellipse $C$ is $\frac{x^{2}}{4}+ \frac{y^{2}}{3}=1$.
(II) Let $... | (\frac{2}{7},0) | Geometry | 6b048d2b-0eaa-5526-bd47-cfbf98ead1e8 |
The lateral area of a cone is $6\pi$, and the radius of the base circle is $2$. Find the length of the generatrix of the cone. | To find the length of the generatrix ($l$) of the cone, we start with the formula for the lateral area of a cone, which is given by the product of $\pi$, the radius of the base ($r$), and the length of the generatrix ($l$). In this problem, the lateral area is given as $6\pi$, and the radius ($r$) is $2$. Thus, we have... | 3 | Geometry | 40d1074b-d23b-58f1-bcaa-84da3e95a09f |
Given $f(x)=ae^{x}+ \frac {1}{ae^{x}}+b$ where $a > 0$.
(1) Find the minimum value of $f(x)$ on $[0,+\infty)$.
(2) If the tangent line equation of the curve $y=f(x)$ at the point $(2,f(2))$ is $3x-2y=0$, find the values of $a$ and $b$. | (1) Let $t=e^{x}$ where $t\geqslant 1$,
then $y=at+ \frac {1}{at}+b \Rightarrow y'=a- \frac {1}{at^{2}}= \frac {a^{2}t^{2}-1}{at^{2}}$.
$\quad\quad$ (i) When $a\geqslant 1$, $y' > 0 \Rightarrow y=at+ \frac {1}{at}+b$ is increasing on $t\geqslant 1$,
$\quad\quad\quad\quad$ thus the minimum value of $f(x)$ is $a+ \frac ... | a= \frac {2}{e^{2}}, b= \frac {1}{2} | Calculus | 8aad8bc7-0779-5d57-88ae-d6ec1b030cd9 |
Given that function $f(x)$ satisfies $f(n^2) = f(n) + 2$ for $n \geq 2$ and $f(2) = 1$, what is the value of $f(256)$? | We know that $256 = 16^2$, $16 = 4^2$, and $4 = 2^2$. Given this, and the recursive relationship $f(n^2) = f(n) + 2$ for $n \geq 2$, we can calculate $f(256)$ step by step:
First, let's apply the relationship to $f(4)$:
$$f(4) = f(2^2) = f(2) + 2 = 1 + 2 = 3.$$
Now let's use that result to calculate $f(16)$:
$$f(16) ... | 7 | Algebra | 32fd4520-182d-5eb6-86c1-f8f0a13ade31 |
Given $a-2b=-2$, find the value of $4-2a+4b$. | Given $a-2b=-2$, we want to find the value of $4-2a+4b$.
Starting from the given equation:
\[a - 2b = -2\]
We can manipulate the expression $4-2a+4b$ by factoring out $-2$ from $-2a+4b$:
\[4 - 2a + 4b = 4 - 2(a - 2b)\]
Substituting the given $a - 2b = -2$ into the equation:
\[4 - 2(-2) = 4 + 4\]
Simplifying the rig... | 8 | Algebra | 0f15605c-f8d3-5720-9b4f-fa49d344d632 |
Given the sets $A=\{x \mid 1 \leq x < 7\}$, and $C = \{x \mid x < a\}$, with the universal set being the set of real numbers $\mathbb{R}$, and $A \cap C \neq \varnothing$, find the range of values for $a$. | The sets $A$ and $C$ have a nonempty intersection, which means there exists at least one element that belongs to both $A$ and $C$.
The set $A$ contains all real numbers $x$ that satisfy $1 \leq x 1$.
Step-by-step, the reasoning can be outlined as follows:
1. Start with the definition of the sets $A$ and $C$ and the... | a > 1 | Algebra | 746e529b-6b90-5b10-a9c1-d1859ebf2050 |
5. In the number $2 * 0 * 1 * 6 * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,7,8,9$ (digits can repeat) so that the resulting 11-digit number is divisible by 12. In how many ways can this be done? | Answer: 1296.
Solution. For a number to be divisible by 12, it is necessary and sufficient that it is divisible by 4 and by 3. To ensure divisibility by 4, we can choose 0, 4, or 8 as the last digit (3 ways).
To ensure divisibility by 3, we proceed as follows. We will choose three digits arbitrarily (this can be done... | 1296 | Combinatorics | 38348021-b86f-5c29-9a02-f3768056ea2b |
In the expansion of $(1-\frac{y}{x})(x+y)^{8}$, the coefficient of $x^{2}y^{6}$ is ______ (provide your answer as a number). | To solve for the coefficient of $x^{2}y^{6}$ in the expansion of $(1-\frac{y}{x})(x+y)^{8}$, we first expand the expression as follows:
\[
(1-\frac{y}{x})(x+y)^{8} = (x+y)^{8} - \frac{y}{x}(x+y)^{8}
\]
Next, we apply the binomial theorem to identify the term $x^{2}y^{6}$ in each part of the expansion:
1. In the firs... | -28 | Algebra | d6460d76-3449-5dc7-892f-aa4caec9a97b |
Given an arithmetic sequence $\left\{a_{n}\right\}$ that satisfies $(a_{3}=7)$, $(a_{5}+a_{7}=26)$, and its sum of the first $n$ terms is denoted as $(S_{n})$.
(1) Find $(a_{n})$ and $(S_{n})$;
(2) Let $(b_{n}=\frac{1}{a_{n}^{2}-1}\left(n∈\mathbb{N}^{*}\right))$, find the sum of the first $n$ terms of the sequence $\l... | (1) Let the common difference of the arithmetic sequence $\left\{a_{n}\right\}$ be $d$.
Given $(a_{3}=7)$ and $(a_{5}+a_{5}=26)$, we have the system of equations:
$$\begin{cases} & a_{1}+2d=7, \\\\ & 2a_{1}+10d=26. \\\\ \end{cases}$$
Solving this system, we find $(a_{1}=3)$ and $(d=2)$.
Thus, $(a_{n}=a_{1}+(n-1)d=3+... | T_{n}=\frac{n}{4(n+1)} | Algebra | 62a90fd3-3d3c-5cf2-a4c9-8abb781ef331 |
Given that $\left\{\begin{array}{l}x=1\\ y=-2\end{array}\right.$ is a solution of the system of equations $\left\{\begin{array}{l}2ax-3y=10-b\\ ax-by=-1\end{array}\right.$, find $\left(b-a\right)^{3}$. | To solve for $a$ and $b$, we substitute $x=1$ and $y=-2$ into the given system of equations:
1. Substituting into the first equation: $2ax - 3y = 10 - b$
\[2a(1) - 3(-2) = 10 - b\]
Simplifying, we get:
\[2a + 6 = 10 - b\]
\[2a = 4 - b\]
\[a = 2 - \frac{b}{2}\]
2. Substituting into the second equation: ... | -125 | Algebra | a0c1baf6-caa9-57da-a8f4-f1c952a7bbac |
13.035. A shoe factory completed $20 \%$ of the monthly plan in the first week, produced $120 \%$ of the amount of products made in the first week in the second week, and produced $60 \%$ of the products made in the first two weeks combined in the third week. What is the monthly production plan for shoes, if it is know... | ## Solution.
Let $x$ be the monthly production plan for shoes. In the first week, the factory produced $0,2 x$ pairs of shoes, in the second week - $1,2 \cdot 0,2 x$; in the third week - $0,6(0,2 x+1,2 \cdot 0,2 x)$ pairs. According to the condition,
$0,2 x+1,2 \cdot 0,2 x+0,6(0,2 x+1,2 \cdot 0,2 x)+1480=x$, from whi... | 5000 | Algebra | ed09537d-785c-5170-ab95-84588644c3c6 |
3. Cubes with side lengths of 1, 2, 3, and so on were placed on the floor next to each other. It turned out that the volume of the resulting staircase is 396900. How many cubes were placed? | Answer: 35
## Examples of answer recording:
45 | 35 | Number Theory | c8261fc1-2761-5de1-8d16-0611c48164fe |
Given the sets $A=\{(x,y)|y=3^{x}\}$ and $B=\{(x,y)|y=2^{-x}\}$, then $A\cap B=$____. | **Analysis**
This question examines the intersection of sets. According to the problem, we set up the system of equations $\begin{cases}y=3^{x} \\ y=2^{-x}\end{cases}$, and solving this system will yield the result.
