LFM Stats And Pure

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gaokao 2018 Q14 5 marks Linear Programming (Optimize Objective over Linear Constraints) View
If $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x - 2 y - 2 \leq 0 , \\ x - y + 1 \geq 0 , \\ y \leq 0 , \end{array} \right.$ then the maximum value of $z = 3 x + 2 y$ is \_\_\_\_
gaokao 2018 Q14 5 marks Linear Programming (Optimize Objective over Linear Constraints) View
Given the constraints $\left\{ \begin{array} { l } x + 2 y - 5 \geq 0 , \\ x - 2 y + 3 \geq 0 , \\ x - 5 \leq 0 , \end{array} \right.$ the minimum value of $z = x + y$ is \_\_\_\_.
gaokao 2018 Q14 5 marks Linear Programming (Optimize Objective over Linear Constraints) View
If $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x + 2 y - 5 \geqslant 0 , \\ x - 2 y + 3 \geqslant 0 , \\ x - 5 \leqslant 0 , \end{array} \right.$ then the maximum value of $z = x + y$ is $\_\_\_\_$.
gaokao 2019 Q1 5 marks Set Operations Using Inequality-Defined Sets View
Let $A = \left\{ x \mid x ^ { 2 } - 5 x + 6 > 0 \right\} , B = \{ x \mid x - 1 < 0 \}$, then $A \cap B =$
A. $( - \infty , 1 )$
B. $( - 2,1 )$
C. $( - 3 , - 1 )$
D. $( 3 , + \infty )$
gaokao 2019 Q1 5 marks Set Operations Using Inequality-Defined Sets View
Given sets $A = \{ - 1,0,1,2 \} , B = \left\{ x \mid x ^ { 2 } \leqslant 1 \right\}$ , then $A \cap B =$
A. $\{ - 1,0,1 \}$
B. $\{ 0,1 \}$
C. $\{ - 1,1 \}$
D. $\{ 0,1,2 \}$
gaokao 2019 Q5 5 marks Linear Programming (Optimize Objective over Linear Constraints) View
If $x , y$ satisfy $| x | \leqslant 1 - y$ and $y \geqslant - 1$, then the maximum value of $3 x + y$ is (A) $- 7$ (B) 1 (C) 5 (D) 7
gaokao 2019 Q5 Set Operations Using Inequality-Defined Sets View
5. Keep the answer sheet clean, do not fold it, do not tear or wrinkle it, and do not use correction fluid, correction tape, or scrapers.
I. Multiple Choice Questions: 12 questions in total, 5 points each, 60 points total. For each question, only one of the four options is correct.

1. Let $A = \left\{ x \mid x ^ { 2 } - 5 x + 6 > 0 \right\} , B = \{ x \mid x - 1 < 0 \}$, then $A \cap B =$
A. $( - \infty , 1 )$
B. $( - 2,1 )$
C. $( - 3 , - 1 )$
D. $( 3 , + \infty )$
2. Let $z = - 3 + 2 \mathrm { i }$. In the complex plane, the point corresponding to $\bar { z }$ is located in
A. the first quadrant
B. the second quadrant
C. the third quadrant
D. the fourth quadrant
3. Given $\overrightarrow { A B } = ( 2,3 ) , \overrightarrow { A C } = ( 3 , t ) , | \overrightarrow { B C } | = 1$, then $\overrightarrow { A B } \cdot \overrightarrow { B C } =$
A. $- 3$
B. $- 2$
C. $2$
D. $3$
