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2021 national-II

12 maths questions

Q1 Complex Numbers Arithmetic Locating Points in the Complex Plane (Quadrant/Axis) View
1. The point corresponding to the complex number $\frac { 2 - \mathrm { i } } { 1 - 3 \mathrm { i } }$ in the complex plane is located in which quadrant?
A. First quadrant
B. Second quadrant
C. Third quadrant
D. Fourth quadrant 【Answer】A 【Solution】 【Analysis】Use complex division to simplify $\frac { 2 - \mathrm { i } } { 1 - 3 \mathrm { i } }$, and thus determine the location of the corresponding point. 【Detailed Solution】 $\frac { 2 - \mathrm { i } } { 1 - 3 \mathrm { i } } = \frac { ( 2 - \mathrm { i } ) ( 1 + 3 \mathrm { i } ) } { 10 } = \frac { 5 + 5 \mathrm { i } } { 10 } = \frac { 1 + \mathrm { i } } { 2 }$, so the point corresponding to this complex number is $\left( \frac { 1 } { 2 } , \frac { 1 } { 2 } \right)$, which is located in the first quadrant. Therefore, the answer is: A.
Q2 Probability Definitions Set Operations View
2. Let $U = \{ 1,2,3,4,5,6 \} , A = \{ 1,3,6 \} , B = \{ 2,3,4 \}$. Then $A \cap \left( \complement_U B \right) =$
A. $\{ 3 \}$
B. $\{ 1,6 \}$
C. $\{ 5,6 \}$
D. $\{ 1,3 \}$
【Answer】B 【Solution】 【Analysis】Use the definitions of intersection and complement to find $A \cap \left( \complement_U B \right)$. 【Detailed Solution】From the given conditions, $\complement_U B = \{ 1,5,6 \}$, so $A \cap \left( \complement_U B \right) = \{ 1,6 \}$, Therefore, the answer is: B.
Q3 Conic sections Eccentricity or Asymptote Computation View
3. For the parabola $y ^ { 2 } = 2 p x ( p > 0 )$, the distance from its focus to the line $y = x + 1$ is $\sqrt { 2 }$. Then $p =$
A. 1
B. 2
C. $2 \sqrt { 2 }$
D. 4
【Answer】B 【Solution】 【Analysis】First determine the coordinates of the focus of the parabola, then use the point-to-line distance formula to find the value of $p$.
【Detailed Solution】The focus of the parabola has coordinates $\left( \frac { p } { 2 } , 0 \right)$. The distance from this point to the line $x - y + 1 = 0$ is: $\quad d = \frac { \left| \frac { p } { 2 } - 0 + 1 \right| } { \sqrt { 1 + 1 } } = \sqrt { 2 }$, Solving: $p = 2$ (we discard $p = -6$). Therefore, the answer is: B.
Q4 Radians, Arc Length and Sector Area View
4. The BeiDou-3 Global Navigation Satellite System is an important achievement of China's space program. In satellite navigation systems, geostationary satellites orbit in the plane of Earth's equator at an orbital altitude of 36,000 km (orbital altitude is the distance from the satellite to Earth's surface). Consider Earth as a sphere with center $O$ and radius $r = 6400$ km. The latitude of a point $A$ on Earth's surface is defined as the angle that $OA$ makes with the equatorial plane. The maximum latitude at which a geostationary satellite can be directly observed from Earth's surface is $a$. The surface area covered by the satellite signal on Earth's surface is $S = 2 \pi r ^ { 2 } ( 1 - \cos a )$ (in $\mathrm { km } ^ { 2 }$). What percentage of Earth's total surface area does $S$ represent? ( )
A. $26 \%$
B. $34 \%$
C. $42 \%$
D. $50 \%$
【Answer】C 【Solution】 【Analysis】From the given information, use the provided surface area formula and the formula for the surface area of a sphere to calculate the result. 【Detailed Solution】From the given information, the percentage of $S$ relative to Earth's surface area is approximately: $\frac { 2 \pi r ^ { 2 } ( 1 - \cos a ) } { 4 \pi r ^ { 2 } } = \frac { 1 - \cos a } { 2 } = \frac { 1 - \frac { 6400 } { 6400 + 36000 } } { 2 } \approx 0.42 = 42 \%$ . Therefore, the answer is: C.