**Solution**
Solve the system $\begin{cases}y=3^{x} \\ y=2^{-x}\end{cases}$,
We get $\begin{cases}x=... | \{(0,1)\} | Algebra | e2cfceb5-e33b-5525-ac50-a633f9d39447 |
Given the function, $f(x)=\sqrt{|x+1|+|x-3|-m}$, its domain is $\mathbb{R}$.
$(1)$ Find the range of the real number $m$;
$(2)$ If the maximum value of the real number $m$ is $n$, and the positive real numbers $a$, $b$ satisfy $\frac{2}{3a+b}+\frac{1}{a+2b}=n$, find the minimum value of $7a+4b$. | Solution: $(1)$ Since the domain of the function is $\mathbb{R}$,
it follows that $|x+1|+|x-3|-m\geqslant 0$ always holds.
Let the function $g(x)=|x+1|+|x-3|$, then $m$ cannot be greater than the minimum value of the function $g(x)$.
Also, $|x+1|+|x-3|\geqslant |(x+1)-(x-3)|=4$, which means the minimum value of $g(x... | \frac {9}{4} | Algebra | acfc7dd5-9978-5b2e-9a71-d6a824ed284c |
Factorize $2x^{2}-4x+2=$____. | To factorize the given quadratic expression $2x^{2}-4x+2$, we start by factoring out the greatest common factor (GCF), which is $2$ in this case. This gives us:
$$2x^{2}-4x+2 = 2(x^{2}-2x+1)$$
Next, we notice that the expression inside the parentheses, $x^{2}-2x+1$, is a perfect square trinomial. It can be factored i... | 2\left(x-1\right)^{2} | Algebra | c2b64ea3-bb72-573c-b393-5e55a1939eac |
5. Find the sum of all natural numbers that have exactly four natural divisors, three of which (of the divisors) are less than 15, and the fourth is not less than 15. | Answer: $\left((2+3+5+7+11+13)^{2}-\left(2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}\right)\right) / 2-$ $6-10-14+27=649$.
Solution: The numbers $N$ specified have exactly 4 divisors either if $N=$ $p^{3}$, or if $N=p q$, where $p$ and $q$ are primes. In the first case, only 27 fits. In the second case, we need to consider ... | 649 | Number Theory | c0815087-02bb-5236-a3f6-09bfc699fb34 |
Given a point $P$ on the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$, if the distance from point $P$ to focus $F\_1$ is $6$, then the distance from point $P$ to the other focus $F\_2$ is _____. | By the definition of an ellipse, we know that the sum of the distances from any point on the ellipse to the two foci is constant and equal to the length of the major axis, which is $2a$. In this case, $2a = 10$.
Given that the distance between point $P$ and focus $F\_1$ is $6$, we can find the distance between point $... | 4 | Geometry | 142391ae-8c0c-5df2-b651-dc6c545e6378 |
Given that one focus of the hyperbola $\frac{x^2}{9} - \frac{y^2}{m} = 1$ is at $(5, 0)$, find the real number $m$. | **Analysis**
This problem tests basic properties of hyperbolas and the student's computational ability. By utilizing the given information that one focus of the hyperbola $\frac{x^2}{9} - \frac{y^2}{m} = 1$ is at $(5, 0)$, we can derive the equation $9 + m = 25$ to find the value of $m$.
**Step-by-step Solution**
1.... | 16 | Algebra | d0e86851-a761-5b40-969d-9196a9f958b5 |
The equation $x^3 + (1-3a)x^2 + 2a^2x - 2ax + x + a^2 - a = 0$ has exactly one root. Find the range of values for $a$. | Since $x^3 + (1-3a)x^2 + 2ax - 2ax + x + a^2 - a = 0$,
we have $x^3 - 3ax^2 + 2a^2x + x^2 - (2a-1)x + a^2 - a = 0$,
thus $x(x-a)(x-2a) + (x-a)[x-(a-1)] = 0$,
which simplifies to $(x-a)[x^2 - (2a-1)x - (a-1)] = 0$,
Given that the equation $x^3 + (1-3a)x^2 + 2a^2x - 2ax + x + a^2 - a = 0$ has exactly one root,
... | \frac{-\sqrt{3}}{2} < a < \frac{\sqrt{3}}{2} | Algebra | b9212b9a-1573-5470-9af5-179690a4b59a |
If the equation $x^{2}-2x+k=0$ has two equal real roots with respect to $x$, then the value of $k$ is ______. | To solve for the value of $k$ in the equation $x^{2}-2x+k=0$ given that it has two equal real roots with respect to $x$, we use the discriminant formula for a quadratic equation, which is $\Delta = b^{2} - 4ac$. For two equal real roots, the discriminant $\Delta$ must be equal to $0$. The equation can be rewritten in t... | 1 | Algebra | fda28b15-de1f-501a-845b-74904eae0708 |
Let's set $S_{n}=1^{3}+\cdots+n^{3}$ and $u_{n}=1+3+5+\cdots+(2 n-1)$ the sum of the first $n$ odd numbers.
1. By evaluating for small $n$, predict and then prove by induction a formula for $u_{n}$.
2. Using the formula found above for the sum of linear terms, find $u_{n}$ again.
3. Can we find a third way to derive a... | 1. Notice that $u_{1}=1, u_{2}=4, u_{3}=9$ and $u_{4}=16$. It seems legitimate to think that $u_{n}=n^{2}$. Let's prove this by induction.
Initialization: For $n=1, u_{1}=1^{2}$
Hereditary: Let $n \in \mathbb{N}^{*}$ such that $u_{n}=n^{2}$. Then
$$
u_{n+1}=u_{n}+(2(n+1)-1)=u_{n}+2 n+1 \underset{\text { H.R. }}{=} n... | S_{n}=(\frac{n(n+1)}{2})^{2} | Algebra | 568dc221-9943-5ce0-ab10-106c10e5f743 |
In the Cartesian coordinate system xOy, the parametric equation of circle C is $$\begin{cases} x=cos\alpha, \ y=1+sin\alpha \end{cases}$$ (where α is the parameter). Establish a polar coordinate system with the coordinate origin O as the pole and the positive semi-axis of x as the polar axis. The polar coordinate equat... | From $$\begin{cases} x=cos\alpha, \ y=1+sin\alpha \end{cases}$$ we obtain the standard equation of circle C: x2 + (y - 1)2 = 1.
From ρcosθ - ρsinθ - 1 = 0 we get x - y - 1 = 0.
Thus, the distance d from the center (0, 1) to the line is given by: d = $$\frac{|0 - 1 - 1|}{\sqrt{1 + 1}}$$ = $$\sqrt{2}$$.
Therefore, the... | \sqrt{2} | Geometry | 7ef1cf66-82ac-5c22-bbcb-67778d50e849 |
Simplify: $\frac{6a^2bc}{3ab}=\_\_\_\_\_\_.$ | To simplify the given expression $\frac{6a^2bc}{3ab}$, we can break down the simplification process as follows:
First, factor out the common terms in the numerator and the denominator:
$$\frac{6a^2bc}{3ab} = \frac{3ab \cdot 2ac}{3ab}$$
Here, we see that $3ab$ is a common term in both the numerator and the denominato... | 2ac | Algebra | a177372b-f6d6-5423-9db9-572b9e0010b8 |
7. Find the area of triangle $ABC$, the vertices of which have coordinates $A(0,0), B(1424233,2848467), C(1424234,2848469)$. Round your answer to the hundredths. | Solution. Note that points $B$ and $C$ lie on the line $y=2x+1$. Their abscissas differ by 1, hence $BC=\sqrt{5}$. The length of the height of triangle $ABC$, drawn from vertex $A$, is equal to the distance $h$ from point $A$ to the line $y=2x+1$, which, in turn, is equal to $1 / \sqrt{5}$. The desired area
$$
S_{ABC}... | 0.50 | Geometry | f24d0fd7-ed05-5904-a413-87b3473f94c8 |
Find the limit, when $n$ tends to the infinity, of $$\frac{\sum_{k=0}^{n} {{2n} \choose {2k}} 3^k} {\sum_{k=0}^{n-1} {{2n} \choose {2k+1}} 3^k}$$ | To find the limit of the given expression as \( n \) tends to infinity, we start by analyzing the sums in the numerator and the denominator using the Binomial Theorem.
