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The interval $(-3, 2]$ is represented as a set by ( )
A: $\{-2, -1, 0, 1, 2\}$
B: $\{x|-3<x<2\}$
C: $\{x|-3<x\leq2\}$
D: $\{x|-3\leq x\leq2\}$ | Based on the relationship between intervals and sets,
the interval $(-3, 2]$ can be represented as: $\{x|-3<x\leq2\}$
Therefore, the correct choice is $\boxed{\text{C}}$ | \text{C} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,973 |
The increasing interval of the function $y={2^{\sqrt{-{x^2}+x+2}}}$ is ____. | To find the increasing interval of the function $y={2^{\sqrt{-{x^2}+x+2}}}$, we first need to analyze the behavior of the function inside the exponent, which is $\sqrt{-{x^2}+x+2}$. Let's denote this inner function as $u$, where $u=\sqrt{-x^{2}+x+2}$.
1. **Rewriting the expression inside the square root in a completed... | \left[-1,\frac{1}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,974 |
Read the material and answer the following questions. By calculation, it is easy to find: ① $\frac{1}{2}-\frac{1}{3}=\frac{1}{2}×\frac{1}{3}$; ② $\frac{1}{4}-\frac{1}{5}=\frac{1}{4}×\frac{1}{5}$; ③ $\frac{1}{6}-\frac{1}{7}=\frac{1}{6}×\frac{1}{7}$; $\ldots \ldots $$(1)$ By observation, calculate the value of $\frac{1}{... | ### Solution:
#### For Part (1):
We start by breaking down the given fractions and rearranging them as follows:
\[
\begin{align*}
\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\cdots+\frac{1}{6×7} &= \left(1-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \left(\frac{1}... | \frac{1002}{2005} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,975 |
The negation of the proposition "For all $x \in \mathbb{R}, x^2 - 2x + 4 \leq 0$" is ( )
A: For all $x \in \mathbb{R}, x^2 - 2x + 4 \geq 0$
B: There exists $x \in \mathbb{R}, x^2 - 2x + 4 > 0$
C: For all $x \notin \mathbb{R}, x^2 - 2x + 4 \leq 0$
D: There exists $x \notin \mathbb{R}, x^2 - 2x + 4 > 0$ | B
Key Points: Universal proposition; negation of a proposition.
Topic: Calculation problem.
Analysis: The proposition in this problem is a universal proposition, and its negation is an existential proposition. The negation of the universal proposition can be written according to the form of negating a universal p... | B | Inequalities | MCQ | Yes | Yes | cn_k12 | false | 548,976 |
The maximum value of the function $y = \sqrt{3}\sin x + \cos x$ is ______. | Given the function $y = \sqrt{3}\sin x + \cos x$, we can rewrite it using the sine of a sum of angles identity.
We have:
\begin{align*}
y &= \sqrt{3}\sin x + \cos x \\
&= 2\left( \frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x \right) \\
&= 2\sin\left(x + \frac{\pi}{6}\right),
\end{align*}
since $\sin\left(\frac{\pi}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,977 |
Given $$a= \frac {1}{\pi } \int_{ -1 }^{ 1 }( \sqrt {1-x^{2}}+\sin x)dx$$, find the constant term in the expansion of the binomial $$(2x- \frac {a}{x^{2}})^{9}$$. | First, we calculate $$a= \frac {1}{\pi } \int_{ -1 }^{ 1 }( \sqrt {1-x^{2}}+\sin x)dx$$
$$= \frac {1}{\pi } \left( \int_{ -1 }^{ 1 } \sqrt {1-x^{2}}dx+ \int_{ -1 }^{ 1 }\sin xdx \right)$$
$$= \frac {1}{\pi } \times \frac {1}{2}\pi \times 1^{2} + \frac {1}{\pi } \times (-\cos x) |_{ -1 }^{ 1 }$$
$$= \frac {1}{2}$$,
The... | -672 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,978 |
Suppose $f(x)$ is an even function on $\mathbb{R}$ and a decreasing function on $[0, +\infty)$. Which of the following conclusions is correct?
A: $f(3)>f(1)>f(-2)$
B: $f(-2)>f(3)>f(1)$
C: $f(1)>f(-2)>f(3)$
D: $f(-2)>f(1)>f(3)$ | Since $f(x)$ is an even function on $\mathbb{R}$, it holds that $f(-2)=f(2)$.
Moreover, as $f(x)$ is a decreasing function on the interval $[0, +\infty)$ and since \(0f(2)>f(3)$. Rewriting the second inequality using the first property, we get $f(1)>f(-2)>f(3)$.
Therefore, the correct answer is \boxed{C}. | C | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,979 |
The equation of the tangent line to the curve $y = -x^3 + 3x^2$ at the point $(1,2)$ is ( )
(A) $y = 3x - 1$ (B) $y = -3x + 5$
(C) $y = 3x + 5$ (D) $y = 2x$ | Since $y' = -3x^2 + 6x$, then $y'|_{x=1} = 3$,
Therefore, the slope of the tangent line to the curve at the point $(1,2)$ is $3$,
Hence, the equation of the tangent line is $y - 2 = 3(x - 1)$, which simplifies to $y = 3x - 1$. Therefore, the correct choice is $\boxed{\text{A}}$. | \text{A} | Calculus | MCQ | Yes | Yes | cn_k12 | false | 548,980 |
Determine the correct conclusion about the complex number $z = \frac{2}{-1 + i}$.
$(\quad)$
$\quad\text{1. } |z| = 2 \text{; } \quad\text{2. } z = 2i \text{; } \quad\text{3. }$ The conjugate of $z$ is $1 + i \text{; } \quad\text{4. }$ The imaginary part of $z$ is $-1$.
A: 1, 2
B: 2, 3
C: 2, 4
D: 3, 4 | First, let's simplify the given complex number $z$. We can rewrite $z$ as:
$$z = \frac{2}{-1 + i} \cdot \frac{-1 - i}{-1 - i} = \frac{-2 - 2i}{(-1)^2 + 1^2} = \frac{-2 - 2i}{2} = -1 - i$$
Now, let's evaluate each statement:
1. The modulus of $z$ is $|z| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \neq 2$. So, statement 1 is f... | \text{D: 3, 4} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,981 |
Which of the following is the condition for the equation $|x|+x^{2}=a$ to have real number solutions?
A: Sufficient but not necessary condition
B: Necessary but not sufficient condition
C: Necessary and sufficient condition
D: Neither sufficient nor necessary condition | To analyze the condition for the equation $|x|+x^{2}=a$ to have real number solutions, we proceed as follows:
1. Consider the specific case when $a=-\frac{1}{4}$. The equation becomes:
\[|x|+x^{2}=-\frac{1}{4}\]
\[x^{2}+|x|+\frac{1}{4}=0\]
\[({|x|+\frac{1}{2}})^{2}=0\]
This implies that there are no real solutions fo... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,982 |
A monotonic increasing interval for $y=\sin (2x- \frac {\pi}{3})-\sin 2x$ is ( )
A: $[- \frac {\pi}{6}, \frac {\pi}{3}]$
B: $[ \frac {\pi}{12}, \frac {7}{12}\pi]$
C: $[ \frac {5}{12}\pi, \frac {13}{12}\pi]$
D: $[ \frac {\pi}{3}, \frac {5\pi}{6}]$ | Simplifying, we get $y=\sin (2x- \frac {\pi}{3})-\sin 2x$
$= \frac {1}{2}\sin 2x- \frac { \sqrt {3}}{2}\cos 2x-\sin 2x$
$=-( \frac { \sqrt {3}}{2}\cos 2x+ \frac {1}{2}\sin 2x)$
$=-\sin (2x+ \frac {\pi}{3})$,
From $2k\pi+ \frac {\pi}{2}\leqslant 2x+ \frac {\pi}{3}\leqslant 2k\pi+ \frac {3\pi}{2}$, we get $k\pi+ ... | B | Calculus | MCQ | Yes | Yes | cn_k12 | false | 548,983 |
If $x \lt y$, which of the following inequalities is true?
A: $x+1 \gt y+1$
B: $2x \gt 2y$
C: $-x \gt -y$
D: $\frac{x}{3} \gt \frac{y}{3}$ | Given $x y+1$**
Starting from the given inequality $x y + 1$. Therefore, option A is incorrect.
**Option B: $2x > 2y$**
Multiplying both sides of the given inequality $x 2y$. Therefore, option B is incorrect.
