Seek the final exam of eighth grade mathematics in Chongqing Qiujing Middle School.

Mathematics Test of Senior Two in Chongqing Qiujing Middle School in 2008-2009 (18)

1 ... It is known that (2x2+4x+3) 6 = A0+a1(x+1) 2+A2 (x+1) 4+...+A6 (x+1).

2. If AB

A.(-∞，- B .)。

9. In a dice-throwing experiment, event A means "even points less than 5 appear" and event B means "points less than 4 appear", so the probability that event A+B- appears in an experiment is about13b.12c.23d.56..

10. As shown in the figure, in the regular triangle ABC, D, E and F are the midpoints of each side, and G, H and I are the midpoints of de, FC and EF respectively. After ABC is folded into a triangular pyramid along DE, EF and FD, the radian number of the angle formed by BG and IH is A.π 6b.π 3c.arccos23d

1 1, e and f are the midpoint of side AB and c 1d 1 d1respectively, a/.

In 12 and cube ABCD-a1b1c1d1,the point p moves on the edge of BCC 1B 1 and its boundary. While keeping AP⊥BD 1, the trajectory of moving point p is a, line segment B 1cb, and part of parabola passes through B 1 and c, line segments d, BC and B 1C 1.

13. In the expansion of (x- )4(2x- 1)3, the coefficient of the x2 term is. -68

14. If n∈N is odd, then 6n+cn16n-1+…+CNN-16-1is divided by 8, and the remainder is. five

15. 1, 2, 3, 4, 5, 6, 7, only two even numbers are adjacent. 2880

1 6.f (x-1) = x+x2+X3+…+xn (x ≠ 0,1), where the coefficient of x is Sn and the coefficient of x3 is Tn, =-

17. It is known that the fifth term of the expansion of (x-)6 is equal to, then (x- 1+x-2+…+x-n)=. 1

18. Tetrahedral ABCD has the following propositions: ① If AC⊥BD, AB⊥CD, then AD ⊥ BC; (2) If E, F and G are the midpoint of BC, AB and CD respectively, then the size of ∠EFG is equal to the size of the angle formed by the straight line AC and BD; (3) If point O is the center of the circumscribed sphere of tetrahedron ABCD, then the projection of O on the curved surface ABD is the outer center of △ABD; ④ If all four faces are congruent triangles and ABCD is a regular tetrahedron, The correct number of propositions is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

19. in the regular hexagonal cone P-ABCDEF, AB=2 and PF=.

Verification: (1) pf ⊥ BD; (2)PF⊥ plane pbd；;

(3) Find the cosine of dihedral angle f-pa-b. 。

20. Both parties flip a coin. Party A flipped a coin for three times, and the number of times when the card face was up was ζ; B toss a coin twice, and the number of times the coin faces up is η. (1), and the mathematical expectations of random variables ζ and η are obtained respectively; (2) if zeta > is specified; When ηA wins. The probability of winning.

2 1. Let an =1+q+Q2+…+qn-1(n ∈ n, q ≠ 1), an = cn1a/kloc-.

(1) When -3, find An (expressed by n and q) (2)

22 .. let fn (x) = f {[f ... f (x)] ...} (nf), (1) to find F2 (x) and F3 (x); (2) Guess fn(x) and prove your conclusion.

19.( 1) Proof: connect FC, pay BD to G, take FC midpoint O and connect PO.

∵ regular hexagonal pyramid p-abcdef, ∴PO is the height of the pyramid, FC⊥BD,

∴PO⊥BD, ∴ flat first class, ∴PF⊥BD.

(2) solution: ∵ABCDEF is a regular hexagon, AB=2,

∴FO=2,FG=3,OG= 1，

Even PG, in the right triangle PFO, PF=, FO=2,

∴PO=. In the right triangle PGO, PO=, OG= 1, ∴PG=

In triangular PGF, PF=, FG=3, pg =；;

∴△ FG2 = PG2+PF2 of PFG is a right triangle.

∴PF⊥PG and PF⊥BD, ∴PF⊥ aircraft PBD.

(3) Point F is FH⊥PA in H, connecting BH and BF.

∴△PFA≌△PBA, ∴BH⊥PA and ∴∠FHB are the plane angles of dihedral angle F-PA-B. 。

Take the point S of FA, and in △PSF, PF=, FS= 1, ∴PS=

∫In△PFA，FH =

At △BFH,

The cosine of dihedral angle f-pa-b is.

20. Solution (1) (Reason) According to the meaning of the question: This test is an independent repeated test, so the random variable conforms to the binomial distribution.

Expectation formula of binomial distribution

=2×0.5= 1.

(Note: Distribution lists can also be listed as definitions. )

(2) There are three situations in which A wins:

① Party A's front face 1 times, and Party B's front face is 0 times;

(2) Party A's frontal increase is twice, and Party B's frontal increase is 0 or 1 time;

③ Party A's frontal increase is 3 times, while Party B's frontal increase is 0 times, 1 times or 2 times.

To sum up, the probability of winning is:

2 1. [Solution] (1)∵an=

∴an=[cn 1( 1-q)+cn2( 1-Q2)+…+CNN( 1-qn)]

=[cn 1+Cn2+…+Cnn-(cn 1q+Cn2q+…+cn 1qn)]

=[(2n- 1)-( 1+q)n+ 1]=[2n-( 1+q)n]