4. On January 3, 2019, the Chang'e-4 probe successfully achieved humanity's first soft landing on the far side of the moon, marking another major achievement in China's space program. A key technical challenge in achieving soft landing on the far side of the moon is maintaining communication between the ground and the probe. To solve this problem, the Queqiao relay satellite was launched, which orbits around the Earth-Moon Lagrange point $L _ { 2 }$. The $L _ { 2 }$ point is an equilibrium point located on the extension of the Earth-Moon line. Let the Earth's mass be $M _ { 1 }$, the Moon's mass be $M _ { 2 }$, the Earth-Moon distance be $R$, and the distance from the $L _ { 2 }$ point to the Moon be $r$. According to Newton's laws of motion and the law of universal gravitation, $r$ satisfies the equation: $\frac { M _ { 1 } } { ( R + r ) ^ { 2 } } + \frac { M _ { 2 } } { r ^ { 2 } } = ( R + r ) \frac { M _ { 1 } } { R ^ { 3 } }$. Let $\alpha = \frac { r } { R }$. Since $\alpha$ is very small, in approximate calculations $\frac { 3 \alpha ^ { 3 } + 3 \alpha ^ { 4 } + \alpha ^ { 5 } } { ( 1 + \alpha ) ^ { 2 } } \approx 3 \alpha ^ { 3 }$. Then the approximate value of $r$ is
A. $\sqrt { \frac { M _ { 2 } } { M _ { 1 } } } R$
B. $\sqrt { \frac { M _ { 2 } } { 2 M _ { 1 } } } R$
C. $\sqrt [ 3 ] { \frac { 3 M _ { 2 } } { M _ { 1 } } } R$
D. $\sqrt [ 3 ] { \frac { M _ { 2 } } { 3 M _ { 1 } } } R$
5. In a speech competition, 9 judges each give an original score to a contestant. When determining the contestant's final score, 1 highest score and 1 lowest score are removed from the 9 original scores, leaving 7 valid scores. Compared with the 9 original scores, the numerical characteristic that remains unchanged for the 7 valid scores is
A. median
B. mean
C. variance
D. range
gaokao 2019 Q11 Quadratic Inequality Holding for All x (or a Restricted Domain) View
11. If the range of the function $f ( x ) = a \cdot \left( \frac { 1 } { 3 } \right) ^ { x } \left( \frac { 1 } { 2 } \leq x \leq 1 \right)$ is a subset of the range of the function $g ( x ) = \frac { x ^ { 2 } - 1 } { x ^ { 2 } + x + 1 } ( x \in \mathbb{R} )$, then the range of positive number $a$ is
A. $(0,2]$
B. $(0,1]$
C. $( 0,2 \sqrt { 3 } ]$
D. $( 0 , \sqrt { 3 } ]$
gaokao 2019 Q11 Linear Programming (Optimize Objective over Linear Constraints) View
11. Let the plane region represented by the system of inequalities $\left\{ \begin{array} { l } x + y \geq 6 , \\ 2 x - y \geq 0 \end{array} \right.$ be $D$ . Proposition $p : \exists ( x , y ) \in D , 2 x + y \geq 9$ ; Proposition $q : \forall ( x , y ) \in D , 2 x + y \leq 12$ . Four propositions are given below:
(1) $p \vee q$
(2) $\neg p \vee q$
(3) $p \wedge \neg q$
(4) $\neg p \wedge \neg q$
The numbers of all true propositions among these four are
A. (1)(3)
B. (1)(2)
C. (2)(3)
D. (3)(4)
gaokao 2019 Q13 Linear Programming (Optimize Objective over Linear Constraints) View
13. If variables $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { l } 2 x + 3 y - 6 \geq 0 , \\ x + y - 3 \leq 0 , \\ y - 2 \leq 0 , \end{array} \right.$ then the maximum value of $z = 3 x - y$ is $\_\_\_\_$ .
gaokao 2019 Q23 10 marks Optimization Subject to an Algebraic Constraint View
Let $x, y, z \in \mathbf{R}$ and $x + y + z = 1$.
(1) Find the minimum value of $(x-1)^2 + (y+1)^2 + (z+1)^2$;
(2) If $(x-2)^2 + (y-1)^2 + (z-a)^2 \geqslant \frac{1}{3}$ holds for all $x, y, z$ satisfying $x + y + z = 1$, prove that $a \leqslant -3$ or $a \geqslant -1$.