Q5 Volumes of Revolution Volume of a 3D Geometric Solid (Pyramid/Tetrahedron) View
5. A right square frustum has upper and lower base edges of lengths 2 and 4 respectively, and lateral edge length 2. Its volume is ( )
A. $20 + 12 \sqrt { 3 }$
B. $28 \sqrt { 2 }$
C. $\frac { 56 } { 3 }$
D. $\frac { 28 \sqrt { 2 } } { 3 }$
【Answer】D 【Solution】 【Analysis】Using the geometric properties of the frustum, calculate the height of the solid and the areas of the upper and lower bases, then use the volume formula for a frustum.
【Detailed Solution】Draw a figure and connect the centers of the upper and lower bases of the right square frustum, as shown in the diagram. [Figure]
Since the upper and lower base edges of the frustum are 2 and 4 respectively, and the lateral edge length is 2, the height of the frustum is $h = \sqrt { 2 ^ { 2 } - ( 2 \sqrt { 2 } - \sqrt { 2 } ) ^ { 2 } } = \sqrt { 2 }$, the area of the lower base is $S _ { 1 } = 16$, and the area of the upper base is $S _ { 2 } = 4$, so the volume of the frustum is $V = \frac { 1 } { 3 } h \left( S _ { 1 } + S _ { 2 } + \sqrt { S _ { 1 } S _ { 2 } } \right) = \frac { 1 } { 3 } \times \sqrt { 2 } \times ( 16 + 4 + \sqrt { 64 } ) = \frac { 28 } { 3 } \sqrt { 2 }$ . Therefore, the answer is: D.
Q6 Normal Distribution Multiple-Choice Conceptual Question on Normal Distribution Properties View
6. The measurement result of a certain physical quantity follows a normal distribution $N \left( 10 , \sigma ^ { 2 } \right)$. Which of the following conclusions is incorrect? ( )
A. The smaller $\sigma$ is, the greater the probability that the physical quantity falls in $( 9.9,10.1 )$ in a single measurement.
B. The smaller $\sigma$ is, the probability that the physical quantity is greater than 10 in a single measurement is 0.5.
C. The smaller $\sigma$ is, the probability that the physical quantity is less than 9.99 equals the probability that it is greater than 10.01 in a single measurement.
D. The smaller $\sigma$ is, the probability that the physical quantity falls in $( 9.9,10.2 )$ equals the probability that it falls in $( 10,10.3 )$ in a single measurement.
【Answer】D
【Solution】
【Analysis】Use the properties of the normal distribution density curve to judge each option. 【Detailed Solution】For option A, $\sigma ^ { 2 }$ is the variance of the data, so the smaller $\sigma$ is, the more concentrated the data is around $\mu = 10$. Therefore, the probability that the measurement result falls in $( 9.9,10.1 )$ increases, so A is correct.
For option B, by the symmetry of the normal distribution density curve, the probability that the physical quantity is greater than 10 in a single measurement is 0.5, so B is correct.
For option C, by the symmetry of the normal distribution density curve, the probability that the physical quantity is greater than 10.01 equals the probability that it is less than 9.99 in a single measurement, so C is correct.
For option D, since the probability that the physical quantity falls in $( 9.9,10.0 )$ does not equal the probability that it falls in $( 10.2,10.3 )$, the probability that the measurement result falls in $( 9.9,10.2 )$ does not equal the probability that it falls in $( 10,10.3 )$, so D is incorrect. Therefore, the answer is: D.