1. **Numerator Analysis:**
The numerator is given by:
\[
\sum_{k=0}^{n} \binom{2n}{2k} 3^k
\]
Using the Binomial Theorem, we can express... | \sqrt{3} | Calculus | aeb26868-b954-5d6f-9151-0d128c9ef74f |
7. (10 points) On the board, 29 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 29 minutes? | Answer: 406.
Solution: Let's represent 29 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connec... | 406 | Number Theory | fb2c03fc-1866-5928-8bec-251f86422b90 |
(1) Given the function $f(x)=2x+2\sin x+\cos x$, the slope of the tangent line at the point $(α,f(α))$ is 2. Find the value of $\frac{\sin (π-α)+\cos (-α)}{2\cos (\frac{π}{2}-α)+\cos (2π-α)}$.
(2) In triangle $△ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given $a=1$ and $a\c... | (1) First, we find the derivative of the function $f(x)$. We have $f'(x)=2+2\cos x-\sin x$. Given $f'(α)=2$, we can deduce that $\tan α=2$. Using this information, we can simplify the expression as follows:
$$
\begin{align}
\frac{\sin (π-α)+\cos (-α)}{2\cos (\frac{π}{2}-α)+\cos (2π-α)} &= \frac{\sin α+\cos α}{2\sin α+... | (2,3] | Algebra | d7851ced-6e5b-5328-8aa3-1c5b05cf1741 |
Given vectors $\overrightarrow{m}=(-1,\cos \omega x+ \sqrt {3}\sin \omega x)$ and $\overrightarrow{n}=(f(x),\cos \omega x)$, where $\omega > 0$, and $\overrightarrow{m}\perp \overrightarrow{n}$. Also, the graph of the function $f(x)$ has a distance of $\frac {3}{2}\pi$ between any two adjacent axes of symmetry.
(Ⅰ) F... | Solution:
(Ⅰ) Since $\overrightarrow{m}=(-1,\cos \omega x+ \sqrt {3}\sin \omega x)$ and $\overrightarrow{n}=(f(x),\cos \omega x)$, and $\overrightarrow{m}\perp \overrightarrow{n}$,
$\therefore \overrightarrow{m}\cdot \overrightarrow{n}=0$,
$\therefore f(x)=\cos \omega x(\cos \omega x+ \sqrt {3}\sin \omega x)= \fr... | - \frac {13 \sqrt {2}}{14} | Algebra | adec285d-9c29-57e6-85a6-b7401a1576b8 |
3. Among all lines passing through the point $(\sqrt{1994}, 0)$, the number of lines that pass through two different rational points is | 3. 1 line.
Obviously, the line $l$ that meets the conditions is not parallel to the $y$-axis, so we can set its equation as $y=k(x-\sqrt{1994})$. Since $l$ passes through two rational points, let's assume they are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$. If $k$ is a rational number, $y_{1}=k\left(x_{1}-\... | 1 | Number Theory | c38b678b-3cda-5bea-9cce-9541b1d4627f |
The function $y=(k+2)x+1$ is an increasing function on the set of real numbers. Find the range of $k$. | Since the function $y=(k+2)x+1$ is an increasing function on the set of real numbers,
when $k+2=0$, $y=1$ is a constant function, which does not meet the condition of the problem,
therefore, $k+2>0$, which implies $k>-2$.
Hence, the answer is $\boxed{(-2, +\infty)}$. | (-2, +\infty) | Algebra | 48b2cc53-b637-5ebf-a2ed-e0983b06a804 |
Given point $P(3m+6, m-3)$, find the coordinates of point $P$ according to the following conditions:$(1)$ Point $P$ lies on the angle bisector in the first and third quadrants;$(2)$ The ordinate of point $P$ is $5$ greater than the abscissa;$(3)$ Point $P$ lies on the line passing through point $A(3, -2)$ and parallel ... | To solve for the coordinates of point $P(3m+6, m-3)$ under the given conditions, we proceed as follows:
**Condition (1):** Point $P$ lies on the angle bisector in the first and third quadrants.
This implies that the abscissa and ordinate of point $P$ are equal. Therefore, we have:
\[3m+6 = m-3\]
Solving for $m$, we g... | (3, -4) | Algebra | cc1a3b6e-a76e-5a9f-95ef-3f442a9a849a |
Calculate: $\sqrt{54} \times \sqrt{\frac{1}{3}} = \_\_\_\_\_\_$. | To solve $\sqrt{54} \times \sqrt{\frac{1}{3}}$, we follow these steps:
1. Combine the square roots into a single square root, since $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$.
2. Calculate the product inside the square root.
Therefore,
\[
\begin{align*}
\sqrt{54} \times \sqrt{\frac{1}{3}} & = \sqrt{54 \times \fr... | 3\sqrt{2} | Algebra | 498e3bc5-04a2-51a9-af39-98d360585670 |
The equation of the tangent line to the curve $y=x^3-x+3$ at the point $(1,3)$ is __. | To find the equation of the tangent line to the curve at a specific point, we first need to determine the derivative of the function, which represents the slope of the tangent line.
Given the function $y=x^3-x+3$, let's calculate its derivative with respect to $x$:
\[ y' = \frac{d}{dx}(x^3) - \frac{d}{dx}(x) + \frac{d... | 2x - y + 1 = 0 | Calculus | 5cf68b4c-ea73-5c5f-93e1-4d984ead213a |
11. (2005 Balkan Mathematical Olympiad) Find the positive integer solutions of the equation $3^{x}=2^{x} y+1$.
untranslated text remains the same as requested. However, if you need any further assistance or a different format, please let me know! | 11. Rewrite the equation as
$$
3^{x}-1=2^{x} y \text {. }
$$
This indicates that if $(x, y)$ is a solution, then $x$ cannot exceed the exponent of the factor 2 in the standard factorization of $3^{x}-1$. Let $x=2^{m}(2 n+1)$, where $m, n$ are non-negative integers. Thus, we have
$$
3^{x}-1=3^{2^{m}(2 n+1)}-1=\left(3^{... | (x,y)=(1,1),(2,2),(4,5) | Number Theory | f2dec61b-6c64-5585-b968-b2e1730dbb36 |
22. Try to write two sets of integer solutions for the indeterminate equation $x^{2}-2 y^{2}=1$ as $\qquad$ . | 22 . $(3,2),(17,12)$ etc. | (3,2),(17,12) | Number Theory | 832b31c2-9d8e-5913-8c30-5866065e18a1 |
Throw a dice twice in succession, what is the probability that the sum of the points facing up is $7$? | To solve this problem, we will apply the classical probability model, which requires determining the number of possible basic outcomes when a dice is thrown twice, and the number of desired outcomes where the sum of the points is $7$.