**Option C: $-x > -y$**
Multiplying both sides of the given inequality $x -y\]
This matches option C e... | C | Inequalities | MCQ | Yes | Yes | cn_k12 | false | 548,984 |
Given the function $f(x)=x-(a+1)\ln x- \frac {a}{x}$ ($a\in\mathbb{R}$, and $a < 1$), and $g(x)= \frac {1}{2}x^{2}+e^{x}-xe^{x}$, if there exists $x_{1}\in[e,e^{2}]$ such that for any $x_{2}\in[-2,0]$, $f(x_{1}) < g(x_{2})$ always holds, then the range of values for $a$ is \_\_\_\_\_\_. | The domain of $f(x)$ is $(0,+\infty)$. The derivative of $f(x)$ is $f′(x)= \frac {(x-1)(x-a)}{x^{2}}$ ($a\in\mathbb{R}$).
When $a \frac {e^{2}-2e}{e+1}$.
Thus, the range of values for $a$ is $\boxed{\left( \frac {e^{2}-2e}{e+1},1\right)}$.
There exists $x_{1}\in[e,e^{2}]$ such that for any $x_{2}\in[-2,0]$, $f(x_{1... | \left( \frac {e^{2}-2e}{e+1},1\right) | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 548,985 |
Simplify first, then evaluate: $({\frac{2}{{x-3}}-\frac{1}{x}})⋅\frac{{{x^2}-3x}}{{{x^2}+6x+9}}$, where $x$ is a root of the equation $x^{2}-2x-3=0$. | To simplify and then evaluate the given expression $({\frac{2}{{x-3}}-\frac{1}{x}})⋅\frac{{{x^2}-3x}}{{{x^2}+6x+9}}$, where $x$ is a root of the equation $x^{2}-2x-3=0$, we follow these steps:
1. **Simplify the given expression:**
The given expression can be rewritten as:
\[
\left(\frac{2}{{x-3}}-\frac{1}{x}... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,986 |
Given $f(x)=\begin{cases} & 2x,x > 0 \\ & f(x+1),x\leqslant 0 \\ \end{cases}$, find the value of $f\left( \frac{4}{3} \right)+f\left( -\frac{4}{3} \right)$.
A: $2$
B: $4$
C: $2$
D: $-4$ | **Analysis**:
This problem involves a piecewise function. Depending on the value of the independent variable, we should choose the correct expression to solve the problem.
**Step 1**:
Evaluate $f\left( \frac{4}{3} \right)$. Since $\frac{4}{3} > 0$, according to the definition of the function, we have $f\left( \frac{4}... | 4 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,987 |
Given that point $P$ is any point on the curve $(x-1)^2+(y-2)^2=9$ with $y \geq 2$, find the minimum value of $x+ \sqrt {3}y$. | The equation $(x-1)^2+(y-2)^2=9$ with $y \geq 2$ represents the upper half of a circle centered at $(1,2)$ with a radius of 3. To find the minimum value of the given expression, we want to find a point on this curve where this value is minimized. This can be addressed as a geometric problem where we are looking for the... | 2\sqrt{3} - 2 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 548,988 |
The condition for "$a=2$" to make the line $ax+2y=0$ parallel to the line $x+y=1$ is __________. | If "$a=2$" holds, then the two lines $x+y=0$ and $x+y=1$ are parallel;
Conversely, when the condition "the line $ax+2y=0$ is parallel to the line $x+y=1$" is met, it can be concluded that $a=2$;
Therefore, "$a=2$" is the necessary and sufficient condition for the line $ax+2y=0$ to be parallel to the line $x+y=1$, ... | \text{necessary and sufficient} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,989 |
Given that the equation of the internal angle bisector $CD$ of triangle $\triangle ABC$ is $x+y=0$, one vertex is $A(2,1)$, and the equation of the median $BE$ on side $AC$ is $5x-2y+10=0$.
$(1)$ Find the coordinates of vertex $C$.
$(2)$ Find the area of $\triangle ABC$. | ### Step-by-Step Solution
#### Part (1): Finding the coordinates of vertex $C$
Given the equation of line $CD$ is $x+y=0$, we assume the coordinates of $C$ to be $(t,-t)$.
The midpoint of $AC$ can be calculated as $\left(\frac{t+2}{2}, \frac{-t+1}{2}\right)$. Since this midpoint lies on the median $BE$ whose equatio... | 9 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 548,990 |
The school plans to purchase 8 projectors and 32 computers, with each projector costing 7500 yuan and each computer costing 3600 yuan. How much money is needed in total? | Calculate as follows: $8 \times 7500 + 32 \times 3600$
$= 60000 + 115200$
$= 175200$ (yuan)
Therefore, a total of $\boxed{175200}$ yuan is needed. | 175200 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,991 |
Given that the function $f(x) = -x^3 + ax^2 - x - 1$ is decreasing on $\mathbb{R}$, the range of the real number $a$ is __.
A: $(-\infty, -\sqrt{3}] \cup [\sqrt{3}, +\infty)$
B: $[-\sqrt{3}, \sqrt{3}]$
C: $(-\infty, -\sqrt{3}) \cup (\sqrt{3}, +\infty)$
D: $(-\sqrt{3}, \sqrt{3})$ | **Analysis**
This question examines the use of derivatives to study the monotonicity of a function.
By deriving the expression for $f(x)$, we find that the derivative is a downward-opening parabola. Since the function is monotonic on $\mathbb{R}$,
the derivative has no intersection points with the x-axis or only one... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,992 |
When $x=5$, the value of the algebraic expression $ax^{5}+bx^{3}+cx-8$ is $6$. Then when $x=-5$, the value of the algebraic expression $ax^{5}+bx^{3}+cx-8$ is ( )
A: $-22$
B: $-2$
C: $2$
D: $22$ | Given the algebraic expression $ax^{5}+bx^{3}+cx-8$, we are told that when $x=5$, the value of this expression is $6$. This gives us our first equation:
1. Substitute $x=5$ into the expression:
\[
ax^{5}+bx^{3}+cx-8 = 6
\]
\[
3125a + 125b + 5c - 8 = 6
\]
\[
3125a + 125b + 5c = 14
\]
2. Now, we need to find the value ... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,993 |
Given that the sequence $1$, $a\_1$, $a\_2$, $a\_3$, $9$ is an arithmetic sequence, and the sequence $-9$, $b\_1$, $b\_2$, $b\_3$, $-1$ is a geometric sequence, find the value of $\frac{b\_2}{a\_1+a\_3}$. | Since the sequence $1$, $a\_1$, $a\_2$, $a\_3$, $9$ is an arithmetic sequence, we know that the difference between consecutive terms is constant. As the first and last terms are $1$ and $9$ respectively, their sum is $1+9=10$. As $a\_1$ and $a\_3$ are also part of this arithmetic sequence, their sum would also be equal... | \frac{b\_2}{a\_1+a\_3}=-\frac{3}{10} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,994 |
Given that the function $y = f(x)$ is an odd function, and $g(x) = f(x) + 2$, it is also known that $g(1) = 1$. Find the value of $g(-1)$. | Since $y = f(x)$ is an odd function, by definition, we have that $f(-x) = -f(x)$ for all $x$ in the domain of $f$.
Thus, for $x = 1$, we have:
$$ f(-1) = -f(1). $$
Now, given that $g(x) = f(x) + 2$, we can express $g(-1)$ and $g(1)$ as follows:
$$ g(-1) = f(-1) + 2, $$
$$ g(1) = f(1) + 2. $$
Given that $g(1) = 1$, w... | 3 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,995 |
Given the function $$f(x)= \begin{cases} \cos{x}, x\in[- \frac {\pi}{2},0] \\ \sqrt {1-x^{2}}, x\in(0,1]\end{cases}$$, find the value of the integral $$∫_{ - \frac {\pi}{2} }^{ 1 }f(x)dx$$. | First, let's focus on the part of the function $f(x) = \sqrt{1 - x^2}$ for $x \in (0, 1]$. Let $y = \sqrt{1 - x^2}$, then $y \geq 0$. Squaring both sides of the equation, we get $y^2 = 1 - x^2$, which simplifies to $x^2 + y^2 = 1$. This is the equation of a circle centered at the origin with a radius of 1. Therefore, t... | 1 + \frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 548,996 |
If $\{a_n\}$ is an arithmetic sequence, then among the following sequences, those that are still arithmetic sequences are ( )
(1) $\{a_n+3\}$; (2) $\{a_n^2\}$; (3) $\{a_{n+1}-a_n\}$; (4) $\{2a_n\}$; (5) $\{2a_n+n\}$.
A: 1
B: 2
C: 3
D: 4 | $\boxed{\text{D}}$
Key point: Determination of arithmetic relationships.