Q7 Laws of Logarithms Compare or Order Logarithmic Values View
7. Given $a = \log _ { 5 } 2 , b = \log _ { 8 } 3 , c = \frac { 1 } { 2 }$, which of the following judgments is correct? ( )
A. $c < b < a$
B. $b < a < c$
C. $a < c < b$
D. $a < b < c$
【Answer】C 【Solution】 【Analysis】Use the monotonicity of logarithmic functions to compare the sizes of $a$, $b$, and $c$, and thus reach the conclusion. 【Detailed Solution】 $a = \log _ { 5 } 2 < \log _ { 5 } \sqrt { 5 } = \frac { 1 } { 2 } = \log _ { 8 } 2 \sqrt { 2 } < \log _ { 8 } 3 = b$, that is, $a < c < b$. Therefore, the answer is: C.
Q8 5 marks Function Transformations View
8. Given that the domain of function $f ( x )$ is $\mathbf { R }$, $f ( x + 2 )$ is an even function, and $f ( 2 x + 1 )$ is an odd function, then ( )
A. $f \left( - \frac { 1 } { 2 } \right) = 0$
B. $f ( - 1 ) = 0$
C. $f ( 2 ) = 0$
D. $f ( 4 ) = 0$
【Answer】B 【Solution】 【Analysis】Derive that $f ( x )$ is a periodic function with period 4. From the given conditions, deduce that $f ( 1 ) = 0$, and combine with the given conditions to reach the conclusion.
【Detailed Solution】Since $f ( x + 2 )$ is an even function, we have $f ( 2 + x ) = f ( 2 - x )$, which gives $f ( x + 3 ) = f ( 1 - x )$. Since $f ( 2 x + 1 )$ is an odd function, we have $f ( 1 - 2 x ) = - f ( 2 x + 1 )$, so $f ( 1 - x ) = - f ( x + 1 )$. Therefore, $f ( x + 3 ) = - f ( x + 1 ) = f ( x - 1 )$, that is, $f ( x ) = f ( x + 4 )$. Thus, $f ( x )$ is a periodic function with period 4. Since $F ( x ) = f ( 2 x + 1 )$ is an odd function, we have $F ( 0 ) = f ( 1 ) = 0$. Therefore, $f ( - 1 ) = - f ( 1 ) = 0$, while the other three options are unknown. Therefore, the answer is: B.
II. Multiple Choice Questions: This section contains 4 questions, each worth 5 points, for a total of 20 points. For each question, there may be multiple correct options. Full marks (5 points) are awarded for selecting all correct options, 2 points for partially correct selections, and 0 points for any incorrect selection.
Q9 Measures of Location and Spread View
9. Which of the following statistics can measure the dispersion of the sample $x _ { 1 } , x _ { 2 } , \cdots , x _ { n }$? ( )
A. The standard deviation of the sample $x _ { 1 } , x _ { 2 } , \cdots , x _ { n }$
B. The median of the sample $x _ { 1 } , x _ { 2 } , \cdots , x _ { n }$
C. The range of the sample $x _ { 1 } , x _ { 2 } , \cdots , x _ { n }$
D. The mean of the sample $x _ { 1 } , x _ { 2 } , \cdots , x _ { n }$
【Answer】AC 【Solution】 【Analysis】Determine which of the given options measure data dispersion and which measure central tendency.