Step 1: Calculate the total number of possible basic outcomes when the dice is throw... | \frac{1}{6} | Combinatorics | 1df1056a-74a7-5e9e-8a89-6fa4e9b40f30 |
8.4 The sequence of numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}, \ldots$ satisfies the relations $\mathrm{a}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}-1} \cdot \mathrm{a}_{\mathrm{n}-3}$ for $\mathrm{n}=4,5,6, \ldots$ Find $\mathrm{a}_{2019}$ if it is known that $\mathrm{a}_{1}=1,... | Solution. It is clear that all members of this sequence are equal to $\pm 1$. We find:
$$
\begin{aligned}
& a_{n}=\left(a_{n-1}\right) \cdot a_{n-3}=\left(a_{n-2} \cdot a_{n-4}\right) \cdot a_{n-3}=\left(a_{n-2}\right) \cdot a_{n-4} \cdot a_{n-3}= \\
& =\left(a_{n-3} \cdot a_{n-4}\right) \cdot a_{n-4} \cdot a_{n-3}=a_... | -1 | Algebra | 055b6813-30b9-5bcc-842e-f00040e88591 |
7.4. In triangle $A B C$, the angles $A$ and $C$ at the base are $20^{\circ}$ and $40^{\circ}$ respectively. It is known that $A C - A B = 5$ (cm). Find the length of the angle bisector of angle $B$. | Answer: 5 cm. Solution. Let $B M$ be the bisector of angle $B$. Mark a point $N$ on the base $A C$ such that $A N=A B$. Then triangle $A B N$ is isosceles and $\angle A B N=\angle A N B=80^{\circ}$. Since $\angle A B M=\frac{180^{\circ}-20^{\circ}-40^{\circ}}{2}=60^{\circ}$, then $\angle B M N=\angle A+\angle A B M=20^... | 5 | Geometry | 10a28ac7-fdc2-5347-a702-5804cfcf9462 |
## Subject 3.
Let $\mathrm{ABCD}$ be a rhombus with $A B=6 \mathrm{~cm}$ and $m(\hat{A})=45^{\circ}$. At point $\mathrm{O}$, the center of the circumcircle of triangle $\mathrm{ABD}$, the perpendicular $\mathrm{OM}$ is raised to the plane of the rhombus, $O M=4 \mathrm{~cm}$. Find:
a) The distance from point O to the... | 3. Let $ABCD$ be a rhombus with $AB=6 \, \text{cm}$ and $m(\hat{A})=45^{\circ}$. At point $O$, the center of the circumcircle of triangle $ABD$, the perpendicular $OM$ is raised to the plane of the rhombus, $OM=4 \, \text{cm}$. Find:
a) The distance from point $O$ to the plane $(MBC)$.
b) The cosine of the angle betw... | \frac{3}{5} | Geometry | 2d990d15-d516-5463-acc3-a710d6ed4d68 |
Select one even number and three odd numbers from the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 to create a four-digit number without any repeating digits. How many such four-digit numbers are possible? | To solve this problem, we need to count the number of ways to form a four-digit number with one even digit and three odd digits without repetition. We need to consider two cases separately: when the even digit is not 0 and when the even digit is 0.
**Case 1: The even digit is not 0.**
- There are 4 choices for the eve... | 1140 | Combinatorics | dd49d163-32e4-53c6-9c39-73bf2fa768ee |
Let line $l$ pass through point $P(-3,3)$, and its angle of inclination is $\frac{5\pi}{6}$.
$(1)$ Write the standard parametric equation of line $l$;
$(2)$ Suppose this line intersects curve $C: \begin{cases}x=2\cos \theta \\ y=4\sin \theta\end{cases}$ (where $\theta$ is the parameter) at points $A$ and $B$, find th... | Solution:
$(1)$ The parametric equation of line $l$ is $\begin{cases}x=-3+t\cos \frac{5\pi}{6}=-3- \frac{ \sqrt{3}}{2}t \\ y=3+t\sin \frac{5\pi}{6}=3+ \frac{1}{2}t\end{cases}$,
$(2)$ Eliminating the parameter $\theta$ from the parametric equations of curve $C$, we get $4x^{2}+y^2-16=0$.
Substituting the parametric... | \frac{116}{13} | Algebra | 8a3579d0-9c2b-5129-b24c-97e77005728e |
A bag contains $5$ balls of the same size, numbered $1$, $2$, $3$, $4$, $5$ respectively. Three balls are randomly drawn from the bag. Let $\xi$ be the highest number among the drawn balls. Calculate $E\xi=$ \_\_\_\_\_\_. | From the given problem, we know that $\xi$ can take on the values $3$, $4$, or $5$. We first calculate the probabilities of each case:
$P(\xi=3) = \dfrac{C_3^3}{C_5^3} = \dfrac{1}{10}$,
$P(\xi=4) = \dfrac{C_3^2 C_1^1}{C_5^3} = \dfrac{3}{10}$,
$P(\xi=5) = \dfrac{C_4^2 C_1^1}{C_5^3} = \dfrac{6}{10}$.
Now, we can comp... | \dfrac{9}{2} | Combinatorics | cbfc0a60-c1f5-5ff8-b16b-94938942421a |
Taxi driver Xiao Shi operated his taxi service on People's Avenue, which runs east to west, one afternoon. If east is defined as positive and west as negative, his driving distances (in kilometers) that afternoon were: $+15$, $-3$, $+14$, $-11$, $+10$, $-12$.
$(1)$ When Xiao Shi dropped off the last passenger at thei... | ### Solution
#### Part 1: Distance from Starting Point
To find out how far Xiao Shi was from the starting point after dropping off the last passenger, we sum up all his driving distances, keeping in mind the direction he drove (east as positive, west as negative):
\[
\begin{align*}
\text{Total Distance} &= +15 + (-3... | 65a \text{ liters} | Algebra | 5bd13322-9061-5942-bd3c-d1ef6d8c2191 |
5. In order to improve students' basic sports skills and enthusiasm for participating in physical exercise, the school has set up 6 training groups: basketball, football, badminton, table tennis, tennis, and track and field. Each student is required to participate in the training of any two groups. To ensure that there... | 5. 【Answer】31 | 31 | Combinatorics | 01a3e44e-29ff-54ac-9c10-b57e49d0287d |
6. Find the minimum value of the function $f(x)=\sqrt{x^{2}-8 x+25}+\sqrt{x^{2}-4 x+13}$. | Solution. By completing the square under each of the radicals, we transform the function into the form
$$
f(x)=\sqrt{(x-4)^{2}+9}+\sqrt{(x-2)^{2}+9}
$$
For this expression, a geometric interpretation can be proposed. Consider each of the radicals as the distance from the point with coordinates $(x ; 0)$ to the point ... | 2\sqrt{10} | Algebra | 5d63e8d3-e5f6-5f1a-b390-2253703ceb2e |
Given that the distance traveled by a free-falling object is $$\frac{1}{2}gt^2$$, the instantaneous velocity of the object at time $t_0$ is ______. | To find the instantaneous velocity of an object at a particular time $t_0$, we need to determine the derivative of the distance with respect to time. The distance $s$ traveled by a free-falling object is given by the formula:
$$s = \frac{1}{2}gt^2$$
Differentiating both sides of the equation with respect to time $t$, ... | gt_0 | Calculus | 7b69a171-27ff-5c5e-a6ea-6adae624dd38 |
Calculate the value of $\sin 63^{\circ}\cos 18^{\circ} + \cos 63^{\circ}\cos 108^{\circ}$. | To solve the given problem, we'll use the sum-to-product identities as well as the values of special angles in trigonometry.
We know that $\cos(90^{\circ} + \theta) = -\sin\theta$. Let's use this to simplify the second term:
\[
\cos 108^{\circ} = \cos (90^{\circ} + 18^{\circ}) = -\sin 18^{\circ}
\]
Now use the sum-t... | \frac{\sqrt{2}}{2} | Algebra | b7a41837-801f-5ae8-8294-8962bcd2903d |
If the slope $k$ of a line satisfies $k \in (0, \sqrt{3})$, then the range of values for the inclination angle $\alpha$ of the line is ______. | To solve for the range of values for the inclination angle $\alpha$ of a line with slope $k$ satisfying $k \in (0, \sqrt{3})$, we follow these steps:
1. **Understanding the Relationship**: We start by acknowledging the relationship between the slope of a line and its inclination angle, which is given by $k = \tan(\alp... | \left(0, \frac{\pi}{3}\right) | Geometry | 830a4528-0c1d-5166-bc6f-6313e1d77081 |
(1) Let $f(x) = \frac{e^x - e^{-x}}{2}$, and $g(x) = \frac{e^x + e^{-x}}{2}$. Prove that $f(2x) = 2f(x) \cdot g(x)$;
(2) If $x\log_{3}4 = 1$, find the value of $4^x + 4^{-x}$. | (1) Proof: Since $f(2x) = \frac{e^{2x} - e^{-2x}}{2}$,
$2f(x)g(x) = 2 \cdot \frac{e^x - e^{-x}}{2} \cdot \frac{e^x + e^{-x}}{2} = \frac{e^{2x} - e^{-2x}}{2}$,
Therefore, $f(2x) = 2f(x) \cdot g(x)$.