gaokao 2020 Q1 5 marks Set Operations Using Inequality-Defined Sets View
Given set $A = \left\{ x \mid x ^ { 2 } - 3 x - 4 < 0 \right\} , B = \{ - 4,1,3,5 \}$ , then $A \cap B =$
A. $\{ - 4,1 \}$
B. $\{ 1,5 \}$
C. $\{ 3,5 \}$
D. $\{ 1,3 \}$
gaokao 2020 Q1 5 marks Set Operations Using Inequality-Defined Sets View
Given sets $U = \{ - 2 , - 1,0,1,2,3 \} , A = \{ - 1,0,1 \} , B = \{ 1,2 \}$ , then $\complement _ { U } ( A \cup B ) =$
A. $\{ - 2,3 \}$
B. $\{ - 2,2,3 \}$
C. $\{ - 2 , - 1,0,3 \}$
D. $\{ - 2 , - 1,0,2,3 \}$
gaokao 2020 Q1 5 marks Set Operations Using Inequality-Defined Sets View
Given sets $A = \{ 1,2,3,5,7,11 \} , B = \{ x \mid 3 < x < 15 \}$, the number of elements in $A \cap B$ is
A. 2
B. 3
C. 4
D. 5
gaokao 2020 Q1 5 marks Set Operations Using Inequality-Defined Sets View
Given the sets $A = \left\{ ( x , y ) \mid x , y \in \mathbf { N } ^ { * } , y \geqslant x \right\} , B = \{ ( x , y ) \mid x + y = 8 \}$ , the number of elements in $A \cap B$ is
A. 2
B. 3
C. 4
D. 6
gaokao 2020 Q1 4 marks Set Operations Using Inequality-Defined Sets View
Given sets $A = \{ 1,2,4 \} , B = \{ 2,3,4 \}$, find $A \cap B =$ $\_\_\_\_$
gaokao 2020 Q2 5 marks Set Operations Using Inequality-Defined Sets View
Let sets $A = \left\{ x \mid x ^ { 2 } - 4 \leqslant 0 \right\}$, $B = \{ x \mid 2 x + a \leqslant 0 \}$, and $A \cap B = \{ x \mid - 2 \leqslant x \leqslant 1 \}$. Then $a =$
A. $- 4$
B. $- 2$
C. 2
D. 4
gaokao 2020 Q7 5 marks Linear Programming (Optimize Objective over Linear Constraints) View
Given $\left\{ \begin{array} { l } x + y \geq 2 \\ y \geq 0 \\ x + 2 y - 3 \leq 0 \end{array} \right.$, find the maximum value of $z = y - 2 x$ as $\_\_\_\_$
gaokao 2020 Q11 5 marks Ordering and Sign Analysis from Inequality Constraints View
If $2 ^ { x } - 2 ^ { y } < 3 ^ { - x } - 3 ^ { - y }$ , then
A. $\ln ( y - x + 1 ) > 0$
B. $\ln ( y - x + 1 ) < 0$
C. $\ln | x - y | > 0$
D. $\ln | x - y | < 0$
gaokao 2020 Q13 5 marks Linear Programming (Optimize Objective over Linear Constraints) View
If $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { l } 2 x + y - 2 \leqslant 0 , \\ x - y - 1 \geqslant 0 , \\ y + 1 \geqslant 0 , \end{array} \right.$ then the maximum value of $z = x + 7 y$ is $\_\_\_\_$
gaokao 2020 Q13 5 marks Linear Programming (Optimize Objective over Linear Constraints) View
If $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x + y \geqslant 0 , \\ 2 x - y \geqslant 0 , \\ x \leqslant 1 , \end{array} \right.$ then the maximum value of $z = 3 x + 2 y$ is $\_\_\_\_$ .
gaokao 2020 Q13 5 marks Linear Programming (Optimize Objective over Linear Constraints) View
If $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x + y \geqslant 0 , \\ 2 x - y \geqslant 0 , \\ x \leqslant 1 , \end{array} \right.$ then the maximum value of $z = 3 x + 2 y$ is $\_\_\_\_$ .
gaokao 2020 Q13 5 marks Identify Always-True Inequality from Options View
Which of the following inequalities always holds? ( )
A. $a ^ { 2 } + b ^ { 2 } \leq 2 a b$
B. $a ^ { 2 } + b ^ { 2 } \geq - 2 a b$
C. $a + b \geq - 2 \sqrt { | a b | }$
D. $a + b \leq 2 \sqrt { | a b | }$
gaokao 2020 Q23 10 marks Absolute Value Inequality View
Given the function $f ( x ) = \left| x - a ^ { 2 } \right| + | x - 2 a + 1 |$ .
(1) When $a = 2$, find the solution set of the inequality $f ( x ) \geqslant 4$;
(2) If $f ( x ) \geqslant 4$ for all $x$, find the range of values of $a$.
gaokao 2022 Q1 Set Operations Using Inequality-Defined Sets View
1. If $M = \{ x \mid \sqrt { x } < 4 \}$ and $N = \{ x \mid 3 x \geqslant 1 \}$, then $M \cap N =$
A. $\{ x \mid 0 \leqslant x < 2 \}$
B. $\left\{ x \left\lvert \, \frac { 1 } { 3 } \leqslant x < 2 \right. \right\}$
C. $\{ x \mid 3 \leqslant x < 16 \}$
D. $\left\{ x \left\lvert \, \frac { 1 } { 3 } \leqslant x < 16 \right. \right\}$