【Detailed Solution】By the definition of standard deviation, standard deviation measures data dispersion. By the definition of median, median measures central tendency. By the definition of range, range measures data dispersion. By the definition of mean, mean measures central tendency. Therefore, the answer is: AC. [Detailed Solution] From the given conditions, $f ( x ) = \left| e ^ { x } - 1 \right| = \left\{ \begin{array} { l } 1 - e ^ { x } , x < 0 \\ e ^ { x } - 1 , x \geq 0 \end{array} \right.$ , then $f ^ { \prime } ( x ) = \left\{ \begin{array} { l } - e ^ { x } , x < 0 \\ e ^ { x } , x > 0 \end{array} \right.$ , Thus point $A \left( x _ { 1 } , 1 - e ^ { x _ { 1 } } \right)$ and point $B \left( x _ { 2 } , e ^ { x _ { 2 } } - 1 \right)$ , $k _ { A M } = - e ^ { x _ { 1 } } , k _ { B N } = e ^ { x _ { 2 } }$ , Therefore $- e ^ { x _ { 1 } } \cdot e ^ { x _ { 2 } } = - 1 , x _ { 1 } + x _ { 2 } = 0$ , So $AM: y - 1 + e ^ { x _ { 1 } } = - e ^ { x _ { 1 } } \left( x - x _ { 1 } \right) , M \left( 0 , e ^ { x _ { 1 } } x _ { 1 } - e ^ { x _ { 1 } } + 1 \right)$ , Thus $| A M | = \sqrt { x _ { 1 } ^ { 2 } + \left( e ^ { x _ { 1 } } x _ { 1 } \right) ^ { 2 } } = \sqrt { 1 + e ^ { 2 x _ { 1 } } } \cdot \left| x _ { 1 } \right|$ , Similarly $| B N | = \sqrt { 1 + e ^ { 2 x _ { 2 } } } \cdot \left| x _ { 2 } \right|$ , Therefore $\frac { | A M | } { | B N | } = \frac { \sqrt { 1 + e ^ { 2 x _ { 1 } } } \cdot \left| x _ { 1 } \right| } { \sqrt { 1 + e ^ { 2 x _ { 2 } } } \cdot \left| x _ { 2 } \right| } = \sqrt { \frac { 1 + e ^ { 2 x _ { 1 } } } { 1 + e ^ { 2 x _ { 2 } } } } = \sqrt { \frac { 1 + e ^ { 2 x _ { 1 } } } { 1 + e ^ { - 2 x _ { 1 } } } } = e ^ { x _ { 1 } } \in ( 0,1 )$ . Thus the answer is: $( 0,1 )$ [Key Point Explanation] The key to solving this problem is to use the geometric meaning of the derivative to transform the condition $x _ { 1 } + x _ { 2 } = 0$ , and after eliminating one variable, the calculation yields the solution.
IV. Answer Questions: This section contains 6 questions totaling 70 points. Solutions should include written explanations, proofs, or calculation steps.
Q17 Arithmetic Sequences and Series Multi-Part Structured Problem on AP View
17. Let $S _ { n }$ denote the sum of the first $n$ terms of an arithmetic sequence $\left\{ a _ { n } \right\}$ with non-zero common difference. If $a _ { 3 } = S _ { 5 }$ and $a _ { 2 } a _ { 4 } = S _ { 4 }$ .
(1) Find the general term formula $a _ { n }$ of the sequence $\left\{ a _ { n } \right\}$ ;
(2) Find the minimum value of $n$ such that $S _ { n } > a _ { n }$ holds.
Answer: (1) $a _ { n } = 2 n - 6$ ; (2) 7 .
[Solution]
[Analysis] (1) From the given conditions, first find the value of $a _ { 3 }$ , then combine with the given conditions to find the common difference of the sequence to determine the general term formula;
(2) First find the expression for the sum of the first $n$ terms, then solve the quadratic inequality to determine the minimum value of $n$ . [Detailed Solution] (1) By the properties of arithmetic sequences, we have $S _ { 5 } = 5 a _ { 3 }$ , thus: $a _ { 3 } = 5 a _ { 3 }$ , therefore $a _ { 3 } = 0$ ,
Let the common difference of the arithmetic sequence be $d$ . Then: $a _ { 2 } a _ { 4 } = \left( a _ { 3 } - d \right) \left( a _ { 3 } + d \right) = - d ^ { 2 }$ , $S _ { 4 } = a _ { 1 } + a _ { 2 } + a _ { 3 } + a _ { 4 } = \left( a _ { 3 } - 2 d \right) + \left( a _ { 3 } - d \right) + a _ { 3 } + \left( a _ { 3 } + d \right) = 4a_3 - 2d = -2d$ ,
Thus: $- d ^ { 2 } = - 2 d$ . Since the common difference is non-zero, we have: $d = 2$ ,
The general term formula of the sequence is: $a _ { n } = a _ { 3 } + ( n - 3 ) d = 2 n - 6$ .