(2) Solution: Since $x\log_{3}4 = 1$, then $x = \log_{4}3$,
By the definition and properties of logarithms, we get... | \frac{10}{3} | Algebra | c371db9c-f7df-5fac-bcc4-0ca791f6c03f |
Given $\log_a 2 = m$ and $\log_a 3 = n$.
(1) Find the value of $a^{2m-n}$;
(2) Express $\log_a 18$ in terms of $m$ and $n$. | (1) Since $\log_a 2 = m$ and $\log_a 3 = n$, we have $a^m = 2$ and $a^n = 3$.
Therefore, the value of $a^{2m-n}$ can be calculated by multiplying $a^{m}$ by itself and dividing by $a^{n}$:
$$a^{2m-n}=a^{2m} \div a^{n}=(a^{m})^2 \div a^{n} = 2^2 \div 3 = \frac{4}{3}.$$
So, we can conclude that $a^{2m-n} = \boxed{\frac{... | m + 2n | Algebra | 59eeedae-843b-5b70-8619-de87f06388b9 |
Given $x=\frac{\sqrt{11}}{2}+\frac{\sqrt{7}}{2}$ and $y=\frac{\sqrt{11}}{2}-\frac{\sqrt{7}}{2}$, find the values of $x+y$ and $xy$. Then, using the results from part $(1)$, find the values of the following algebraic expressions: $①x^{2}y+xy^{2}$ and $②\frac{y}{x}+\frac{x}{y}$. | To solve the given problem, we proceed in a step-by-step manner as follows:
### Part 1: Finding $x+y$ and $xy$
Given:
$$x=\frac{\sqrt{11}}{2}+\frac{\sqrt{7}}{2} \quad \text{and} \quad y=\frac{\sqrt{11}}{2}-\frac{\sqrt{7}}{2}$$
To find $x+y$:
$$x+y=\left(\frac{\sqrt{11}}{2}+\frac{\sqrt{7}}{2}\right)+\left(\frac{\sqrt... | 9 | Algebra | 32b84619-a9b5-529c-88b5-bca7b3ad55fa |
10. (15 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=p, a_{2}$ $=p+1, a_{n+2}-2 a_{n+1}+a_{n}=n-20$, where $p$ is a given real number, $n$ is a positive integer, try to find the value of $n$, such that the value of $a_{n}$ is minimized. | 10. Solution. Let $b_{n}=a_{n+1}-a_{n}, n=1,2, \cdots$
Given $a_{n+2}-2 a_{n+1}+a_{n}=n-20$, we have $b_{n+1}-b_{n}=n-20$, and $b_{1}=1$.
Thus $\sum_{i=1}^{n-1}\left(b_{i+1}-b_{i}\right)=\sum_{i=1}^{n-1}(i-20)$, which means $b_{n}-b_{1}=[1+2+\cdots+(n-1)]-2 n(n-1)$.
$\therefore b_{n}=\frac{(n-1)(n-40)}{2}+1$.
Also, $... | 40 | Algebra | cb83829f-963c-5487-880a-977e688b4835 |
4. Three Rolls (from 6th grade. 1 point). A die is rolled three times. Which event is more likely:
$A$ "some number of points will fall at least twice" or
$B$ "three different numbers of points will fall on three rolls." | Solution. The probability of event $B$ is
$$
\frac{6 \cdot 5 \cdot 4}{6^{3}}=\frac{5}{9}>\frac{1}{2}
$$
Therefore, event $B$ is more likely than event $A=\bar{B}$.
Answer: $B$. | B | Combinatorics | 21fdba64-c1f9-5920-b250-ddb020459f75 |
Given that the sum of the first $n$ terms, $S_n$, and the general term, $a_n$, of a sequence ${a_n}$ satisfy the equation $$S_{n}= \frac {1}{2}(1-a_{n})$$.
1. Find the general formula of the sequence ${a_n}$ and prove that $$S_{n}< \frac {1}{2}$$.
2. Let the function $$f(x)=log_{ \frac {1}{3}}x$$, and $b_n$ is the sum... | 1. For $n\geq2$, we have $S_{n-1}= \frac {1}{2}(1-a_{n-1})$, and $a_n=S_n-S_{n-1}$. Thus,
$$a_{n}= \frac {1}{2}(1-a_{n})- \frac {1}{2}(1-a_{n-1})=- \frac {1}{2}a_{n}+ \frac {1}{2}a_{n-1}$$
Simplifying, we get $2a_n=-a_n+a_{n-1}$, which leads to $$\frac {a_{n}}{a_{n-1}}= \frac {1}{3}$$
For $n=1$, we have
$$S_{1}=a_{... | \frac{2n}{n+1} | Algebra | 398c2040-8e77-5253-84ba-5dd1b25af162 |
90. There and Back. Colonel Crackham asserts that his friend, Mr. Wilkinson, walks from his country house to the nearest town at a speed of 5 km/h, and on the way back, a little tired, he covers the same distance at a speed of 3 km/h. The round trip takes him exactly 7 hours.
How far from the town is Mr. Wilkinson's h... | 90. The distance is \(13 \frac{1}{8}\) km; so that Mr. Wilkinson walks to the city in \(2 \frac{5}{8}\) hours, and returns in \(4 \frac{8}{3}\) hours, spending a total of 7 hours on the journey. | 13\frac{1}{8} | Algebra | 1e0367dd-a223-54bc-9eae-978d8ff05ce2 |
# Problem 7. (4 points)
In an $8 \times 8$ table, some cells are black, and the rest are white. In each white cell, the total number of black cells on the same row or column is written; nothing is written in the black cells. What is the maximum value that the sum of the numbers in the entire table can take? | Answer: 256
## Solution:
The number in the white cell consists of two addends: a "horizontal" and a "vertical" one. Consider the sum of all "horizontal" addends and the sum of all "vertical" addends separately across the entire table. If we maximize each of these two sums separately, the total sum will also be the gr... | 256 | Combinatorics | f20934b6-5cdf-5c11-8bc7-154a277b863b |
Given the function $f(x) = x^2(x + 1)$, find $f'(-1)$. | First, we simplify the given function:
$$f(x) = x^2(x + 1) = x^3 + x^2.$$
Next, we compute the derivative of the function with respect to $x$:
$$f'(x) = \frac{d}{dx}(x^3 + x^2) = 3x^2 + 2x.$$
Now, we find the value of the derivative at $x = -1$:
$$f'(-1) = 3(-1)^2 + 2(-1) = 3 - 2 = \boxed{1}.$$ | 1 | Calculus | b9a5f2c3-7881-5c66-9f8d-2306731bfca7 |
Given $\sin\alpha= \frac {2}{3}$, with $\alpha\in\left(\frac {\pi}{2},\pi\right)$, and $\cos\beta= -\frac {3}{5}$, with $\beta\in\left(\pi, \frac {3\pi}{2}\right)$, find the value of $\sin(\alpha+\beta)$. | Since $\sin\alpha= \frac {2}{3}$, with $\alpha\in\left(\frac {\pi}{2},\pi\right)$, and $\cos\beta= -\frac {3}{5}$, with $\beta\in\left(\pi, \frac {3\pi}{2}\right)$,
Therefore, $\cos\alpha= -\sqrt {1-\sin^{2}\alpha}= -\frac { \sqrt {5}}{3}$, and $\sin\beta= -\sqrt {1-\cos^{2}\beta}= -\frac {4}{5}$,
Thus, $\sin(\alpha+... | \frac {4 \sqrt {5}-6}{15} | Algebra | 16e64519-03c9-5a25-bb82-7cdff3325b49 |
Find all real numbers $p$ such that all three roots of the cubic equation $5x^3 - 5(p+1)x^2 + (71p - 1)x + 1 = 66p$ are positive integers. | Solution: $x=1$ is a root of the equation. Therefore, we only need to consider the quadratic equation
\[5x^2 - 5px + 66p - 1 = 0\]
whose two roots are positive integers.
