Senior one math winter vacation homework 1 reference answer:
I.1~ 5cabcb6 ~10cbbcc1~12bb
Second, 13,
14( 1); (2){ 1，2，3 } N； (3){ 1}; (4)0; 15- 1 16 or; ;
Or ...
Three. 17.{0.- 1, 1}; 18.； 19.( 1)a2-4b=0(2)a=-4，b=320 ..
Senior one math winter vacation homework 2 reference answer:
I.1~ 5cdbd6 ~10cccca1~12bb
Two. 13.( 1, +∞) 14. 13 15 16,
Three. 17. 18 is omitted, which can be proved by definition. The maximum value of f(x) is: and the minimum value is:
19. Solution: (1) Set any sum.
That is, adding functions in the world.
⑵
20. Solution: Even function in the world, monotonically decreasing in the world.
Add functions again.
,
allow
The solution set is.
Senior one mathematics winter vacation homework 3 reference answer.
First, multiple-choice questions:
1.B2 . C3 . C4 . a5 . C6 . a7 . A8 . d9 . a 10。 B 1 1。 B 12。 C
Second, fill in the blanks:
13. 14. 12 15.； 16.4-a，
Third, answer questions:
17. Omit
18. Omit
19. Solution: (1) The opening is downward; The symmetry axis is; Vertex coordinates are;
(2) The maximum value of the function is1; There is no minimum value;
(3) Functions are increasing in the world and decreasing in the world.
20.Ⅰ、Ⅱ、
Senior one mathematics winter vacation homework 4 reference answer.
I.1~ 8 cbcdaacc9-12 bbcd
2. 13, where the upper part is a decreasing function and the interval is increasing function. ∴ when, that is, x=log3.
When x=0
Senior one math winter vacation homework 6 answer:
First, multiple-choice questions:
1.D2 . C3 . D4 . C5 . a6 . C7 . D8 . a9 . c 10。 A 1 1。 D 1。 B
Second, fill in the blanks
13.(-2,8),(4, 1) 14.[- 1, 1] 15.(0,2/3)∪( 1,+∞) 16.[0.5, 1)
17. Omit 18. leave out
19. Solution: Even function in the world, increasing function in the world.
and
,
allow
The solution set is.
20.( 1) or (2) at that time, which may be:. Solve them separately to obtain;
Senior one mathematics winter vacation homework 7 reference answer.
First, multiple-choice questions:
1.B2 B3 . D4 . D5 . B6 . a7 . b8 . a9 . d 10。 B 1 1。 D 12。 D
Second, fill in the blanks
13. 14
15. 16
Third, the answer: 17.
18 solution: (1)
(2)
19.–2 tanα
20T=2×8= 16=，=，A=
Let the abscissa of the point near the origin in the intersection of the curve and the X axis be, then 2-=6-2 means =-2.
∴=–=,y=sin()
When =2kл+, that is, x= 16k+2, the maximum value of y =
When =2kл+, that is, x= 16k+ 10, y is minimum =-
As can be seen from the figure, the rising range is [16k-6, 16k+2] and the falling range is [16k+2, 16k+00] (k ∈ z).