Topic: Arithmetic and geometric sequences.
Analysis: By using the definition of an arithmetic sequence, for each option's sequence, it is sufficient to prove that the difference between the $(n+1)$-th term and the $n$-th term is a constant. ... | \text{D} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 548,997 |
Define a new operation: $a \triangle b = a \times b + (a - b) + 6$. For example, $6 \triangle 2 = 6 \times 2 + (6 - 2) + 6 = 22$. Find the value of $50 \triangle 40$. | According to the definition, $a \triangle b$ equals the product of two numbers plus the difference between these two numbers, and then add 6. We just need to substitute the numbers into the formula for calculation.
This problem tests the mixed operation of rational numbers: perform multiplication first, then division,... | 2016 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 548,998 |
The solution set of the inequality $4x^2-x-5\leq0$ is __________. | To solve the inequality $4x^2-x-5\leq0$, it can be transformed into:
$(4x-5)(x+1)\leq0$,
The real roots of the corresponding equation are $\frac{5}{4}$ and $-1$,
Therefore, the solution set of the inequality is $\left[-1, \frac{5}{4}\right]$.
Hence, the answer is $\boxed{\left[-1, \frac{5}{4}\right]}$.
By ide... | \left[-1, \frac{5}{4}\right] | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false | 548,999 |
Given a geometric sequence $\{a_n\}$, and $a_6+a_8=4$, then the value of $a_8(a_4+2a_6+a_8)$ is __( ).
A: $2$
B: $4$
C: $8$
D: $16$ | From the given, we have: $a_8(a_4+2a_6+a_8)=a_8a_4+2a_8a_6+a_8^2$,
Since $a_6+a_8=4$,
Thus, $a_8a_4+2a_8a_6+a_8^2=(a_6+a_8)^2=16$.
Therefore, the correct choice is $\boxed{D}$.
Expanding the expression "$a_8(a_4+2a_6+a_8)$" and using the property of geometric sequences: if $m$, $n$, $p$, $q \in \mathbb{N}^*$ and $m... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,000 |
For any $x \in \mathbb{R}$, the necessary and sufficient condition for the function $f(x) = ax^3 + ax^2 + 7x$ to have no extreme points is ( )
A: $0 \leq a \leq 21$
B: $0 21$
D: $a = 0$ or $a = 21$ | Since $f(x) = ax^3 + ax^2 + 7x$, we have $f'(x) = 3ax^2 + 2ax + 7$.
If $a = 0$, then $f'(x) = 7 > 0$ always holds, and $f(x)$ is an increasing function on $\mathbb{R}$, which meets the condition.
If $a \neq 0$, then when $\Delta = 4a^2 - 84a \leq 0$, which is equivalent to $0 < a \leq 21$, $f'(x) \geq 0$ always holds... | A | Calculus | MCQ | Yes | Yes | cn_k12 | false | 549,001 |
In the vast and lonely universe, the closest celestial body to Earth is the moon. How far is the moon from us? The answer is approximately $384000$ km, which can be expressed in scientific notation as ( ).
A: $38.4\times 10^{4}$
B: $3.84\times 10^{5}$
C: $0.384\times 10^{6}$
D: $3.84\times 10^{6}$ | To express $384000$ km in scientific notation, we start by identifying the significant figures and the appropriate power of $10$ that will allow us to represent this number succinctly.
The number $384000$ can be broken down as follows:
- First, we identify the non-zero digits, which are $3, 8, 4$.
- Then, we place a... | B | Other | MCQ | Yes | Yes | cn_k12 | false | 549,002 |
If \\(i\\) is the imaginary unit, then the complex number \\( \dfrac {1-i}{1+i} \\) equals \\((\\) \\()\\)
A: \\(-1\\)
B: \\(-i\\)
C: \\(1\\)
D: \\(i\\) | Solution: \\( \dfrac {1-i}{1+i} = \dfrac {(1-i)(1-i)}{(1+i)(1-i)} = \dfrac {-2i}{2} = -i \\).
Therefore, the correct answer is B.
According to the rules of division for complex numbers, multiplying both the numerator and the denominator by the conjugate of the denominator, \\(1-i\\), simplifies the calculation.
T... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,003 |
Given that the two foci of ellipse $C$ are $F\_1(-\sqrt{3},0)$ and $F\_2(\sqrt{3},0)$, and the two endpoints of the minor axis are $B\_1$ and $B\_2$,
(1) If $\triangle F\_1B\_1B\_2$ is an equilateral triangle, find the equation of ellipse $C$;
(2) Under the condition of (1), a line $l$ passing through point $F\_2$ in... | (1) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $(a > b > 0)$.
Given that the foci of the ellipse are $F\_1(-\sqrt{3},0)$ and $F\_2(\sqrt{3},0)$, we have $c = \sqrt{3}$.
By the definition of an ellipse, we know that $B\_1F\_1 = a$ and $B\_1B\_2 = 2b$. Since $\triangle F\_1B\_1B\_... | \frac{8}{5} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,004 |
Given the ellipse ${x}^{2}+\frac{{y}^{2}}{2}=1$, a line $l$ passing through the point $P(\frac{1}{2},1)$ intersects the ellipse at points $A$ and $B$, where $P$ is the midpoint of the line segment $AB$. What is the slope $k$ of the line $AB$?
A: $-1$
B: $-\frac{1}{4}$
C: $1$
D: $4$ | To find the slope $k$ of the line $AB$ that passes through the point $P(\frac{1}{2},1)$ and intersects the ellipse ${x}^{2}+\frac{{y}^{2}}{2}=1$ at points $A$ and $B$, we proceed as follows:
1. Let's denote the coordinates of points $A$ and $B$ as $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$, respectively. Given that point... | \text{A} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,005 |
Given two sequences $\{a_n\}$ and $\{b_n\}$ satisfying: $a_n+b_n=1$, $b_{n+1}= \frac {b_n}{(1-a_n)(1+a_n)}$, and $a_1$, $b_1$ are the roots of the function $f(x)=16x^2-16x+3$ $(a_1 < b_1)$.
(1) Find $a_1$, $b_1$, $b_2$;
(2) Let $c_n= \frac {1}{b_n-1}$, prove that the sequence $\{c_n\}$ is an arithmetic sequence, an... | (1) Solving $16x^2-16x+3=0$ yields: $x_1= \frac {1}{4}, x_2= \frac {3}{4}$,
$\therefore a_1= \frac {1}{4}, b_1= \frac {3}{4}$.
From $a_n+b_n=1, b_{n+1}= \frac {b_n}{(1-a_n)(1+a_n)}$, we get $b_{n+1}= \frac {b_n}{b_n(2-b_n)}= \frac {1}{2-b_n}$,
Substituting $b_1= \frac {3}{4}$ into the equation yields $b_2= \frac ... | (-\infty,1] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,006 |
Intelligent trash cans are divided into several compartments such as "hazardous waste, recyclable waste," etc. Residents can automatically open the compartments for disposal by swiping cards, using mobile numbers, facial recognition, and other identification methods, and put different types of garbage into different co... | To find the median and mode of the points converted by 7 families in a week, we first arrange the data in ascending order and identify the most frequently occurring value.
**Step 1: Arrange the data in ascending order:**
\[23, 23, 25, 25, 25, 27, 30\]
**Step 2: Identify the mode:**
The value $25$ appears $3$ times, w... | \text{D: }25, 25 | Other | MCQ | Yes | Yes | cn_k12 | false | 549,007 |
Given the function $f(x) = \log_a |x|$ is monotonically increasing on the interval $(0, +\infty)$, then ( )
A: $f(3) < f(-2) < f(1)$
B: $f(1) < f(-2) < f(3)$
C: $f(-2) < f(1) < f(3)$
D: $f(3) < f(1) < f(-2)$ | According to the problem, it is easy to deduce that $f(x) = f(-x)$, which means $f(x)$ is an even function.
Therefore, we have $f(-2) = f(2)$.
Given that the function $f(x)$ is monotonically increasing on the interval $(0, +\infty)$,
we have $f(1) < f(2) < f(3)$.
Since $f(-2) = f(2)$,
we thus have $f(1) < f(-... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,008 |
Given that the sum of the interior angles of a regular polygon is greater than the sum of the exterior angles by $360^{\circ}$, find the number of sides of this regular polygon and the degree measure of each exterior angle. | To solve this problem, let's start by denoting the number of sides of the regular polygon as $n$. The sum of the interior angles of any polygon is given by the formula $180^{\circ} \times (n-2)$. According to the problem statement, this sum is greater than the sum of the exterior angles by $360^{\circ}$. However, the s... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,009 |
Given a sequence $\{a\_n\}$ with its first $n$ terms sum $S\_n$, where $a_1=3$ and $2S_n = a_{n+1} + 2n$.