(2) From the general term formula, we have: $a _ { 1 } = 2 - 6 = - 4$ , thus: $S _ { n } = n \times ( - 4 ) + \frac { n ( n - 1 ) } { 2 } \times 2 = n ^ { 2 } - 5 n$ . The inequality $S _ { n } > a _ { n }$ becomes: $n ^ { 2 } - 5 n > 2 n - 6$ . Simplifying: $( n - 1 ) ( n - 6 ) > 0$ ,
Solving: $n < 1$ or $n > 6$ . Since $n$ is a positive integer, the minimum value of $n$ is 7 . [Key Point] The solution of basic quantities in arithmetic sequences is a fundamental problem in arithmetic sequences. The key to solving such problems is to master the relevant formulas of arithmetic sequences and apply them flexibly.
Q18 Sine and Cosine Rules Multi-step composite figure problem View
18. In $\triangle A B C$ , the sides opposite to angles $A$ , $B$ , $C$ are $a$ , $b$ , $c$ respectively, with $b = a + 1$ , $c = a + 2$ .
(1) If $2 \sin C = 3 \sin A$ , find the area of $\triangle A B C$ ;
(2) Does there exist a positive integer $a$ such that $\triangle A B C$ is an obtuse triangle? If it exists, find the value of $a$ ; if not, explain the reason.
Answer: (1) $\frac { 15 \sqrt { 7 } } { 4 }$ ; (2) Yes, and $a = 2$ .
[Solution]
[Analysis] (1) By the law of sines, we can obtain $2 c = 3 a$ . Combined with the known conditions, find the value of $a$ , and further obtain the values of $b$ and $c$ . Use the law of cosines and the fundamental trigonometric identity to find $\sin B$ , then use the area formula of a triangle to obtain the result;
(2) Analyze that angle $C$ is obtuse. From $\cos C < 0$ combined with the triangle inequality, find the value of the integer $a$ . [Detailed Solution] (1) Since $2 \sin C = 3 \sin A$ , by the law of sines we have $2 c = 3 a$ . Thus $2 ( a + 2 ) = 3 a$ , so $a = 4$ . Therefore $b = 5$ , $c = 6$ . By the law of cosines: $\cos C = \frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 2 a b } = \frac { 16 + 25 - 36 } { 2 \times 4 \times 5 } = \frac { 5 } { 40 } = \frac { 1 } { 8 }$ . Since $C$ is acute, $\sin C = \sqrt { 1 - \cos ^ { 2 } C } = \sqrt { 1 - \frac { 1 } { 64 } } = \frac { 3 \sqrt { 7 } } { 8 }$ .
Therefore, $S _ { \triangle A B C } = \frac { 1 } { 2 } a b \sin C = \frac { 1 } { 2 } \times 4 \times 5 \times \frac { 3 \sqrt { 7 } } { 8 } = \frac { 15 \sqrt { 7 } } { 4 }$ ;
(2) Clearly $c > b > a$ . If $\triangle A B C$ is an obtuse triangle, then $C$ is obtuse. By the law of cosines: $\cos C = \frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 2 a b } = \frac { a ^ { 2 } + ( a + 1 ) ^ { 2 } - ( a + 2 ) ^ { 2 } } { 2 a ( a + 1 ) } = \frac { a ^ { 2 } + a ^ 2 + 2a + 1 - a ^ 2 - 4a - 4 } { 2 a ( a + 1 ) } = \frac { a ^ { 2 } - 2 a - 3 } { 2 a ( a + 1 ) } < 0$ , Solving: $- 1 < a < 3$ , thus $0 < a < 3$ . By the triangle inequality: $a + a + 1 > a + 2$ , we get $a > 1$ . Since $a \in \mathbb{Z}$ , we have $a = 2$ .