Let these two positive integer roots be $u$ and $v$. Then, by Vieta's formulas, we have
\[5uv - 66(u + v) = -1.\] Multiplying by $5$ gives: $5^2uv -... | p = 76 | Algebra | 1915d408-1680-579b-a55a-38b6955dda5f |
Given a sequence of positive terms $\{a_{n}\}$ with the sum of the first $n$ terms denoted as $S_{n}$, for all positive integers $n$, the point ${P}_{n}({a}_{n},\sqrt{{S}_{n}})$ lies on the graph of the function $f(x)=\frac{x+1}{2}$.
$(1)$ Find the general formula for the sequence $\{a_{n}\}$.
$(2)$ Let $\{b_{n}\}$ b... | ### Solution:
#### Part (1): Finding the general formula for the sequence $\{a_{n}\}$
Given that the point ${P}_{n}({a}_{n},\sqrt{{S}_{n}})$ lies on the graph of the function $f(x)=\frac{x+1}{2}$, we have the relationship:
$$\sqrt{{S}_{n}}=\frac{{a}_{n}+1}{2}$$
- **Base Case ($n=1$):** When $n=1$, we substitute $n=1... | \lambda \in (-\infty,\frac{29\sqrt{15}}{15}) | Algebra | fe256cbe-0479-59d6-a2b0-6ed3ee6f3449 |
403*. Solve the equation:
$$
\sqrt[4]{1-x^{2}}+\sqrt[4]{1-x}+\sqrt[4]{1+x}=3
$$ | $\triangle$ The domain of the equation is the interval $[-1 ; 1]$. In this domain, we can apply the inequality between the geometric mean and the arithmetic mean of two non-negative numbers to each of the radicals in the left-hand side:
$\sqrt[4]{1-x^{2}}=\sqrt{\sqrt{1+x} \cdot \sqrt{1-x}} \leq \frac{\sqrt{1+x}+\sqrt{... | 0 | Algebra | 3f14b7de-111d-5f2e-9bf9-9ce5b1daa35d |
It is known that the point $(1,2)$ lies on the graph of the function $f(x) = a^x$ (where $a>0$ and $a\neq 1$). The sum of the first $n$ terms of the sequence $\{a_n\}$ is given by $S_n = f(n) - 1$.
(1) Find the general formula for the $n$-th term of the sequence $\{a_n\}$;
(2) Let $b_n = \log_a (a_{n+1})$, find the sum... | (1) By substituting the point $(1,2)$ into the function $f(x)=a^x$, we find $a=2$. Therefore, the sum of the first $n$ terms of the sequence $\{a_n\}$ is$S_n=f(n)−1 = 2^n-1$. When $n=1$, $a_1 = S_1 = 1$; for $n\geq 2$, we have $a_n = S_n - S_{n-1} = 2^n - 2^{n-1} = 2^{n-1}$. This formula is also valid for $n=1$, theref... | T_n = (n-1)\cdot2^n + 1 | Algebra | c4e36b36-e396-5fc5-a0f6-95113ba600ca |
33. Among all polynomials of the form $x^{2}+a x+b$, find the one for which the maximum modulus on the interval $[-1 ; 1]$ is minimal. | 65.33. Answer. $x^{2}-1 / 2$. | x^{2}-\frac{1}{2} | Algebra | f24ffb09-e645-5885-9afb-4d2e2282d8ea |
Given an arithmetic sequence $\{a_n\}$, the sum of the first $n$ terms is denoted as $S_n$. If $a_4=4$, then $S_7=$ ______. | Since the sum of the first $n$ terms of the arithmetic sequence $\{a_n\}$ is $S_n$, and $a_4=4$,
we have $S_7= \frac{7}{2}(a_1+a_7)=7a_4=28$.
Therefore, the answer is $\boxed{28}$.
From the given information, we can derive $S_7= \frac{7}{2}(a_1+a_7)=2a_4$, which allows us to find the result.
This problem tests ... | 28 | Algebra | d9e9632f-c432-5e01-85de-d6748c823c3d |
Given two arithmetic sequences $ \{ a_n \} $ and $ \{ b_n \} $ with the sum of the first $ n $ terms denoted as $ S_n $ and $ T_n $ respectively, and satisfying $ \frac {S_{n}}{T_{n}} = \frac {7n+2}{n+3} $, find the ratio $ \frac {a_{5}}{b_{5}} $. | Let's choose $ n = 9 $ and substitute it into the given ratio to find the values of $ S_{9} $ and $ T_{9} $:
$$ \frac {S_{9}}{T_{9}} = \frac {7 \times 9 + 2}{9 + 3} = \frac {65}{12} $$
Now, using the formula for the sum of the first $ n $ terms of an arithmetic sequence, we can express $ S_9 $ and $ T_9 $ as follows:
... | \frac {65}{12} | Algebra | a9549d0d-38e8-576c-adb2-4bcf8590668c |
In an opaque box, there are $3$ red balls and $2$ yellow balls. Except for the color, all the balls are the same. If $3$ balls are randomly drawn from the box, then the event "drawing at least one red ball" is ____ (Fill in "certain event", "random event", or "impossible event"). | To determine whether drawing at least one red ball from a box containing $3$ red balls and $2$ yellow balls (a total of $5$ balls) when drawing $3$ balls is a "certain event", "random event", or "impossible event", let's analyze the possibilities.
First, let's consider the opposite event of drawing at least one red ba... | \text{certain event} | Combinatorics | 189702b6-d737-5960-9596-1be10b11a6f0 |
6. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same... | Solution. Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time at the 53rd minute, $C$ and $D$ will al... | 53 | Algebra | 4524e653-abdb-5fef-b03c-02aca7407ee0 |
15. Find the sum: $1+2+3+\ldots+1999$. | 15. Let $S=1+2+3+\ldots+1999$. Then: $S=1999+1998+\ldots+3+2+1$ and $2 S=(1+1999)+(2+1998)+\ldots+(1999+1)=2000 \cdot 1999$. Therefore: $S=1000 \cdot 1999$. Answer: 1999000. | 1999000 | Algebra | 70e770b2-202d-5269-a18b-27155eb4f7b0 |
Four. (20 points) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with an inscribed $\triangle A B C$ where sides $A B$ and $A C$ pass through the left and right foci $F_{1}$ and $F_{2}$ respectively, and the left and right vertices of the ellipse are $D$ and $E$ respectively. The lines $D B$ and $C E$ intersect ... | As shown in Figure 5, let \( A(2 \cos \theta, \sqrt{3} \sin \theta), B(2 \cos \alpha, \sqrt{3} \sin \alpha), C(2 \cos \beta, \sqrt{3} \sin \beta) \), where \(\alpha, \beta, \theta\) are distinct. The line passing through points \(A\) and \(B\) is given by
\[
l_{A E}: y=\frac{\sqrt{3}(\sin \theta-\sin \alpha)}{2(\cos \t... | \frac{x^{2}}{4}+\frac{y^{2}}{27}=1 | Geometry | f96a2a1e-f83e-54a9-9d20-236f5aaa8583 |
Let $n$ be a positive integer. Let $(a, b, c)$ be a random ordered triple of nonnegative integers such that $a + b + c = n$, chosen uniformly at random from among all such triples. Let $M_n$ be the expected value (average value) of the largest of $a$, $b$, and $c$. As $n$ approaches infinity, what value does $\frac{... | 1. **Define the problem and variables:**
We are given a positive integer \( n \) and need to find the expected value \( M_n \) of the largest of three nonnegative integers \( a, b, \) and \( c \) such that \( a + b + c = n \). We are interested in the limit of \( \frac{M_n}{n} \) as \( n \) approaches infinity.
2. ... | \frac{11}{18} | Combinatorics | b102aaad-c719-51d7-9aed-745e85e6ba0b |
Observe the following equations
$$(1+x+x^{2})^{1} = 1+x+x^{2},$$
$$(1+x+x^{2})^{2} = 1+2x+3x^{2}+2x^{3}+x^{4},$$
$$(1+x+x^{2})^{3} = 1+3x+6x^{2}+7x^{3}+6x^{4}+3x^{5}+x^{6},$$
$$(1+x+x^{2})^{4} = 1+4x+10x^{2}+16x^{3}+19x^{4}+16x^{5}+10x^{6}+4x^{7}+x^{8},$$
...