1. Find $a_2$.
2. Find the general formula for the $n$-th term $a_n$ of the sequence $\{a_n\}$.
3. Let $b_n = (2n - 1)(a_n - 1)$, find the sum $T_n$ of the first $n$ terms of the sequence $\{b_n\}$. | 1. When $n=1$, we have $2S_1 = 2a_1 = a_2 + 2$, which implies $a_2 = 4$.
2. When $n \geq 2$, we have $2a_n = 2S_n - 2S_{n-1} = a_{n+1} + 2n - a_n - 2(n-1) = a_{n+1} - a_n + 2$. This implies $a_{n+1} = 3a_n - 2$. Consequently, we have $a_{n+1} - 1 = 3(a_n - 1)$. Thus, $\frac{a_{n+1} - 1}{a_n - 1} = 3$ for $n \geq 2$. So... | T_n = (n - 1) \cdot 3^n + 2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,010 |
Throw a fair dice twice. Let $A=\{$both numbers are odd$\}$ and $B=\{$the sum of the two numbers is $6\}$. Find $P\left(B|A\right)$.
A: $\frac{1}{12}$
B: $\frac{1}{3}$
C: $\frac{2}{9}$
D: $\frac{2}{3}$ | To solve for $P(B|A)$, which is the probability of event $B$ given that event $A$ has occurred, we follow these steps:
1. Calculate the total number of outcomes for event $A$, which is the event where both numbers are odd when a fair dice is thrown twice. Since a dice has $3$ odd numbers ($1, 3, 5$), and we are throwi... | \text{C. } \frac{2}{9} | Combinatorics | MCQ | Yes | Yes | cn_k12 | false | 549,011 |
Given $|\vec{a}|=|\vec{b}|=2$ and $(\vec{a}+2\vec{b})\cdot(\vec{a}-\vec{b})=-2$, find the angle between $\vec{a}$ and $\vec{b}$. | **Analysis**
This problem examines the dot product of two vectors in a plane and the angle between them. From the given condition, we obtain $\cos \theta= \frac{1}{2}$. Then, considering the range of the angle between vectors, $\theta \in [0,\pi]$, we can find the answer.
**Solution**
Let the angle between $\vec{a}$... | \frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,012 |
Let $x_{0}$ be the extreme point of the function $f(x)=\frac{1}{2}x^2+mx+\ln x$ $(x>0)$. If the real number $x_{0}$ satisfies the inequality $\frac{1}{2} \leq x_{0} \leq 3$ and there is only one such real number, then the range of the real number $m$ is ( ).
A: $(-\frac{10}{3}, -\frac{5}{2})$
B: $[-\frac{10}{3}, -\fr... | To find the range of the real number $m$ given the conditions on $x_0$, we first need to understand the behavior of the derivative of the function $f(x) = \frac{1}{2}x^2 + mx + \ln x$ for $x > 0$. The derivative of $f(x)$ is given by:
\[f'(x) = x + \frac{1}{x} + m.\]
Given that there is only one real number $x_0$ sat... | \text{B: } [-\frac{10}{3}, -\frac{5}{2}) | Calculus | MCQ | Yes | Yes | cn_k12 | false | 549,013 |
Given circle C: x^2+y^2=5-m and circle E: (x-3)^2+(y-4)^2=16, find the value of m if they have three common tangent lines. | If two circles have three common tangent lines, it is equivalent to saying that they are externally tangent to each other.
Circle E is centered at (3,4) with a radius R=4, and circle C is centered at (0,0) with a radius r=$\sqrt{5-m}$.
The distance between the centers of the two circles is given by $|EC|=4+\sqrt{5-m}=\... | m=4 | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,014 |
Given lines $a$ and $b$ are skew lines, the number of correct statements among the following four is ( )
① There is at least one plane through $a$ that is parallel to $b$; ② There is at least one plane through $a$ that is perpendicular to $b$;
③ There is at most one line that is perpendicular to both $a$ and $b$; ④... | Since $a$ and $b$ are skew lines,
thus, by taking any point $P$ on line $a$, a plane $\alpha$ determined by $P$ and line $b$ can be found. In plane $\alpha$, a line $c$ through $P$ and parallel to $b$ can be drawn,
and lines $a$ and $c$ determine plane $\beta$, with $b \parallel \beta$, hence statement ① is correct... | \text{C} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,015 |
Given: Using 2 type A cars and 1 type B car can transport 10 tons of goods at a time; using 1 type A car and 2 type B cars can transport 11 tons of goods at a time. Based on the above information, answer the following questions:
1. How many tons of goods can be transported by 1 type A car and 1 type B car respectively ... | ### Step-by-Step Solution:
#### Part 1: Finding the capacity of 1 type A car and 1 type B car
Let's denote:
- The capacity of 1 type A car as $x$ tons.
- The capacity of 1 type B car as $y$ tons.
From the given information, we have two equations:
1. $2x + y = 10$ (Using 2 type A cars and 1 type B car can transport 1... | \text{The logistics company needs to transport 50 tons of goods.} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,016 |
Given that the graph of the function $f(x)=ax+\frac{b}{x}$ passes through points $A(1,1)$ and $B(2,-1)$.
1. Find the analytical expression of the function $f(x)$;
2. Determine the monotonicity of the function $f(x)$ on $(0,+\infty)$ and prove it using the definition;
3. Find the range of $f(x)$ on the interval $[\frac{... | 1. Since the graph of $f(x)$ passes through points $A(1,1)$ and $B(2,-1)$, we have the following system of equations:
$$\begin{cases} a + b = 1 \\ 2a + \frac{b}{2} = -1 \end{cases}$$
Solving this system of equations yields $a=-1$ and $b=2$. Therefore, the function is given by $f(x)=-x+\frac{2}{x}$.
2. Proof: Let $x_1$... | [1, \frac{31}{4}] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,017 |
In the Cartesian coordinate system $(xOy)$, it is known that the point $P(a, b)$ in the first quadrant lies on the line $x + 2y - 1 = 0$. Find the minimum value of $\frac{4}{a + b} + \frac{1}{b}$. | This problem tests our ability to apply basic inequalities and common transform techniques. The key lies in replacing "1" and setting up a relationship between $a$ and $b$ using the equation of the line.
First, substitute the coordinates of point $P$ into the line equation, we get $a + 2b - 1 = 0$, or $(a + b) + b = 1... | 9 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,018 |
Place 5 different math books on the same shelf. How many different arrangements are there?
A: 50
B: 60
C: 120
D: 90 | **Answer:** For 5 different math books, we consider all permutations. Therefore, there are $A_5^5 = 120$ ways.
Thus, the correct choice is $\boxed{C}$.
**Analysis:** This problem is a permutation problem, and we can solve it by considering all possible permutations. | C | Combinatorics | MCQ | Yes | Yes | cn_k12 | false | 549,019 |
For any point on the graph of the function $y = \cos x$, the inclination angle of the tangent line at that point is $\alpha$. Determine the range of possible values for $\alpha$.
A: $(0,\pi)$
B: $\left[0, \frac{\pi}{4}\right]$
C: $\left[0, \frac{\pi}{4}\right] \cup \left[\frac{3\pi}{4}, \pi\right)$
D: $\left[0, \frac{... | The derivative of the function $y = \cos x$ is given by $y' = -\sin x$.
We know that for any point on the graph, the slope of the tangent line is $k = -\sin x$. Considering the range of the sine function, we have $-1 \leq k \leq 1$.
Relating this to the tangent of the inclination angle $\alpha$ (where $0 \leq \alpha ... | C | Calculus | MCQ | Yes | Yes | cn_k12 | false | 549,020 |
Given vectors $\overrightarrow{a} = (\cos\theta, \sin\theta)$ and $\overrightarrow{b} = (\sqrt{3}, -1)$, find the maximum value of $|2\overrightarrow{a} - \overrightarrow{b}|$. | Consider the given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$. By applying the operations of vector addition and scalar multiplication, we can find an expression for $2\overrightarrow{a} - \overrightarrow{b}$:
$$
2\overrightarrow{a} - \overrightarrow{b} = 2(\cos\theta, \sin\theta) - (\sqrt{3}, -1) = (2\cos\... | 4 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,021 |
Solve the equations:
$(1) \frac{5x-1}{4}=\frac{3x+1}{2}-\frac{2-x}{3}$;
$(2) \frac{3x+2}{2}-1=\frac{2x-1}{4}-\frac{2x+1}{5}$. | ### Problem 1:
Solve the equation: $\frac{5x-1}{4}=\frac{3x+1}{2}-\frac{2-x}{3}$.
