20 1 1 Anhui college entrance examination science mathematics reference answer
Multiple choice questions1.a2.c3.a4.b5.c6.c7.d8.B9.c10.b.
Fill in the blanks11.1512.013.60015. ( 1), (3), (5)
answer the question
16. Solution: (1)f' (x)= When a=, make f' (x)=0 to get x= or x=.
When x, f' (x) >; 0; When x, f' (x)
When x, f' (x) >; 0, so f(x) gets the maximum value when x= and the minimum value when x=.
(2)f'(x) is always greater than or equal to zero, or f'(x) is always less than or equal to zero, if it is a monotonic function over,
Because of a>0, δ = (-2a) 2-4a ≤ 0, and the solution is 0.
17. Solution: (1) Take the midpoint of OA and OD respectively, and connect M and N with MC, MB, NF and NE. Then MC∨NF, MB∨NE
So the plane MBC∨NEF plane, so BC∨EF.
(2)S quadrilateral OBED=, h= so VF-OBED=
18. Solution: (1) makes C 1= 1, Cn+2= 100.
Then tn2 = (c1cn+2) (c2cn+1) ... (cn+2c1) =100n+2, so
Tn=, so an=n+2
(2)bn = tan(n+2)tan(n+3)= 1-tan(-n-2)tan(n+3)- 1
= tan(-n-2+n+3)(tan(-n-2)+tan(n+3))- 1 = tan 1(tan(n+3)-tan(n+2))- 1
So sn = b1+B2++bn = tan1((tan4-tan3)+(tan5-tan4)++(tan (n+3)-tan (n+2))-n.
=tan 1 (tan(n+3)-tan3)-n
19. Prove (1) Prove the original inequality, as long as you prove x2y+xy2+ 1≤x+y+x2y2, the following is proved by the difference method:
(x+y+x2 y2)-(x2y+xy2+ 1)=(xy- 1)(x- 1)(y- 1)>0
So the original inequality was proved.
(2)∫logab logbc = logac∴ The original inequality is transformed into
logab+logbc+≤++logac
Let logab=x≥ 1, logbc=y≥ 1, ∴ (1) show that the inequality is established.
20. solution: (1) p = p1+(1-p1) p2+(1-p/-p2).
= p 1+P2+P3-p 1p 2-P2P 3-p3p 1+p 1p2p 3
The probability that the task can be completed remains unchanged.
(2)X= 1，2，3
x
1
2
three
P
q 1
( 1-q 1)q2
( 1-q 1)( 1-q2)
ex = q 1q 2+3-2q 1-Q2 =(2-Q2)( 1-q 1)+ 1
(3) when q 1 >; Q2(q 1q 2+3-2q 1-Q2)-(q 1q 2+3-2q 2-q 1)= Q2-q 1
Start A, then B, and finally C.
2 1. solution: let Q(x, y)B(x0, y0)∴=(x-x0, y-y0)=( 1-x, 1-y).
∫∴x-x0=( 1-x) and y-y0=( 1-y)
∴x0=x-( 1-x) and y0 = y-( 1-y)∫y0 = x02.
∴ y-(1-y) = (x-(1-x)) 2 is the trajectory equation of point Q.
Let P(x, y)Q(x0, y0) then M(x, x2)∴=(0, x2-y0) =(0, y-x2).
∵∴x=x0 and x2-y0=(y-x2)∴x0=x and y0=x2-(y-x2) generation.
y0-( 1-y0)=(x0-( 1-x0))2，y=-2x-
∴p's trajectory equation is y=-2x-