Mathematics (Zhejiang Volume) Reference Answer
1. Multiple-choice question: This big question is a * *12 small question, with 5 points for each small question and 60 points for * * *.
1.D 2。 A 3。 B 4。 C 5。 A six. A seven. C 8。 B 9。 D 10。 D 1 1。 B 12。 D
Fill-in-the-blank question: This big question has 4 small questions, with 4 points for each small question, with a full score of 16.
13. 14. 14 - 25 15.5 16.
3. Solution: There are 6 small questions in this big question, with a full score of 74 points.
17. (The full mark of this question is 12)
Solution: (1)
=
=
=
=
(Ⅱ) ∵
∴ ,
It's also VIII
∴
If and only if b=c= and bc=, the maximum value of bc is.
(18) (in 12)
Solution: (1) According to the meaning of the question, the values of random variables are 2, 3, 4, 6, 7, 10.
The probability distribution list of random variables is as follows
2 3 4 6 7 10
p 0.09 0.24 0. 16 0. 18 0.24 0.09
Mathematical expectation of random variables
=2×0.09+3×0.24+4×0. 16+6×0. 18+7×0.24+ 10×0.09=5.2.
(19) (in 12)
Method one
Solution: (1) remember that the intersection of AC and BD is O, connecting OE,
O and m are the midpoint of AC and EF respectively, and ACEF is a rectangle.
∴ Quadrilateral AOEM is a parallelogram,
∴AM∥OE.
Plane BDE, plane BDE,
∴AM∥ plane BDE.
(ii) AS⊥DF ii)AFD plane, connecting BS,
∵AB⊥AF，AB⊥AD，
∴AB⊥ Aircraft ADF,
∴AS is the projection of BS on the plane ADF,
⊥ Direction finding obtained from three perpendicular theorems.
∴∠BSA is the plane angle of dihedral angle A-DF-B.
In rt Δ ASB,
∴
∴ dihedral angle A-DF-B is 60? .
(iii) Let CP=t(0≤t≤2) and there is PQ⊥AB in Q, then pq∑ad,
∵PQ⊥AB,PQ⊥AF，，
Pq ⊥ aircraft base, aircraft base,
∴PQ⊥QF.
In rt δ pqf, ∠FPQ=60? ,
PF=2PQ。
Σ Δ PAQ is an isosceles right triangle,
∴
Σ Δ PAF is a right triangle,
∴ ,
∴
So t= 1 or t=3 (truncation)
That is, point p is the midpoint of AC.
Method 2
(1) Establish a spatial rectangular coordinate system as shown in the figure.
Set, connect network elements,
Then the coordinates of point N and point E are (,(0,0, 1) respectively.
∴ ,
The coordinates of point A and point M are respectively
( )、(
∴
And NE and AM are not * * * lines,
∴NE∥AM.
There are also: plane BDE, plane BDE,
∴AM∥ aircraft BDF
(ⅱ)∵af⊥ab,ab⊥ad,af
∴AB⊥ Plane ADF ..
∴ is the normal vector of the plane DAF.
∵ ? =0,
∴ ? =0
, ,
∴ is the normal vector of the plane BDF.
∴
The angle between ∴ and is 60 degrees? .
That is, the dihedral angle A-DF-B is 60? .
(iii) Let P(t, t, 0)(0≤t≤)
∴
What is the angle between PF and CD? .
∴
Solve or (give up),
That is, point p is the midpoint of AC.
(20) (in 12)
Solution: (1) Because
So the slope of the tangent is
Therefore, the tangent equation is.
(ii) Let y=0 to get x=t+ 1,
And make x=0.
So S(t)= 1
=
therefore
∫ When (0, 1), > 0,
When (1, +∞), < 0,
So the maximum value of S(t) is S( 1)= 1
(2 1) (in 12)
Solution: (1) Get the equation of straight line AP from conditions.
that is
Because the distance from m point to straight AP is 1,
∵
Namely.
∵
∴
The solution is+1≤m≤3 or-1 ≤ m ≤ 1-.
The value range of ∴m is
(ii) Based on hyperbolic equation.
Yes
And because m is the heart of δAPQ, the distance from m to AP is 1, so ∠MAP=45? The straight line AM is the bisector of ∠PAQ, and the distance from m to AQ and PQ is 1. Therefore, (we set p in the first quadrant).
The linear PQ equation is.
Equation y of straight AP = x-1,
The coordinate of ∴p is (2+, 1+), and it is substituted into the coordinate of point p.
So the hyperbolic equation is
that is
(22) (in 14)
Solution: (1) Because,
So, you can see from the meaning of the question.
∴
=
=
∴ is a series of constants.
∴
(ii) divide both sides of the equation by 2 to obtain
It's also VIII
∴
(Ⅲ)∵
It's also VIII
∴: This is the geometric series of the common ratio.