**Step-by-Step Solution:**
1. Start by clearing the denominators to simplify the equation:
\[
3(5x-1)=6(3x+1)-4(2-x)
\]
2. Expand the equation:
\[
15x-3=18x+6-8+4x
\]
3. Rearrange the terms to bring like terms to on... | x=-\frac{9}{28} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,022 |
A travel agency rented 12 boats for 58 people, filling them exactly. Each big boat seats 6 people, and each small boat seats 4 people. The number of big boats rented was ____, and the number of small boats rented was ____. | Let's assume all 12 boats rented were big boats. Then, the number of small boats would be:
$(6 \times 12 - 58) \div (6 - 4)$
$= 14 \div 2$
$= 7$ (boats)
Therefore, the number of big boats would be: $12 - 7 = 5$ (boats)
Answer: There were 5 big boats and 7 small boats.
Thus, the answer is: $\boxed{5, 7}$.
... | 5, 7 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,023 |
Given two non-zero vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ that satisfy $\overrightarrow{a} \cdot (\overrightarrow{a} - \overrightarrow{b}) = 0$ and $2|\overrightarrow{a}| = |\overrightarrow{b}|$, find the angle $$.
A: $30^{\circ}$
B: $60^{\circ}$
C: $120^{\circ}$
D: $150^{\circ}$ | According to the problem, $\overrightarrow{a} \cdot (\overrightarrow{a} - \overrightarrow{b}) = 0$, thus $\overrightarrow{a} \cdot \overrightarrow{a} = \overrightarrow{a} \cdot \overrightarrow{b}$, which means $|\overrightarrow{a}|^2 = \overrightarrow{a} \cdot \overrightarrow{b}$.
Given $2|\overrightarrow{a}| = |\over... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,024 |
After consuming a small amount of alcohol, a person's blood alcohol content (BAC) quickly rises to 0.3mg/mL. After they stop drinking, their BAC decreases at a rate of 25% per hour. To ensure traffic safety, according to the "Road Traffic Safety Law" of a certain place, the BAC of a driver must not exceed 0.09 mg/mL. T... | To find out how many hours it takes for the BAC to decrease to 0.09 mg/mL or below, we can use the formula for exponential decay, which in this case is $BAC = 0.3 \times (0.75)^t$, where $t$ is the time in hours.
We need to solve the inequality $0.3 \times (0.75)^t \leq 0.09$.
Let's divide both sides by 0.3 to simpl... | 5 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,025 |
3.66666 is a repeating decimal. \_\_\_\_\_\_． (Judge true or false) | Solution: 3.66666 is a finite decimal, not a repeating decimal.
Therefore, the statement in the question is incorrect.
Hence, the answer is: $\boxed{\times}$.
A repeating decimal is an infinite decimal, while 3.66666 is a finite decimal, not a repeating decimal.
This question tests the concept of repeating deci... | \times | Number Theory | math-word-problem | Yes | Yes | cn_k12 | false | 549,026 |
If $\tan(\alpha + \beta) = \frac{2}{5}$ and $\tan(\beta - \frac{\pi}{4}) = \frac{1}{4}$, then what is the value of $\tan(\alpha + \frac{\pi}{4})$?
A: $\frac{1}{6}$
B: $\frac{3}{22}$
C: $\frac{13}{22}$
D: $\frac{13}{18}$ | [Analysis]
This problem tests the understanding of the trigonometric identity for the tangent of a sum and difference of angles. Familiarity with these formulas is essential to solving this problem.
[Solution]
First, let's rewrite the desired expression as a difference of angles:
$$\tan\left(\alpha + \frac{\pi}{4}\ri... | \frac{3}{22} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,027 |
The absolute value of $-\sqrt{6}$ is ____. | To find the absolute value of $-\sqrt{6}$, we follow the definition of absolute value, which is the distance of a number from $0$ on the real number line, regardless of direction. The absolute value of a number is always non-negative.
Starting with the given expression:
$$|-\sqrt{6}|$$
By the definition of absolute v... | \sqrt{6} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,028 |
Given vectors $\overrightarrow{a} = (-2, 3)$ and $\overrightarrow{b} = (1, m- \frac{3}{2})$, and $\overrightarrow{a}$ is parallel to $\overrightarrow{b}$, find the value of $m$.
Options:
A: 3
B: 0
C: $\frac{13}{6}$
D: $\frac{3}{2}$ | Since we know that vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel ($\overrightarrow{a} \parallel \overrightarrow{b}$), they must satisfy the property that their corresponding components are proportional. That is, the ratio of corresponding components of two parallel vectors should be equal:
$$
\fra... | 0 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,029 |
The graph of the function $y=\sin 2x$ is translated to the left by $\phi (\phi > 0)$ units. If the resulting graph passes through the point $(\frac{\pi}{6}, \frac{\sqrt{3}}{2})$, find the minimum value of $\phi$. | When the graph of the function $y=\sin 2x$ is translated to the left by $\phi (\phi > 0)$ units, the corresponding function expression becomes $y=\sin 2(x+\phi)$.
Given that the resulting graph passes through the point $(\frac{\pi}{6}, \frac{\sqrt{3}}{2})$, we have $\sin 2(\frac{\pi}{6}+\phi) = \frac{\sqrt{3}}{2}$. To... | \frac{\pi}{6} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,030 |
There are seven animals living on Bear Ridge: Big Bear, Little Bear, Gigi, Momo, Bouncy, Radish Head, and Tootoo. To better protect the forest, they need to elect two animals as group leaders. What is the probability that at least one among Big Bear and Little Bear is chosen as a group leader? | To solve this problem, we will consider the total number of ways two animals can be chosen as group leaders and the event where neither Big Bear nor Little Bear is chosen.
The total number of ways to select two animals out of the seven is given by the combinatorial number $C_{7}^{2}$. This represents the total possibl... | \frac{11}{21} | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false | 549,031 |
The polynomial $x^{2}+x-6$ can be factored into $\left(x+a\right)\left(x+b\right)$, where $a$ and $b$ are both integers. Then the value of $\left(a+b\right)^{2023}$ is ( )
A: $-1$
B: $1$
C: $-2023$
D: $2023$ | To factor the polynomial $x^{2}+x-6$, we can look for two numbers that multiply to $-6$ and add to $1$ (the coefficient of $x$). These numbers are $-2$ and $3$. Thus, we can rewrite the polynomial as follows:
\begin{align*}
x^{2}+x-6 &= x^{2}+\left(-2+3\right)x+\left(-2\times 3\right) \\
&= x^{2}-2x+3x-6 \\
&= (x^{2}-... | \text{B: }1 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,032 |
The imaginary part of the complex number $z= \frac {2i^{2}+4}{i+1}$ is ( )
A: $-3$
B: $-1$
C: $1$
D: $2$ | Since $z= \frac {2i^{2}+4}{i+1}= \frac {2}{1+i}= \frac {2(1-i)}{(1+i)(1-i)}=1-i$,
the imaginary part of the complex number $z= \frac {2i^{2}+4}{i+1}$ is $-1$.
Therefore, the correct option is $\boxed{B}$.
This solution directly utilizes the algebraic operations of multiplication and division in complex numbers to... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,033 |
Given points $A(4\sin \theta,6\cos \theta)$ and $B(-4\cos \theta,6\sin \theta)$, when $\theta$ is any real number, the trajectory of the midpoint of segment $AB$ is ( )
A: Line
B: Circle
C: Ellipse
D: Hyperbola | **Analysis**
This question examines the equation of a trajectory, focusing on the application of parametric equations, which is quite basic. First, use the midpoint coordinate formula to establish the relationship between points $A$, $B$, and point $M$, deriving the parametric equations that fit their coordinates. Fin... | \text{C} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,034 |
If $m=x^2+2x+3$ ($x\in \mathbb{R}$), and $n=2$, then the relationship between $m$ and $n$ is \_\_\_\_\_\_. | Solution: $m-n=x^2+2x+3-2=x^2+2x+1=(x+1)^2\geq0$, hence $m\geq n$.
Therefore, the answer is: $m\geq n$.
This problem can be solved by comparing the difference between $m$ and $n$.
The main point of this problem is to compare the sizes of two numbers using the comparison method, which is a basic question.
