2016年模擬考試答案(理科數(shù)學(xué))發(fā)排稿

n

n

nn

1

n

n

2016

2-2

9

（I）

a

n2nan-,......................................................................................................

2a

n-1an=na

n_2

n1

n，1a-na

nn-1

(n)

a=2a=2

a=2n,

1

n(n'1)

11.................................................................................

nn1

b

n

2n(2n2)

2

*

*

1

1

,

T

n

*b?|l(

b

n

+

ni-

n

1

11

n1

1n

n1n1

01

n1

1....................................................................................................

1

fnN

n+1

1

1

n+1

N

1

n=1

T

n

2

：：：

1............................................................................................................

I)

M

AB=ACDBCBC_AD

NABAC

MNBC........................................................................1

MN_AD

AA

ABCMNABCAA^MN.....................................................

AD,AA

ADDA

AD

AA

MN

ADD

1

(n)

AP

A

AE_AP

E

E

EF_AMFAF

(I)

MN

AEA

1

AE

AM

AEAAMN

AMNAM_AE

AEFAM_AF

AFE

A-AM-N

1

AA

1

AB=AC=2AABAC=120”BAD=60[AB=2,AD=11

PADVMAB

AP=—AM=1

2'

11

1

1

1

1

=11,7

222

/5

RtLAAFAPRtLAAMAM

2

2

—

5

AAi

A

AEBC

A

A|DAA

x

y

z

O-xyz0

A

A0,0,0A0,0,1

PAD

M,NAB,AC

M

I22

3

1

AA

=（

0,0,1

）

,

NM

=

（

V3,0,0

）

.

=0,

r

n

AAM

—*

r—AM，n1?AM=0,

nAA,n

=0,

Z,

（

=0.

X,0,0,1）

1

0,

x^1

=0.

m3,0AA|M

AMN

n^'x,y,z

n

22

2

2

2

2

3

n2MM",

2?NM

=0,

r

"1

i（X,V,Z,-,1

222^—

I22

X,y,Z.3,0,0=0.222*

=0,

31

Xy

22

Z=0,

y

>/3x=0.

nh[0,2,

-1A|MN

A-AM-N

.15

5

A-AM-N

1,-3,0?0,2,-1

101

R=P（40：：：X:::80）

505

35751

P

2

P（80X120）P=P（X120）

50105010

....................................................3

P=c0（1-P）4c（1-P）3P

4

A4

10

1010

9477

10000

0.9477...................................................................................................................

（n）

Y

Y=5000

=0,

EY500015000

40：：：X::80Y=5000-800=4200

23

2

3

P(Y=4200)=P(40vXc80)=R=0.2

X_80

Y=50002=10000

P(Y=10000)=P(X_80)=FP=0.8Y

Y

P

EY=42000.2100000.8=8840

?.....................................................................................................9

40

：：：

X:

：

:80

Y=5000-80023400

p(Y=3400)=p(40cXc80)=R=0.2

80^X^120Y=50002-800=9200P(Y=9200)=p(80X120)=p=0.7

X120y=50003=15000

P(Y=15000)=P(X120)=p=0.1

Y

Y

P

EY=34000.292000.7150000.1=8620

)

(I)

e

2

a

2

-b

2

=4b2

3

a2

4

22

1,x2

4b2b2

■4y2

=4b2

M(x,y)

UMQ|=J(x_0)

2+(y_3)2=J4b2_4y2+(y_3)2

y-_1|MQ|

，4b2?12=4

=-3y2-6y4b29=-3(y1)24b212

12

2

2

y-_1|MQ|

，4b2?12=4

b

2

=1a2

=4

2

Gy2

4

=1.............................................................................................

(n)

C

y=x2

N(t,t2)yJ2x

BC

22

y-t=2t(x-t),y=：2tx-t

①....................................

2

4

y2

-1

(116t2)x2

-16t

3x4t4

-4

=0...............................................................................................................

322442

16t

3

4t4

-4

116t

2

x?x

mx

1-16t

2"

-(16t)-4(116t)(4t-4)=16(-t16t1)

|BC|14t

2|x|14t2

2

x)2-4

2

2

4.14t2

—t4

16t2

1

-1+16t2

ABCd

1+16t2

d

—

2

16,14t2

114j1+4t2J-4+16t2+11+16t2

=ABCS|BC|d

2121+16t2

16j1+4t2

J—t4

16t2

1-(t2

-8)2

65

65

88'8

^ABC

8

f(x)=e

x-axf'(x)ex-a

y-_1|MQ|

f(0)=1-a1a=2

，

4b2?12=4

f(x)=e

x

-2xf'(x)=ex

-2

0

0

0

0

0

f'(x)=0

xdn2

x=In2

f'(x)<0,f(x)

xIn2

f'(x).0,f(x)

x=ln2

f(x)f(In2)=e

ln2-2In2=2-In4,f(x)

.....................................4

(n)

g(x)ex_x2

g'(x)e

x_2x

g'(x)f(x)_f(In2).0g(x)

R

x0g(x)g(0)=10x2:ex?.........................................................................................6c_1ex<cex

(n)

x=0

x0x2:::ex

x2

：：：

x0x2

：：：

cex

cex

1

c

1

x2::cex

ex-kx2

exkx2

x-In(kx2)

k1

x2Inx?Ink

c

x2

2x—2h(x)=x-2Inx-lnkh'(x)=1xx

h'(x)0,h(x)(2,：：)

x=16k16h(x)(x

)

x7k

h(x)=16k_2In(16k)-Ink=8(k-In2)3(k-Ink)5kkInk,kIn2,5k0

::)

h(x)0x(x^

c

cx

cex

x2cex

(x,

x2

...................................

：：：

f'(x)=0

x=In2

(n)

x0

ex

x

xx

ex=e2Le2

o

0

xX

22c2c

222

XxIx4x12

ex

cx

0

（x0,?：：）

x

2

：：：

cex

1

1

3

x‘：：：

ex

hxx

3

ex

U

hx=x

2

ex

（n）

x0

exx,

2

hx：0hx0,

j

1

hx:::h00x

3

：：：ex3

.......................................................................................9

x

0

c

xx

0

-x2::-x3xc3

cX

0

x

?（

x,

x

2

：：：cex

.................

)

(I)

ACB

ACB=90‘,CD_ABD

,

CD

2

ADDB...........................................................................................................................2

CDO

CD

2

CECB..........................................................................................................4

CECBADDB

..................................................................................................................5

（n）

ON_NFNF=.OF2_ON2

OFON

........................................................................

NBEFBE

1

NF

max

BE

=2........................................................................................................................................................................