If $(1+x+x^{2})^{6} = a_{0}+a_{1}x+a_{2}x^{2}+\l... | From the given equations, we can observe and analyze the pattern:
The third term on the right side of the expanded equations are respectively: 1, 3, 6, 10, ...
These are obtained by summing up consecutive numbers: 1, 1+2, 1+2+3, 1+2+3+4, ...
For the nth (where n is a positive integer) equation, $a_{2}$ can be ded... | 21 | Algebra | c95463e3-8c01-54dc-9d17-354e0f9cbc64 |
Given that $a$ and $b$ are distinct positive numbers, $A$ is the arithmetic mean of $a$ and $b$, and $G$ is the positive geometric mean of $a$ and $b$. The relationship between $A$ and $G$ is ______. | **Analysis**
This question mainly examines the application of sequences, where familiarity with the properties of geometric and arithmetic sequences is key to solving the problem. It is a basic question.
**Solution**
Given the problem, we have $A= \frac{a+b}{2}$ and $G= \pm \sqrt{ab}$. By the basic inequality, we ha... | A > G | Algebra | d29a8a77-b00a-5b62-9010-8f6bb9ba8f06 |
Given the sequence $\{a\_n\}$ with the first term $a\_1= \frac {3}{2}$, the sum of the first $n$ terms is $S\_n$, and it satisfies $2a_{n+1}+S_{n}=3$ ($n∈N^{*}$). Find the sum of all $n$ that satisfy $\frac {18}{17} < \frac {S_{2n}}{S_{n}} < \frac {8}{7}$. | Since $2a_{n+1}+S_{n}=3$,
we have $2a_{n+2}+S_{n+1}=3$,
subtracting the two equations, we get $2a_{n+2}+S_{n+1}-2a_{n+1}-S_{n}=0$,
which simplifies to $2a_{n+2}=a_{n+1}$,
when $n=1$, $2a_{2}+a_{1}=3$,
so $a_{2}= \frac {3}{4}$, and $2a_{2}=a_{1}$,
thus, $2a_{n+1}=a_{n}$, and $\frac {a_{n+1}}{a_{n}}= \frac {1}{2}$,... | 7 | Algebra | 4add2f2e-888f-5ba9-85ed-a1706a6d7cc9 |
The coefficient of the $x^{2}$ term in the expansion of the binomial $(2+x)(1-2x)^{5}$ is \_\_\_\_\_. (Answer with a number) | To solve, $(2+x)(1-2x)^{5}=(2+x)(1- C_{5}^{1}⋅2x+ C_{5}^{2}⋅4x^{2}+\ldots)$,
Therefore, in the expansion of the binomial $(2+x)(1-2x)^{5}$,
the terms containing $x^{2}$ are $-10x^{2}+2×40x^{2}=70x^{2}$,
Thus, its coefficient is $70$.
Hence, the answer is: $\boxed{70}$.
By expanding $(1-2x)^{5}$, we find the c... | 70 | Algebra | 271a1ba1-f58d-5923-a73a-f5fe78654b8c |
12.87 Determine all integer solutions of the indeterminate equation
$$
x^{3}+x^{2} y+x y^{2}+y^{3}=8\left(x^{2}+x y+y^{2}+1\right)
$$
(Luxembourg and four other countries' International Mathematical Competition, 1980) | 【Solution】Given the equation can be transformed into
$$
\left(x^{2}+y^{2}\right)(x+y-8)=8 x y+8 \text {. }
$$
If $x$ and $y$ are one odd and one even, then the left side of (1) is odd, and the right side is even, which is impossible.
Therefore, $x$ and $y$ have the same parity.
Thus, $x+y-8$ is even.
(1) If $x+y-8 \ge... | (x,y)=(8,2)\text{or}(2,8) | Algebra | 3b3c02c6-8d10-59ef-8f91-2a885a71a8ab |
Given that the radius of the base of a cone and the radius of a sphere are both $1cm$, if the volume of the cone is exactly equal to the volume of the sphere, then the lateral surface area of this cone is _______ $cm^{2}$. | From the given information, we know the volume of the sphere is: $\dfrac{4\pi}{3} \times 1^{3} = \dfrac{4\pi}{3}cm^{3}$,
The volume of the cone is: $\dfrac{1}{3} \times \pi \times 1^{2} \times h = \dfrac{\pi}{3}hcm^{3}$,
Since the volume of the cone is exactly equal to the volume of the sphere,
Thus $\dfrac{4\pi}{3}... | \sqrt{17}\pi | Geometry | 79669efc-df86-55ef-b447-a64c7d8c9671 |
In a WeChat group, members A, B, C, D, and E simultaneously grab 4 red envelopes, each person can grab at most one red envelope, and all red envelopes are grabbed. Among the 4 red envelopes, there are two worth 2 yuan and two worth 3 yuan (red envelopes with the same amount are considered the same). The number of situa... | **Analysis**
This problem examines the principles of classified counting and the comprehensive application of permutations and combinations. According to the properties of the red envelopes, we classify the situations. If A and B grab one 2-yuan and one 3-yuan red envelope, or if both grab 2-yuan or both grab 3-yuan r... | 18 | Combinatorics | 9d84136c-e89b-5c8a-9b5a-c1e3d4c0b184 |
Problem 3. The equations $x^{2}+2019 a x+b=0$ and $x^{2}+2019 b x+a=0$ have one common root. What can this root be, given that $a \neq b$? | Answer: $\frac{1}{2019}$.
Solution. Let the common root of the given equations be $r$. Then
$$
r^{2}+2019 a r+b=0=r^{2}+2019 b r+a
$$
From this, we obtain that $2019 r(a-b)=(a-b)$. Since $a \neq b$, it follows that $r=\frac{1}{2019}$
Criteria
4 p. A complete and justified solution is provided.
In the absence of a... | \frac{1}{2019} | Algebra | 82f5fa17-49bd-5d1d-9cee-3dae621b53c6 |
1. A single section at a stadium can hold either 7 adults or 11 children. When $N$ sections are completely filled, an equal number of adults and children will be seated in them. What is the least possible value of $N$ ? | Answer: The least common multiple of 7 and 11 is 77 . Therefore, there must be 77 adults and 77 children. The total number of sections is $\frac{77}{7}+\frac{77}{11}=11+7=18$. | 18 | Number Theory | 3c3dc846-7882-5b1c-8366-c266f1269880 |
Given a finite arithmetic sequence $\{{a}_{n}\}:{a}_{1},{a}_{2},⋯,{a}_{m}(m≥3,m∈{N}^{*})$ with a common difference $d$ greater than zero.$(1)$ Prove that $\{a_{n}\}$ is not a geometric sequence;$(2)$ Does there exist an exponential function $y=f\left(x\right)$ such that the tangent line of $y=f\left(x\right)$ at $x=a_{... | **Solution:**
**(1)** To prove that $\{a_{n}\}$ is not a geometric sequence, we examine the relationship between three consecutive terms in the sequence. Specifically, we look at ${a}_{2}^{2}-{a}_{1}{a}_{3}$:
\[
\begin{align*}
{a}_{2}^{2}-{a}_{1}{a}_{3} & = {a}_{2}^{2}-({a}_{2}-d)({a}_{2}+d) \\
& = {a}_{2}^{2} - ({a}... | m = 3 | Algebra | 469e0f83-9ee1-5c0c-8383-7ff88a172d5a |
The parabola $y=-(x+3)(2x+a)$ intersects the x-axis at points A and B, and intersects the y-axis at point C. If $\angle ACB = 90^\circ$, then the value of $a$ is. | When $y=0$, we have $-(x+3)(2x+a)=0$,
Solving this, we get $x_1=-3$, $x_2=-\frac{a}{2}$,
Therefore, the coordinates of the intersection points with the x-axis are $(-3,0)$ and $\left(-\frac{a}{2},0\right)$,
When $x=0$, $y=-3a$,
Therefore, the coordinate of the intersection point with the y-axis is $C(0,-3a)$,
Sinc... | -\frac{1}{6} | Algebra | 647ff15b-b09d-5be6-a1dc-62056e8a504c |
Calculate the following indefinite integrals.