So, the fi... | m\geq n | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,035 |
Given that the complex number $z$ satisfies the equation $(1+i)z=2i$ (where $i$ is the imaginary unit), the correct conclusion is ( ):
A: $|z|=2$
B: The imaginary part of $z$ is $i$
C: $z^2=2$
D: The conjugate of $z$ is $1-i$ | From $(1+i)z=2i$, we get $z=\frac{2i}{1+i}=\frac{2i(1-i)}{(1+i)(1-i)}=1+i$.
Thus, $|z|=\sqrt{2}$, the imaginary part of $z$ is $1$, $z^2=(1+i)^2=2i$, and the conjugate of $z$ is $1-i$.
Therefore, the correct answer is $\boxed{\text{D}}$.
To solve this problem, we first rearrange the given equation, then simplify it ... | \text{D} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,036 |
In 2019, our county built 4 million square meters of new housing, of which 2.5 million square meters are mid-to-low-priced houses. It is expected that in the coming years, the average annual increase in the area of new housing in our county will be 8% higher than the previous year. In addition, the area of mid-to-low-p... | To solve this problem, we will break it down into two parts as given in the problem statement.
### Part 1: Cumulative Area of Mid-to-Low-Priced Houses
Let's denote the area of mid-to-low-priced houses built each year as a sequence $\{a_{n}\}$, where $n$ represents the year, starting from 2019 as the first year.
Give... | 2024 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,037 |
At a school, $4$ students are going to $3$ communities to participate in a garbage classification promotion activity. Each student can only go to $1$ community, and each community must have at least $1$ student. The number of different arrangements is ( ).
A: $6$
B: $12$
C: $24$
D: $36$ | To solve this problem, we break it down into two main steps, following the given solution closely:
**Step 1: Dividing the 4 students into 3 groups.**
Given that each community must have at least 1 student, the only way to divide 4 students into 3 groups is to have two students in one group and one student in each of ... | D | Combinatorics | MCQ | Yes | Yes | cn_k12 | false | 549,038 |
A moving point M on a plane has a distance from a fixed point F(3,0) that is 2 units greater than its distance to the line $l: x+1=0$. The equation satisfied by the moving point M is ( )
A: $x^2=6y$
B: $x^2=12y$
C: $y^2=6$
D: $y^2=12$ | From the given conditions, the distance from the moving point P to the fixed point A(3,0) is equal to its distance to the fixed line $x+3=0$. Therefore, the trajectory of P is a parabola with point A as its focus and the line $x+3=0$ as its directrix, and $p=6$. Thus, the equation of the parabola is $y^2=12x$. Therefor... | D | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,039 |
If the quadratic equation $x(kx+1)-x^{2}+3=0$ has real roots, then the largest integer value of $k$ is ( ).
A: $2$
B: $1$
C: $0$
D: $-1$ | To solve the problem, we first simplify the given quadratic equation:
\[
x(kx + 1) - x^2 + 3 = 0
\]
This simplifies to:
\[
kx^2 + x - x^2 + 3 = 0
\]
Further simplification gives:
\[
(k - 1)x^2 + x + 3 = 0
\]
For a quadratic equation $ax^2 + bx + c = 0$ to have real roots, the discriminant $b^2 - 4ac$ must be grea... | \text{B: }1 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,040 |
The coefficient of the $x^2$ term in the expansion of $x^2(1+x+x^2)(x- \frac {1}{x})^{6}$ is __________. | To find the coefficient of the $x^2$ term in the expansion, we only need to find the constant term in $(1+x+x^2)(x- \frac {1}{x})^{6}$.
The constant term in $(x- \frac {1}{x})^{6}$ is the fourth term: $-C_{6}^{3} = -20$. The constant term in $1+x+x^2$ is $1$.
The coefficient of the $x^{-2}$ term in $(x- \frac {1}{x})... | -5 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,041 |
We know that any positive real number $N$ can be represented as $N=a\times 10^{n}$ $(1\leqslant a \lt 10, n\in Z)$, then $\lg N=n+\lg a$ $(0\leqslant \lg a \lt 1)$. When $n \gt 0$, $N$ has $n+1$ digits. Given $\lg 3\approx 0.4771$, then the number of digits in $3^{100}$ is ( )
A: $46$
B: $47$
C: $48$
D: $49$ | To solve this problem, we start by applying the logarithmic property that allows us to bring the exponent in front of the logarithm. This gives us:
$$\lg 3^{100} = 100 \lg 3$$
Given that $\lg 3 \approx 0.4771$, we can substitute this value into our equation:
$$100 \lg 3 = 100 \times 0.4771 = 47.71$$
According to th... | C | Number Theory | MCQ | Yes | Yes | cn_k12 | false | 549,042 |
In a job interview, there are three questions, and the candidate has three chances to answer. Once the candidate answers a question correctly, they pass the interview; otherwise, they keep drawing questions until the third attempt. If the probability of the candidate Xiao Wang answering each question correctly is $0.7$... | To solve this problem, let's break down the probability of Xiao Wang passing the interview into three separate scenarios, each corresponding to the attempt in which he answers a question correctly:
1. **First Attempt:** The probability of Xiao Wang answering the question correctly on his first attempt is directly give... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,044 |
Given that $a$ and $b$ are real numbers, $a \neq b$, and $a + b = 2$, the inequality that holds true is ( )
A: $1 > ab < \frac{a^2 + b^2}{2}$
B: $ab < 1 < \frac{a^2 + b^2}{2}$
C: $ab < \frac{a^2 + b^2}{2} < 1$
D: $\frac{a^2 + b^2}{2} < ab < 1$ | **Analysis**:
This problem tests fundamental knowledge and basic skills related to the properties of real numbers, important inequalities, and basic inequality properties. It is a moderately difficult question.
**Step-by-step Solution**:
1. Given that $a + b = 2$ and $a \neq b$, we can infer that $ab 0$ (as $a \neq ... | \frac{a^2 + b^2}{2} > 1 > ab | Inequalities | MCQ | Yes | Yes | cn_k12 | false | 549,045 |
(1) The maximum value of the function $f(x)=x(1-x)$, where $x\in(0,1)$, is _______.
(2) In the sequence $\{a_n\}$, the sum of the first $n$ terms $S_n=3\cdot2^n+k$. If the sequence $\{a_n\}$ is a geometric sequence, then the value of the constant $k$ is _______.
(3) In $\triangle ABC$, where $A=60^{\circ}$, $b=1$, an... | (1) **Analysis**
This question examines the properties and graph of a quadratic function. From the quadratic function's expression, we know the graph opens downwards, and the axis of symmetry is $x= \frac{1}{2}$. This leads to the determination of the function's maximum value.
**Solution**
Since $f(x)=x(1-x)=-x^2+x$... | a_n=3\cdot2^{n-2} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,046 |
Determine the function among the following that has the same graph as $y=x$ ( )
A: $y={(\sqrt{x})^{2}}$
B: $y=\sqrt{x^{2}}$
C: $y=e^{\ln x}$
D: $y=\ln e^{x}$ | **Analysis**
This question primarily tests the ability to determine whether two functions are identical and the understanding of function concepts, function domains, etc. It is a basic question.
To find a function with the same graph as $y=x$, we only need to consider whether their domains and expressions are the sam... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,047 |
The opposite of $-5$ is
A: $-\frac{1}{5}$
B: $-5$
C: $\frac{1}{5}$
D: $5$ | The opposite of a number is defined as the number that, when added to the original number, results in zero. For the number $-5$, we can find its opposite by considering the addition that equals zero:
$$-5 + x = 0$$
To solve for $x$, we add $5$ to both sides of the equation:
$$-5 + x + 5 = 0 + 5$$
This simplifies to... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,048 |
Given a regular tetrahedron ABCD (with all six edges equal) with edge length 1, and edge AB is parallel to plane $\alpha$, the range of the area of the shape formed by the projections of all points on the regular tetrahedron onto plane $\alpha$ is
A: $[ \frac {1}{2},1]$
B: $[ \frac { \sqrt {2}}{4}, \frac {1}{2}]$
C: $[... | From the problem statement, when the side edge opposite to AB is parallel to the projection plane, the projection is the largest. In this case, the projection consists of two isosceles triangles that are symmetrical about the line segment AB. Since the edge length of the regular tetrahedron is 1, the area of the projec... | [ \frac { \sqrt {2}}{4}, \frac {1}{2}] | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,049 |
In triangle $\triangle ABC$, $AC=2$, $D$ is the midpoint of $AB$, $CD=\frac{1}{2}BC=\sqrt{7}$, $P$ is a point on $CD$, and $\overrightarrow{AP}=m\overrightarrow{AC}+\frac{1}{3}\overrightarrow{AB}$. Find $|\overrightarrow{AP}|$.
