Reference answer of science mathematics

First, multiple-choice questions:

Volume I: DCCC

Volume B: ABCDC CAADB database

Second, fill in the blanks:

( 13)56. 19 ( 14)48 ( 15)√2- 1 ( 16)√ 1 1

Third, answer questions:

(17) solution:

∴s 1+2s 2+…+NSN = 2[(n- 1)2n+ 1+2]-n(n+ 1)=(n- 1)[2(。

(19) solution:

(I)∫SD⊥ plane ABCD, ∴ plane SAD⊥ plane ABCD,

∵AB⊥AD, ∴AB⊥ The plane is sad, ∴ De ⊥ ab.

Sd = ad, e is the midpoint of SA, ∴DE⊥SA,

∫ab∩sa = a, ∴DE⊥ plane SAB

∴ Aircraft Company BED⊥ Aircraft Company ... 4 points

(2) Establish the coordinate system D-XYZ as shown in the figure, and let AD = 2, then

D(0，0，0)，A(2，0，0)，B(2，2，0)，

C(0，2，0)，S(0，0，2)，E( 1，0， 1)。

DB→=(2，2，0)，DE→=( 1，0， 1)，CB→=(2，0，0)，CS→=(0，-2，2)。

Let m = (x 1, y 1, z 1) be the normal vector of the surface layer, then

m？ DB→=0，m？ De → = 0, that is, 2x 1+2y 1 = 0, x 1+z 1 = 0,

Therefore, m = (- 1, 2, 1)...8 points can be taken.

Let n = (x2, y2, z2) be the normal vector of SBC, then

n？ CB→=0，n？ CS→ = 0, namely 2x2 = 0, -2Y2+2Z2 = 0,

Therefore, n = (0 0,2, 1)... 10.

Because? m，n？ =m？ n|m||n|=323=32，

Therefore, the dihedral angle formed by the plane bed and the plane SBC is 30? ... 12 point

(20) Solution:

(I)F(p ^ 2，0)，C(-p ^ 2，0)，

Let A(x? 1, y 1), B(x2, y2), and the linear l equation is x = my+p 2.

Substituting into the parabolic equation y2 = 2px, we get y2-2pmy-p2 = 0.

Y 1+Y2 = 2pm, Y 1Y2 =-P2...4 points.

Let y 1>0 > 0 and y2 < 0, then

tan∠ACF = y 1x 1+p2 = y 1y 2 12p+p2 = 2py 1y 2 1+p2 = 2py 1y 2 1-y 1 y2 = 2py 1-y2，

tan∠BCF =-y2 x2+p2 =-2py 2-y 1，

∴ tan ∠ ACF = tan ∠ BCF, so ∠ ACF = ∠ BCF...8 points.

(ii) if y 1 > 0, tan ∠ ACF = 2py 1+p2 ≤ 2py12py1=1if and only if y/kloc-

At this time, ∠ACF takes the maximum π 4, ∠ ACB = 2 ∠ ACF takes the maximum π 2,

And A( p 2, p), B( p 2, -p), | ab | = 2p... 12.

(2 1) solution:

(ⅰ)f(x)=-lnx-ax2+x，

f？ (x) =-1x-2ax+1=-2ax2-x+1x ... 2 points.

Let δ = 1-8a.

When a≥ 1 8 and δ≤ 0, f? (x)≤0, and f(x) monotonically decreases at (0, +∞) ... 4 points.

When 0 < a < 18, δ > 0, the equation 2ax2-x+ 1 = 0 has two unequal positive roots x 1, x2,

Let's say x 1 < x2,

So when x∈(0, x 1)∩(x2, +∞), f? (x) < 0, when x∈(x 1, x2), f? (x)>0，

F(x) is not a monotonic function at this time.

To sum up, the range of a is [1 8, +∞)...6 points.

(ii) According to (i), if and only if a ∈ (0, 1 8), f(x) has a minimum point x 1 and a maximum point x2.

And x 1+x2 = 12a, x 1x2 = 12a.

f(x 1)+f(x2)=-lnx 1-ax 2 1+x 1-lnx 2-ax22+x2

=-(lnx 1+ln x2)- 1 2(x 1- 1)- 1 2(x2- 1)+(x 1+x2)

=-ln (x1x 2)+12 (x1+x2)+1= ln (2a)+14a+1... 9 points.

Let g (a) = ln (2a)+14a+1,a ∈ (0, 1 8),

So when a ∈ (0, 1 8), g? (a) =1a-14a2 = 4a-14a2 < 0, and g(a) monotonically decreases at (0, 18).

So g (a) > g (18) = 3-2LN2, that is, f (x1)+f (x2) > 3-2ln2 ...12 points.

(22) Proof:

(I) ∵ FG, the circle O is tangent to the G point, ∴ FG2 = FD? Law,

EF = FG，EF2=FD？ FA,∴EFFD=FAEF，

∫∠EFD =∠AFE, ∴△ EFD ∽△ AFE...5 points.

(ii) From (i), there is ∠ Federal Reserve = ∠ FAE,

∠BCDFAE and∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠,

∴ef∨BC... 10 point

(23) Solution:

(1) The general equation of curve C in rectangular coordinate system is X2 16+Y24 = 1,

The polar coordinate equation is ρ 2cos2θ16+ρ 2sin2θ 4 =1,

Substitute θ = π 4 and θ =-π 4 to get | OA | 2 = | OB | 2 = 325.

Because ∠ AOB = π 2, the area of △AOB is S =12 | OA ||| OB | =165 ... 5 minutes.

(ii) Substituting the parametric equation of L into the ordinary equation of curve C to obtain (t-22) 2 = 0,

∴ t = 22, substituted into the parameter equation l, x = 22, y = 2,

So the coordinate of the intersection of curve C and straight line L is (22,2) 2)... 10/0.

(24) Solution:

(ⅰ)f(x)= | x+ 1 |+| x- 1 | =-2x，x 1。

When x

F (x) = 2 < 4 when-1 ≤ x ≤ 1;

When x > 1, from 2x < 4, we get 1 < x < 2.

So m = (-2,2) ... 5 points.

(ii) when a, b∈M is -2 < a, b < 2,

∫4(a+b)2-(4+ab)2 = 4(a2+2ab+B2)-( 16+8ab+a2 B2)=(a2-4)(4-B2)< 0，

∴4(a+b)2