1.BDCAA？ BBABC？ Civil band
Second, 13. ( 1)(2) 14. 15.5 16.(2, 19/8)
Third, answer questions.
17.( 1) The scatter diagram is as follows.
What is the linear regression equation of ∴? ............................, eight.
(3) For the question: y= 1 15, 6.5x+17.5 =115, x =15 ...
18.( 1)f(x)？ =
=sin(2x+π/6)+3/2，
∴ f (the smallest positive period ∴f(x) is π...6 points.
(2) The maximum value of ∵ y = sin (2x+π/6) is 1, and the minimum value is-1.
∴f(x)==sin(2x+π/6)+3/2 The maximum value is 5/2, and the minimum value is 1/2.
19.( 1) The frequencies with a score in the range of [120, 130] are
1-(0.01+0.015+0.015+0.025+0.005) ×10 =1-0.7 = 0.3.
The frequencies with scores between [135, 150] are,
The frequency is 0. 175? ×? 60= 10.5≈ 1 1,
Therefore, the number of students with excellent grades is 1 1. ..........., eight.
The average score is 95× 0.1+105× 0.15× 0.15+125× 0.3+135× 0.25.
So it is estimated that the average score of mathematics in the whole grade of this exam is 12 1. ............ is 12.
20.2 1. certificate: (1) link a? C, straight triangular prism a? b？ c？ -in -ABC, quadrilateral AA? c？ C is a rectangle,
∴ Point F is in A? C, and f is a? The midpoint of C.
At △A? In BC, ∫E and f are a respectively? b，A？ The midpoint of c, so ef∨BC.
∵BC and plane ABC, EF? Aircraft ABC? EF∨ Plane ABC...........4 points
(II) in triangular prism a? b？ c？ -Yes. -ABC, b? B⊥ Airplane ABC, ∴? b？ B⊥EF？
If (i) knows ef∑BC and AB⊥BC, then EF⊥AB,
∵? b？ B∩AB=B,∴？ EF⊥ Aircraft ABB? One,
∵EF and AEF, ∴ AEF⊥ ABB? A..........？ 8 points
㈢？ ..... 12 point
22. solution: (1)│CR│=2, and the equation of the circle is x? +(y-3)？ =4..........3 points.
(2) Method 1: ① When k does not exist,
X=- 1, then P (- 1, 3-√ 3), Q (- 1, 3+√ 3), M (- 1, 3),
Obviously there is AC am = 9,
② when k exists, let y=k(x+ 1),
The equation of ∴l is y=kx+k,
P(x？ ，y？ )，Q(x？ ，y？ )，M(x？ ，y？ ),
∴,
Do you have an X? + 1+3y？ =9, that is, simultaneous y=kx+k, x? +(y-3)？ =4,
Substitute into the equation: x? + 1+3y？ =9, so:
Solution: k=4/3,
To sum up, the equation of L is x=- 1 or 4x-3y+4=0. .............. 12 point
Method 2: ∫M is the midpoint of PQ line,
According to the vertical diameter theorem, that is, CM⊥PQ, that is, am MC = 0,
(1) if k exists, let the straight line l be y-0=k(x+ 1), that is, y=kx+k,?
Distance from the center of the circle c (0 0,3) to the straight line l
The solution is k=4/3,
The equation of the straight line is 4x-3y+4=0,
② If k does not exist, the straight line through a (- 1, 0) is x=- 1,
It is also satisfied that the distance from c (0 0,3) to the straight line x=- 1 is 1.
To sum up, the equation of a straight line is 4x-3y+4=0 or x=- 1.
Method 3: A (- 1, 0), C (0 0,3), set point M(x? ，y？ ), then:
From the meaning of the question: AC am = x? + 1+3y？ =9，x？ = 8-3 years old ①
Because m is the midpoint of the chord PQ, AM⊥CM,
AM CM=x？ (x？ + 1)+y？ (y？ -3)=0, and substitute it into Formula ① to get:
(8-3y？ )(9-3y？ )+y？ (y？ -3)=0, sorting:
(y？ -3)( 10y？ -24)=0, and the solution is: y? =3 or y? = 12/5
The coordinate of m is (-1, 3) or (4/5, 15/5), so the equation of straight line L is 4x-3y+4=0 or x=- 1.