A: $\frac{{\sqrt{31}}}{4}$
B: $\frac{{\sqrt{13}}}{3}$
C: $\frac{{\sqrt{13}}}{2}$
D: $\fr... | Given the problem setup, let's break down the solution step by step:
1. **Midpoint Relation and Vector Equation**:
Since $D$ is the midpoint of $AB$, we can express $\overrightarrow{CD}$ in terms of $\overrightarrow{CA}$ and $\overrightarrow{CB}$ as follows:
\[
\overrightarrow{CD} = \frac{1}{2}(\overrightarro... | \text{D: } \frac{2\sqrt{13}}{3} | Geometry | MCQ | Yes | Yes | cn_k12 | false | 549,050 |
In the geometric sequence $\{a_n\}$, $a_2=6$ and $a_3=-18$. Find the sum of the first four terms, $a_1+a_2+a_3+a_4$.
A: 26
B: 40
C: 54
D: 80 | Since we have a geometric sequence $\{a_n\}$ with $a_2=6$ and $a_3=-18$, we can find the common ratio ($q$) as follows:
$$q = \frac{a_3}{a_2} = \frac{-18}{6} = -3.$$
Then, we can determine the first term ($a_1$) using the common ratio:
$$a_1 = \frac{a_2}{q} = \frac{6}{-3} = -2.$$
To find the fourth term ($a_4$), we use... | B: 40 | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,051 |
Given: $x - y = 1$, prove: $x^3 - 3xy - y^3 = 1$ | **Proof:**
We start with the expression $x^3 - 3xy - y^3$ and manipulate it as follows:
$$x^3 - 3xy - y^3 = x^3 - y^3 - 3xy$$
This can be rewritten using the factorization formula for the difference of cubes, and then simplifying:
$$= (x - y)(x^2 + xy + y^2) - 3xy$$
Substituting $x - y = 1$ into the equation, we g... | 1 | Algebra | proof | Yes | Yes | cn_k12 | false | 549,052 |
Due to the needs of epidemic prevention and control, Xiaoming's father bought $3$ thermometers at a unit price of $a$ yuan each, and $b$ masks at a unit price of $1$ yuan each, spending a total of ______ yuan. (Express using algebraic expressions involving $a$ and $b$). | To solve the problem, we need to calculate the total cost of buying thermometers and masks. The cost of buying thermometers can be calculated by multiplying the number of thermometers bought by the unit price of each thermometer. Similarly, the cost of buying masks can be calculated by multiplying the number of masks b... | 3a + b | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,053 |
Given a geometric sequence $\{a_n\}$, it is known that for any natural number $n$, the sum of the first $n$ terms satisfies $a_1 + a_2 + \dots + a_n = 2^n - 1$. Determine the value of $a_1^2 + a_2^2 + \dots + a_n^2$. | As the sequence $\{a_n\}$ is geometric and for any natural number $n$, we have $a_1 + a_2 + \dots + a_n = 2^n - 1$,
Let's start by finding the first term $a_1$ and the common ratio $r$:
We know that for $n=1$, the sum of the first term is just $a_1$ itself, hence:
$$a_1 = 2^1 - 1 = 1.$$
For $n \geq 2$, the $n$-th ter... | \frac{1}{3}(4^n - 1) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,054 |
In the expansion of $({{x^2}-\frac{2}{x}})^5$, the coefficient of $x$ is ( ).
A: $80$
B: $-80$
C: $40$
D: $-40$ | To find the coefficient of $x$ in the expansion of $({{x^2}-\frac{2}{x}})^5$, we start by identifying the general term of the expansion. The general term for any binomial expansion of the form $(a+b)^n$ is given by ${C}_{n}^{r}a^{n-r}b^{r}$. Applying this to our given expression, we have:
1. The general term is ${C}_{... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,056 |
If the curve $y = kx + \ln x$ has a tangent at the point $(1, k)$ that is parallel to the line $y = 2x + 15$,
then $k = \_\_\_\_\_\_\_\_\_\_$. | The slope of the tangent line to a curve at a given point is found by taking the derivative of the curve with respect to $x$ and evaluating it at that point.
For the curve $y = kx + \ln x$, let's find the derivative:
$$ y' = \frac{d}{dx}(kx + \ln x) $$
$$ y' = k + \frac{1}{x} $$
Given that the tangent line at the po... | k = 1 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,058 |
The proposition $P$: The equation $x^{2}+mx+1=0$ with respect to $x$ has two different negative real roots. The proposition $q$: The equation $4x^{2}+4(m-2)x+1=0$ with respect to $x$ has no real roots. If $P$ or $q$ is true, and $P$ and $q$ are both false, find the range of the real number $m$. | Proposition $P$: $\begin{cases} \Delta _{1}=m^{2}-4 > 0 \\ x_{1}+x_{2}=-m 0 \end{cases}$, solving for $m$ gives $m > 2$.
Proposition $Q$: $\Delta _{2}=16(m-2)^{2}-16 2 \\ m\geqslant 3 \text{ or } m\leqslant 1 \end{cases}$, so $m\geqslant 3$.
$\bullet$ If $P$ is false and $q$ is true, then $\begin{cases} m\leqslant ... | m\in(1,2]\cup[3,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,059 |
Which of the following real numbers is irrational?
A: $\sqrt{2}$
B: $\frac{1}{3}$
C: 3.1415
D: -5 | To determine which of the given real numbers is irrational, we analyze each option as follows:
- **Option A: $\sqrt{2}$**
- The square root of 2 cannot be expressed as a fraction of two integers, which is a defining characteristic of irrational numbers. Therefore, $\sqrt{2}$ is irrational.
- **Option B: $\frac{1}{3... | \text{A} | Number Theory | MCQ | Yes | Yes | cn_k12 | false | 549,060 |
Given an odd function $f(x)=\frac{ax-b}{{x}^{2}+1}$ defined on $(-1,1)$, and $f(\frac{1}{3})=\frac{3}{10}$.
$(1)$ Find the analytical expression of the function $f(x)$;
$(2)$ Determine the monotonicity of $f(x)$ and prove your conclusion;
$(3)$ Solve the inequality $f(2t)+f(3t-1) \lt 0$. | ### Step-by-Step Solution
#### Part 1: Analytical Expression of $f(x)$
Given that $f(x)$ is an odd function and $f\left(\frac{1}{3}\right)=\frac{3}{10}$, we can set up the following equations based on the definition of the function $f(x)=\frac{ax-b}{{x}^{2}+1}$:
1. Since $f(0) = \frac{-b}{1} = 0$, we have $b = 0$.
2... | (0, \frac{1}{5}) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,061 |
Given the function $f(x)=|2x+1|+|2x-3|$.
$(1)$ Find the solution set of the inequality $f(x) \leqslant 6$;
$(2)$ If the solution set of the inequality $f(x) < |a-1|$ with respect to $x$ is non-empty, find the range of the real number $a$. | Solution: $(1)$ Since $f(x)=|2x+1|+|2x-3|$,
The inequality $f(x) \leqslant 6$ is equivalent to $\begin{cases} x \dfrac{3}{2} \\ 2x+1+2x-3 \leqslant 6 \end{cases}③$.
Solving $①$ yields $-1 \leqslant x 4$, i.e., $a-1 > 4$ or $a-1 5$ or $a 5 \text{ or } a < -3\}}$. | null | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false | 549,062 |
Given the function $f(x)=\cos{x}\sin{x}-\sqrt{3}\cos^2{x}+\frac{\sqrt{3}}{2}, x \in \mathbb{R}$.
$(1)$ Find the smallest positive period of $f(x)$;
$(2)$ Find the minimum value of $f(x)$ on the closed interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$ and the corresponding value of $x$. | ### Solution Detailed Steps
#### Part 1: Smallest Positive Period of $f(x)$
Given the function $f(x)=\cos{x}\sin{x}-\sqrt{3}\cos^2{x}+\frac{\sqrt{3}}{2}$, let's simplify it using trigonometric identities:
1. First, recall the double angle identities:
- $\sin{2x} = 2\sin{x}\cos{x}$
- $\cos{2x} = 2\cos^2{x}-1$
... | -\frac{\pi}{12} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,063 |
A batch of equipment is worth $a$ ten thousand yuan. Due to wear and tear from use, its value decreases by $b\%$ every year. The value of this batch of equipment after $n$ years is ( ).
A: $na(1-b\%)$
B: $a(1-nb\%)$
C: $a(1-b\%)^{n}$
D: $a[1-(b\%)^{n}]$ | According to the problem, we know that the value after the first year is $a(1-b\%)$, and the value after the second year is $a(1-b\%)^{2}$.
