Mathematics (Literature and History) (Beijing Volume) Reference Answer
First, multiple-choice questions (this big question is ***8 small questions, each with 5 points and ***40 points)
( 1)D (2)A (3)A (4)C
(5)B (6)A (7)C (8)B
2. Fill in the blanks (6 small questions in this big question, 5 points for each small question, 30 points for * * *)
(9) ( 10)|x|x 0，
therefore
The solution is ω= 1.
(2) Derived from (1)
Because 0≤x≤,
So ≤≤
So ≤ ≤ 1.
So 0 ≤≤≤≤that is, the range of f(x) is [0,].
(16)(* 14)
Solution 1:
(I) take point d of AB and connect PD and CD.
AP = BP，
∴PD⊥AB.
AC = BC。
∴CD⊥AB.
∫PD∩CD = d。
∴AB⊥ Flat panel display.
∫PC plane PCD,
∴PC⊥AB.
(ⅱ)∵AC = BC，AP=BP，
∴△APC≌△BPC.
And PC⊥AC,
∴PC⊥BC.
And ∠ ACB = 90, that is, AC⊥BC,
And AC∩PC=C,
∴AB=BP，
∴BE⊥AP.
∫EC is the projection of BE on the plane PAC,
∴CE⊥AP.
∴∠BEC is the plane angle of dihedral angle b-AP-C.
In △BCE, ∠ BCE = 90, BC = 2, Be =,
∴sin∠BEC=
The dihedral angle B-AP-C is an angle.
Solution 2:
(ⅰ)∵AC = BC，AP=BP，
∴△APC≌△BPC.
PC⊥AC again.
∴PC⊥BC.
∫AC∩BC = C，
∴PC⊥ airplane ABC.
∫AB plane ABC,
∴PC⊥AB.
(2) As shown in the figure, establish a spatial rectangular coordinate system C-xyz with C as the origin.
Then c (0 0,0,0), a (0 0,2,0) and b (2 2,0,0).
Let P(0, 0, t),
∵|PB|=|AB|=2，
∴t=2,P(0,0,2).
Take the midpoint e of AP and connect BE and CE.
∵|AC|=|PC|，|AB|=|BP|，
∴CE⊥AP,BE⊥AP.
∴∠BEC is the plane angle of dihedral angle b-AP-C.
∫E(0， 1， 1)，
∴cos∠BEC=
The dihedral angle B-AP-C is the arc angle.
(17)(* 13)
Solution: (i) Because the function g(x)=f(x)-2 is odd function,
Therefore, for any x∈R, g(-x)=-g(x), that is, f(-x)- 2=-f(x)+2.
And f(x)=x3+ax2+3bx+c,
so-x3+ax2-3bx+c-2 =-x3-ax2-3bx-c+2。
therefore
The solution is a=0 and c = 2.
(ii) f (x) from (i) = x3+3bx+2.
So f'(x)=3x2+3b(b≠0).
When b < 0, x =+from f'(x)= 0.
When x changes, f'(x) changes as follows:
x
(-∞,- )
-
(- , )
( ,+∞)
f′(x)
+
-
+
Therefore, when b < 0, the function f (x) monotonically increases at (-∞,-), monotonically decreases at (-,) and monotonically increases at (,+∞).
F ′ (x) > 0 when b > 0. Therefore, the function f (x) monotonically increases in (-∞,+∞).
(18)(* * 13)
Solution:
(1) Please remember that both Party A and Party B participated in the service of Station A at the same time as the EA incident, then
p(EA)= 1
That is, the probability that both Party A and Party B participate in the latter service at the same time is
(2) Remember that both Party A and Party B participate in the same post service at the same time as the E event, then
P(E)
Therefore, the probability that Party A and Party B do not work in the same post is
P()= 1-P(E)= 0
(19)(* * 14)
Solution: (i) Because both AB∨l and AB edge pass through point (0,0), the equation of the straight line where AB is located is y = x. 。
Let the coordinates of A and B be (x 1, y 1) and (x2, y2) respectively.
allow
therefore
And because the height h of the AB side is equal to the distance from the origin to the straight line L,
therefore
(Ⅱ) Let the equation of the straight line where AB is located be y = x+m. 。
allow
Because a and b are on the ellipse,
therefore
Let the coordinates of A and B be (x 1, y 1) and (x2, y2) respectively.
rule
therefore
And because the length of BC is equal to the distance from the point (0, m) to the straight line L, that is
therefore
So when m=- 1, the AC side is the longest.
At this time, the equation of the straight line where AB is located is y=x- 1.
(20)(* * 13)
Solution: (i) Because a 1= 1,
Therefore, when a2=- 1, we get,
therefore
therefore
(ii) The sequence {an} cannot be arithmetic progression. The evidence is as follows:
From a 1= 1, we get
If it exists, let {{an}} be arithmetic progression, then a3-a2=a2-a 1, that is.
The solution is =3.
therefore
This contradicts that {an} is arithmetic progression, so {an} can't be arithmetic progression for anyone.
(iii) According to the meaning of the question, b 1 2 and N*), there is always N* at this time, and BN > 0； when n≥n0 is satisfied; When n≤n0- 1, BN < 0.
So it can be seen from an+ 1=bnan and a 1 = 1 > 0 that if n0 is even, then when n > n0.
When an < 0; If n0 is odd, then when n > n0, an > 0.
So "there are m N*, when n > m, there is always a < 0" if and only if no is even,
Remember no = 2k (k = 1, 2, ...), and that's enough.
Therefore, the range of values is 4k2+2k(k N*).