[1] $\int \sin x\sin 2x dx$
[2] $\int \frac{e^{2x}}{e^x-1}dx$
[3] $\int \frac{\tan ^2 x}{\cos ^2 x}dx$
[4] $\int \frac{e^x+e^{-x}}{e^x-e^{-x}}dx$
[5] $\int \frac{e^x}{e^x+1}dx$ | ### Problem 1: $\int \sin x \sin 2x \, dx$
1. Use the product-to-sum identities to simplify the integrand:
\[
\sin x \sin 2x = \frac{1}{2} [\cos(x - 2x) - \cos(x + 2x)] = \frac{1}{2} [\cos(-x) - \cos(3x)] = \frac{1}{2} [\cos x - \cos 3x]
\]
2. Split the integral:
\[
\int \sin x \sin 2x \, dx = \frac{1}{... | \ln |e^x + 1| + C | Calculus | 627126e9-0a0e-53b1-a7fd-80fbc3fb1c4d |
Given the function $f(x)=\cos \frac {π}{3}x$, find the sum of $f(1)+f(2)+f(3)+…+f(2016)+f(2017)$. | The function $f(x)=\cos \frac {π}{3}x$ has a period of $T= \frac {2π}{ω} = \frac {2π}{\frac {π}{3}}=6$.
We can calculate the values of $f(x)$ for the first six integers:
- $f(1)=\cos \frac {π}{3} = \frac {1}{2}$
- $f(2)=\cos \frac {2π}{3} = -\frac {1}{2}$
- $f(3)=\cos π = -1$
- $f(4)=\cos \frac {4π}{3} = -\frac {1}{2}... | \frac {1}{2} | Algebra | 245bdbce-7760-54bd-8180-55aa5d6cc8d7 |
Given the ellipse $$C: \frac{x^2}{a^2} + \frac{y^2}{3} = 1 (a > \sqrt{3})$$ has an eccentricity of $$\frac{\sqrt{2}}{2}$$, and its right vertex is A.
(Ⅰ) Find the equation of ellipse C;
(Ⅱ) If a line $l$ passes through the left focus $F_1$ of C and intersects C at points B and D, find the maximum area of $\triangle... | Solution:
(Ⅰ) The eccentricity is $$\frac{\sqrt{2}}{2}$$, which equals $$\frac{c}{a} = \frac{\sqrt{2}}{2}$$,
Given $b = \sqrt{3}$, and $a^2 - b^2 = c^2$, we solve to find $a = \sqrt{6}$,
Thus, the equation of the ellipse is $$\frac{x^2}{6} + \frac{y^2}{3} = 1$$;
(Ⅱ) Let the equation of the line passing through ... | x = -\sqrt{3} | Geometry | 1c9d0a42-5beb-57e9-8885-02165a4e2f76 |
Given $f(x)=x\ln x$ and $g(x)=-x^2+ax-3$.
(1) Find the minimum value of the function $f(x)$ in the interval $[t, t+2]$ where $t>0$.
(2) For all $x \in (0, +\infty)$, if $2f(x) \ge g(x)$ always holds, find the range of real number $a$.
(3) Prove that for all $x \in (0, +\infty)$, the inequality $\ln x > \frac{1}{e... | (1) Since $f(x) = x\ln x$,
its derivative is $f'(x)=\ln x + 1$.
When $x \in (0, \frac{1}{e})$, we have $f'(x) 0$, thus $f(x)$ is strictly increasing.
Therefore, we can deduce the following:
When $0 0$.
Let $h(x) = x + 2\ln x + \frac{3}{x}$.
The first derivative of $h(x)$ is $h'(x) = 1 + \frac{2}{x} - \fra... | (-\infty, 4] | Calculus | 371f7393-34e4-5436-bcc3-b6a6185bffe2 |
Simplify first, then evaluate: $(1-\frac{1}{x})\div \frac{{x}^{2}-2x+1}{{x}^{2}}$, where $x=\sqrt{2}+1$. | To simplify and then evaluate the given expression $(1-\frac{1}{x})\div \frac{{x}^{2}-2x+1}{{x}^{2}}$ for $x=\sqrt{2}+1$, we proceed as follows:
First, we simplify the expression:
\begin{align*}
Original &= \frac{1-\frac{1}{x}}{\frac{{x}^{2}-2x+1}{{x}^{2}}} \\
&= \frac{x-1}{x}\cdot \frac{{x}^{2}}{(x-1)^{2}} \\
&= \fra... | 1+\frac{\sqrt{2}}{2} | Algebra | df29d3f2-3099-5e47-ad72-16a7ceaf6c5b |
A book has 500 pages, numbered 1, 2, 3, ..., how many times does the digit 1 appear in the page numbers? | The numbers 1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51, 61, 71, 81, 91 each contain the digit 1, making a total of 20 occurrences. From pages 100 to 199, there are $100 + 20 = 120$ occurrences of the digit 1.
For the remaining pages from 200 to 500, there are $3 \times 20 = 60$ occurrences of the digit... | 200 | Number Theory | 95c214d1-a628-5ce5-90c9-fbd2d8c7542d |
16 Given that $\alpha+\beta=17$ and $\alpha \beta=70$, find the value of $|\alpha-\beta|$. | 16. Answer: 3
Note that $\alpha$ and $\beta$ are the roots of the equation $x^{2}-17 x+70=0$. Solving we have $x=7$ or 10 . Hence $|\alpha-\beta|=3$. | 3 | Algebra | 5ca09d06-72ec-5fdb-95e1-65917632a348 |
Using the numbers $1$, $2$, $3$, $4$, $5$, $6$, $7$ to form a four-digit number without repeating digits, and at most one digit is even, the total number of such four-digit numbers is ______. (Provide your answer using numbers) | To solve this problem, we must consider two separate scenarios based on the conditions given. Let's break it down step by step.
**Case 1: The four-digit number has no even digits.**
In this case, we can only use the odd numbers $1, 3, 5, 7$. Since we are forming a four-digit number without repeating any digits and w... | 312 | Combinatorics | 279a4ce0-1a88-564e-ba3c-4d56222301cf |
The function $f(x)=ax^{3}-x-\ln x$ takes an extreme value at $x=1$. Determine the value of the real number $a$. | From the problem, we know that the domain of the function $f(x)$ is $(0,+\infty)$.
Differentiate the function to obtain $f′(x)=3ax^{2}-1-\frac{1}{x}$.
Given that the function $f(x)=ax^{3}-x-\ln x$ takes an extreme value at $x=1$, we have:
$$f′(1)=3a-1-1=0$$
Solving for $a$, we get:
$$a=\frac{2}{3}$$
To make the solu... | a=\frac{2}{3} | Calculus | d046eb54-595d-579f-8e7b-c26b1b409442 |
Throw a fair six-sided die (with faces labeled with the numbers 1 to 6),
Find:
(1) The probability that the numbers facing up are different when the die is thrown twice in a row;
(2) The probability that the sum of the numbers facing up is 6 when the die is thrown twice in a row;
(3) The probability that exactly three... | (1) Let $A$ represent the event "Throwing the die twice with different numbers facing up". There are a total of $6 \times 6 = 36$ possible outcomes when throwing the die twice. Since for the first throw there are 6 possible outcomes, and for the second throw there are 5 remaining outcomes that are different from the fi... | \frac{5}{16} | Combinatorics | b1dfb4bc-e6ad-522d-95f5-3ef6e45697df |
## Problem Statement
Calculate the indefinite integral:
$$
\int(1-6 x) e^{2 x} d x
$$ | ## Solution
$$
\int(1-6 x) e^{2 x} d x=
$$
Let's denote:
$$
\begin{aligned}
& u=1-6 x ; d u=-6 d x \\
& d v=e^{2 x} d x ; v=\frac{1}{2} e^{2 x}
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\begin{aligned}
& =(1-6 x) \cdot \frac{1}{2} e^{2 x}-\int \frac{1}{2} e^{2 ... | (2-3x)e^{2x}+C | Calculus | 74bb4902-0a81-5d9a-a45f-5124f8fe5689 |
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