Following this pattern, we can see that the value each year forms a geometric sequence, with the first term being $a(1-b\%)$ and the common ratio being $(1-b\%)$.
Therefore, the ... | \text{C} | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,064 |
Given the universal set $U=\{1,2,3,4,5\}$, subsets $M$ and $N$ of $U$ satisfy $M=\{1,4\}$, $M \cap N=\{1\}$, and $N \cap (\complement_U M)=\{3,5\}$. Determine the set $N$.
A: $\{1,3\}$
B: $\{3,5\}$
C: $\{1,3,5\}$
D: $\{1,2,3,5\}$ | Since the universal set is $U=\{1,2,3,4,5\}$, and the set $M=\{1,4\}$, we can find the complement of $M$ in $U$, which is $\complement_U M=\{2,3,5\}$ because the complement consists of all elements in $U$ not in $M$.
Given that $N \cap (\complement_U M)=\{3,5\}$, we can infer that both $3$ and $5$ must be elements of ... | \{1,3,5\} | Logic and Puzzles | MCQ | Yes | Yes | cn_k12 | false | 549,065 |
Let $\{a_n\}$ be a positive arithmetic sequence with the sum of the first $n$ terms denoted as $S_n$, and it satisfies $a_2+a_5=7$ and $a_3 a_4=12$.
$(1)$ Find the general formula for the $n$-th term of $\{a_n\}$.
$(2)$ If $b_n = \frac{a_{n+1}}{a_{n+2}} + \frac{a_{n+2}}{a_{n+1}} - 2$, calculate the sum of the first $n$... | $(1)$ Given the conditions:
$$\begin{cases} a_2 + a_5 = a_3 + a_4 = 7, \\ a_3 \cdot a_4 = 12, \\ d > 0. \end{cases}$$
We can solve this system of equations. By using the property of an arithmetic sequence: $a_2 + a_5 = 2a_3 + 2d$ and $a_3 + a_4 = 2a_3 + d$. Since both expressions equal $7$, we deduce that $d = a_4 - a... | T_n = \frac{n+1}{2n+4} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,066 |
Given $\sin\theta= \frac {5}{13}$, and $\theta$ is an angle in the second quadrant, then $\tan\theta= \_\_\_\_\_\_$. | Since $\theta$ is an angle in the second quadrant, we have $\cos\theta < 0$.
Given $\sin\theta= \frac {5}{13}$, we can find $\cos\theta$ as $-\sqrt{1-(\frac{5}{13})^2} = -\frac{12}{13}$.
Therefore, $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{5}{13}}{-\frac{12}{13}} = -\frac{5}{12}$.
Hence, the answe... | -\frac{5}{12} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,067 |
Given that the sum of the polynomials $A$ and $B$ is $12x^{2}y+2xy+5$, where $B=3x^{2}y-5xy+x+7$. Find:$(1)$ The polynomial $A$;$(2)$ When $x$ takes any value, the value of the expression $2A-\left(A+3B\right)$ is a constant. Find the value of $y$. | ### Step-by-Step Solution
#### Part 1: Finding the Polynomial $A$
Given that the sum of the polynomials $A$ and $B$ is $12x^{2}y+2xy+5$, and $B=3x^{2}y-5xy+x+7$, we can find $A$ by subtracting $B$ from the sum:
\[
\begin{align*}
A &= 12x^{2}y+2xy+5 - (3x^{2}y-5xy+x+7) \\
&= 12x^{2}y+2xy+5 - 3x^{2}y + 5xy - x - 7 \\
... | \frac{2}{11} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,068 |
In $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are denoted as $a$, $b$, and $c$ respectively. If $a^{2}=b^{2}+ \sqrt {3}bc+c^{2}$, then $A=$ ______. | Given: $a^{2}=b^{2}+ \sqrt {3}bc+c^{2}$,
Therefore, $b^{2}+c^{2}-a^{2}=- \sqrt {3}bc$,
By applying the cosine rule, we get: $\cos A= \frac {b^{2}+c^{2}-a^{2}}{2bc}= \frac {- \sqrt {3}bc}{2bc}=- \frac { \sqrt {3}}{2}$.
Since $A \in (0,\pi)$,
We solve to find: $A= \frac {5\pi}{6}$.
Hence, the answer is: $\boxed{\fra... | \frac {5\pi}{6} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false | 549,069 |
Find the interval $(n, n+1)$ (where $n$ is a natural number) within which the root of the function $f(x) = 3x - 7 + \ln(x)$ lies. | To find the root of the function $f(x) = 3x - 7 + \ln(x)$, we estimate the function's values at two consecutive natural numbers:
Let's start with $f(2)$:
$$
f(2) = 3(2) - 7 + \ln(2) = 6 - 7 + \ln(2) = -1 + \ln(2)
$$
Since $\ln(2) 1$, we conclude that:
$$
f(3) > 2 + 1 = 3 > 0
$$
Given that $f(2) 0$, and considering ... | 2 | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,070 |
Given points $A(3,0)$, $B(0,3)$, and $C(\cos α, \sin α)$, with $O$ as the origin.
(1) If $\overrightarrow{OC}$ is parallel to $\overrightarrow{AB}$, find the value of $\tan α$;
(2) If $\overrightarrow{AC}$ is perpendicular to $\overrightarrow{BC}$, find the value of $\sin 2α$;
(3) If $|\overrightarrow{OA} + \overrighta... | (1) Given $A(3,0)$, $B(0,3)$, and $C(\cos α, \sin α)$,
we have $\overrightarrow{OC} = (\cos α, \sin α)$ and $\overrightarrow{AB} = (-3, 3)$.
Since $\overrightarrow{OC}$ is parallel to $\overrightarrow{AB}$, we have $3\cos α + 3\sin α = 0$. Solving for $\tan α$, we get $\tan α = -1$.
(2) From the given problem, we have... | \frac{π}{6} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,071 |
Given $f(x) = \max\{x^2 - ax + a, ax - a + 1\}$, where $\max\{x, y\} = \begin{cases} y, & x \leq y \\ x, & x > y \end{cases}$.
(Ⅰ) If for any $x \in \mathbb{R}$, we always have $f(x) = x^2 - ax + a$, find the value of the real number $a$;
(Ⅱ) If $a > 1$, find the minimum value $m(a)$ of $f(x)$. | Solution:
(Ⅰ) Since for any $x \in \mathbb{R}$, we always have $f(x) = x^2 - ax + a$,
$\therefore$ For $x \in \mathbb{R}$, $x^2 - ax + a \geq ax - a + 1$ always holds,
which means $x^2 - 2ax + 2a - 1 \geq 0$ always holds.
$\therefore \Delta = 4a^2 - 4(2a - 1) \leq 0$, i.e., $(a - 1)^2 \leq 0$,
$\therefore a =... | m(a) = \begin{cases} -\frac{a^2}{4} + a, & 1 2 \end{cases} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,072 |
If the function $f(x)=kx-\ln x$ is monotonically increasing on the interval $(1,+\infty)$, then the range of values for $k$ is ______. | First, we find the derivative of $f(x)$: $f'(x)=k- \dfrac {1}{x}$.
Since the function $f(x)=kx-\ln x$ is monotonically increasing on the interval $(1,+\infty)$, it follows that $f'(x) \geqslant 0$ holds true on the interval $(1,+\infty)$.
Therefore, we have $k \geqslant \dfrac {1}{x}$.
Given that $y= \dfrac {1}{x}$ ... | [1,+\infty) | Calculus | math-word-problem | Yes | Yes | cn_k12 | false | 549,073 |
After organizing a set of data into four groups, the frequencies of the first, second, and third groups are 0.1, 0.3, and 0.4, respectively. If the frequency count of the first group is 5, determine the frequency count of the fourth group. | Let's denote the total number of data points as $N$. According to the problem, the frequencies for the first three groups are given, and the frequency count for the first group is known to be $5$.
The frequency ($f$) of a group is calculated by dividing the frequency count ($n$) of the group by the total count ($N$):
... | 10 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,074 |
According to reports, in 2010, the relevant departments of the Suzhou municipal government will complete the comprehensive renovation work of 1.3 million square meters of old residential areas in the urban area. The number 1.3 million (i.e., 1,300,000) can be represented in scientific notation as
A: $1.3 \times 10^4$
B... | 1.3 million = "1" 300 000 = $1.3 \times 10^6$.
Therefore, the correct answer is $\boxed{C}$. | C | Algebra | MCQ | Yes | Yes | cn_k12 | false | 549,075 |
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