gaokao

2015 national-II-arts

16 maths questions

Q1 Probability Definitions Set Operations Using Inequality-Defined Sets View
1. Given sets $A = \{ x \mid - 1 < x < 2 \} , B = \{ x \mid 0 < x < 3 \}$, then $A \cup B =$
A. $( - 1,3 )$
B. $( - 1,0 )$
C. $( 0,2 )$
D. $( 2,3 )$
2. If $a$ is a real number and $\frac { 2 + a i } { 1 + i } = 3 + i$, then $a =$
A. $- 4$
B. $- 3$
C. $3$
D. $4$
Q3 Data representation View
3. Based on the bar chart of China's annual carbon dioxide emissions (in units of 10,000 tons) from 2004 to 2013 shown below, which of the following conclusions is incorrect? [Figure]
A. Year-on-year comparison shows that 2008 had the most significant effect in reducing carbon dioxide emissions
B. China's efforts to control carbon dioxide emissions showed results in 2007
C. Since 2006, China's annual carbon dioxide emissions have shown a decreasing trend
D. Since 2006, China's annual carbon dioxide emissions are positively correlated with the year
Q4 Vectors Introduction & 2D Dot Product Computation View
4. Given $\vec { a } = ( 0 , - 1 ) , \vec { b } = ( - 1,2 )$, then $( 2 \vec { a } + \vec { b } ) \cdot \vec { a } =$
A. $- 1$
B. $0$
C. $1$
D. $2$
Q5 Arithmetic Sequences and Series Compute Partial Sum of an Arithmetic Sequence View
5. Let $S _ { n }$ be the sum of the first $n$ terms of an arithmetic sequence $\left\{ a _ { n } \right\}$. If $a _ { 1 } + a _ { 3 } + a _ { 5 } = 3$, then $S _ { 5 } =$ [Figure] [Figure]
A. $5$
B. $7$
C. $9$
D. $11$
7. Given three points $A ( 1,0 ) , B ( 0 , \sqrt { 3 } ) , C ( 2 , \sqrt { 3 } )$, the distance from the circumcenter of $\triangle A B C$ [Figure]
to the origin is
A. $\frac { 5 } { 3 }$
B. $\frac { \sqrt { 21 } } { 3 }$
C. $\frac { 2 \sqrt { 5 } } { 3 }$
D. $\frac { 4 } { 3 }$
Q9 Geometric Sequences and Series Finite Geometric Sum and Term Relationships View
9. Given a geometric sequence $\left\{ a _ { n } \right\}$ satisfying $a _ { 1 } = \frac { 1 } { 4 } , a _ { 3 } a _ { 5 } = 4 \left( a _ { 4 } - 1 \right)$, then $a _ { 2 } =$
A. $2$
B. $1$
C. $\frac { 1 } { 2 }$
D. $\frac { 1 } { 8 }$
11. As shown in the figure, the rectangle's edge, is the midpoint, point moves along the edge, and moves with, denote, express the sum of distances from the moving point to two points as a function of, then the graph of is approximately
[Figure]
A.
[Figure]
B.
[Figure]
C.
[Figure]
D.
12. Let $f ( x ) = \ln ( 1 + | x | ) - \frac { 1 } { 1 + x ^ { 2 } }$. The range of $x$ for which $f ( x ) > f ( 2 x - 1 )$ holds is
A. $\left( \frac { 1 } { 3 } , 1 \right)$
B. $\left( - \infty , \frac { 1 } { 3 } \right) \cup ( 1 , + \infty )$
C. $\left( - \frac { 1 } { 3 } , \frac { 1 } { 3 } \right)$
D. $\left( - \infty , - \frac { 1 } { 3 } \right) \cup \left( \frac { 1 } { 3 } , + \infty \right)$
II. Fill-in-the-Blank Questions: This section contains 4 questions, 5 points each, 20 points total
Q13 Exponential Functions Parameter Determination from Conditions View
13. Given that the graph of function $f ( x ) = a x ^ { 3 } - 2 x$ passes through the point $( - 1,4 )$, then $a = $ $\_\_\_\_$ .
14. If $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { c } x + y - 5 \leq 0 \\ 2 x - y - 1 \geq 0 \\ x - 2 y + 1 \leq 0 \end{array} \right.$, then the maximum value of $z = 2 x + y$ is $\_\_\_\_$ .
15. A hyperbola passes through the point $( 4 , \sqrt { 3 } )$ and has asymptote equations $y = \pm \frac { 1 } { 2 } x$. The standard equation of this hyperbola is $\_\_\_\_$ .
Q16 Tangents, normals and gradients Determine parameters from function or curve conditions View
16. The tangent line to the curve $y = x + \ln x$ at the point $( 1,1 )$ is tangent to the curve $y = a x ^ { 2 } + ( a + 2 ) x + 1$. Then $a = $ $\_\_\_\_$ .
III. Solution Questions
17 (This question is worth 12 points). In $\triangle A B C$, $D$ is a point on $BC$, $AD$ bisects $\angle B A C$, and $B D = 2 D C$.
(1) Find $\frac { \sin \angle B } { \sin \angle C }$; (II) If $\angle B A C = 60 ^ { \circ }$, find $\angle B$.
Q18 Sine and Cosine Rules Compute area of a triangle or related figure View
18. (This question is worth 12 points). A company randomly surveyed 40 users from regions $A$ and $B$ respectively to understand user satisfaction with its products. Based on user satisfaction scores, a frequency distribution histogram of user satisfaction scores in region $A$ and a frequency distribution table of user satisfaction scores in region $B$ were obtained.
Frequency Distribution Histogram of User Satisfaction Scores in Region A [Figure]
Satisfaction Score Interval\multicolumn{5}{|c|}{Frequency Distribution Table of User Satisfaction Scores in Region B}
Frequency$[ 50,60 )$$[ 60,70 )$$[ 70,80 )$$[ 80,90 )$$[ 90,100 ]$
2814106

(I) Draw the frequency distribution histogram of user satisfaction scores in region $B$ on the answer sheet, and compare the mean and dispersion of satisfaction scores in the two regions through this histogram (no need to calculate specific values, just provide conclusions); [Figure] (II) Based on user satisfaction scores, user satisfaction levels are divided into three categories:
Satisfaction ScoreBelow 7070 to 89At least 90
Satisfaction LevelDissatisfiedSatisfiedVery Satisfied

Estimate which region has a higher probability that a user's satisfaction level is dissatisfied, and explain the reasoning.
Q19 Vectors 3D & Lines Multi-Part 3D Geometry Problem View
19. (This question is worth 12 points). In the rectangular prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$, $A B = 16 , B C = 10 , A A _ { 1 } = 8$. Points $E$ and $F$ are on $A _ { 1 } B _ { 1 }$ and $D _ { 1 } C _ { 1 }$ respectively, with $A _ { 1 } E = D _ { 1 } F = 4$. A plane $\alpha$ passes through points $E$ and $F$ and intersects the faces of the rectangular prism, with the intersection lines forming a square. [Figure] (I) Draw this square in the figure (no need to explain the method or reasoning); (II) Find the ratio of the volumes of the two parts that plane $\alpha$ divides the rectangular prism into.
20. (This question is worth 12 points). The ellipse $C : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$ has eccentricity $\frac { \sqrt { 2 } } { 2 }$, and the point $( 2 , \sqrt { 2 } )$ lies