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1、【名师简评】（teacher.）teacher.The whole paper is compared with previous years, relatively stable, no tricky question and strange questions, based on the basic knowledge of the examination, also examined the students ability to solve the problem of flexible use of knowledge. There are not many Chinese char
2、acters in the questions, but they are more concise. But there are also some innovative questions, like the multiple-choice questions twelfth questions, 16 questions, answer questions twenty-second questions, in addition to other questions remain in style, the question is simple, good start, but it i
3、s not so easy. On the whole, the examination questions are graded from easy to difficult, and most of the questions are suitable for students to answer, reflecting the double base, examining the use of the four major ideas of the students, it is a good test paper.First, multiple-choice questions1, c
4、omplex =A 2+I B 2-I C 1+2i D 1- 2I2, the known set A = 1.3., B = 1, m, A B = A, then m=A 0 or B 0 or 3 C 1 or D 1 or 33 the center of the ellipse at the origin, a focal length of 4 line x=-4, the elliptic equationA + =1 B + =1C + =1 D + =13.Cproposition this question mainly examines the elliptic equ
5、ation and the use of properties. The focus position is determined by the alignment equation, then with the help of the focal length and alignment parameters a, B, C, and elliptic equation.resolution because4 in the known four prism ABCD- A1B1C1D1, AB=2, CC1= E is the midpoint of CC1, then the distan
6、ce between line AC1 and plane BED isA 2 B C D 1(5) the known arithmetic sequence an before the N and Sn, a5=5, S5=15, is the first 100 series and(A) (B) (C) (D)(6) ABC, AB and high CD, b=0, |a|=1, if a? |b|=2, then(A) (B) (C) (D)6 Dproposition this question mainly examines the use of geometric meani
7、ng of vector addition and subtraction, combined with the use of special right triangle to solve the position of point D.resolution because(7) known alpha is second quadrant angle, sin alpha + sin beta =, then Cos2 =(A) (B) (C) (D)(8) F1 F2, known as the left and right focus hyperbola C:x2-y2=2, P on
8、 C, |PF1|=|2PF2|, cos / F1PF2=(A) (B) (C) (D)(9) known x=ln PI, y=log52, then.(A) x y Z (B) Z x y (C) Z y x (D) y Z 0) have a common point, and at A, the tangent of the two curve is the same line L.(I) seeking r;(II) let m and n be the two straight lines which are different from L and tangent to C a
9、nd M. The intersection point of M and N is D, and the distance from D to L is obtained.21 proposition intention this question examines the parabola and the circle equation, as well as the two curve common point tangent application, and on this foundation solves the point to the straight line the men
10、ts the examination questions differently than usual, because involves two intersection of the two curves, and tangent to two curves at the study points out that the tools of analytic geometry and the derivative of the combination is the innovation of the test. In addition, in the second questions is
11、 more difficult, there are two other public tangent, such a problem for our future learning is also a need to practice direction.(the only 22 out of 12) (Note: in the papers on the answer is invalid)The function f (x) =x2-2x-3 defines the sequence xn as follows: x1=2, xn+1 is the abscissa of the intersection of the line PQn and the X axis of the two points P (4,5), Qn (xn, f (xn).(I) proved: 2 xn xn+1 3;(II) finding the general term formula of sequence xn.