Q19 Vectors 3D & Lines Multi-Part 3D Geometry Problem View
19. In the quadrangular pyramid $Q - A B C D$ , the base $A B C D$ is a square with $A D = 2$ , $Q D = Q A = \sqrt { 5 }$ , $Q C = 3$ . [Figure]
(1) Prove: plane $Q A D \perp$ plane $A B C D$ ;
(2) Find the cosine of the dihedral angle $B - Q D - A$ . Answer: (1) See proof below; (2) $\frac { 2 } { 3 }$ .
[Solution]
[Analysis] (1) Let $O$ be the midpoint of $A D$ , and connect $Q O$ and $C O$ . We can prove that $Q O \perp$ plane $A B C D$ , thus obtaining plane $Q A D \perp$ plane $A B C D$ .
(2) In plane $A B C D$ , through $O$ draw $O T \parallel C D$ , intersecting $B C$ at $T$ . Then $O T \perp A D$ . Establish a coordinate system as shown in the figure. After finding the normal vectors of planes $Q A D$ and $B Q D$ , we can find the cosine of the dihedral angle.
[Detailed Solution]
[Figure]
(1) Let $O$ be the midpoint of $A D$ , and connect $Q O$ and $C O$ . Since $Q A = Q D$ and $O A = O D$ , we have $Q O \perp A D$ . Since $A D = 2$ and $Q A = \sqrt { 5 }$ , we have $Q O = \sqrt { 5 - 1 } = 2$ . In square $A B C D$ , since $A D = 2$ , we have $D O = 1$ , thus $C O = \sqrt { 1 + 4 } = \sqrt { 5 }$ . Since $Q C = 3$ , we have $Q C ^ { 2 } = 9 = 4 + 5 = Q O ^ { 2 } + O C ^ { 2 }$ , so $\triangle Q O C$ is a right triangle with $Q O \perp O C$ . Since $O C \cap A D = O$ , we have $Q O \perp$ plane $A B C D$ . Since $Q O \subset$ plane $Q A D$ , we have plane $Q A D \perp$ plane $A B C D$ .
(2) In plane $A B C D$ , through $O$ draw $O T \parallel C D$ , intersecting $B C$ at $T$ . Then $O T \perp A D$ . Combined with $Q O \perp$ plane $A B C D$ from part (1), we can establish a coordinate system as shown in the figure. [Figure]
Then $D ( 0,1,0 ) , Q ( 0,0,2 ) , B ( 2 , -1,0 )$ , so $\overrightarrow { B Q } = ( -2,1,2 ) , \overrightarrow { B D } = ( -2,2,0 )$ . Let the normal vector of plane $Q B D$ be $\vec { n } = ( x , y , z )$ . Then $\left\{ \begin{array} { l } \vec { n } \cdot \overrightarrow { B Q } = 0 \\ \vec { n } \cdot \overrightarrow { B D } = 0 \end{array} \right.$ , i.e., $\left\{ \begin{array} { l } - 2 x + y + 2 z = 0 \\ - 2 x + 2 y = 0 \end{array} \right.$ . Taking $x = 1$ , we get $y = 1 , z = \frac { 1 } { 2 }$ , Thus $\vec { n } = \left( 1,1 , \frac { 1 } { 2 } \right)$ . The normal vector of plane $Q A D$ is $\vec { m } = ( 1,0,0 )$ . Therefore $\cos \langle \vec { m } , \vec { n } \rangle = \frac { |\vec{m} \cdot \vec{n}| } { |\vec{m}| \cdot |\vec{n}| } = \frac