Đê đáp án Sat hạch Toán 11 D lan 2 2011. Lê Xoay

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XOay
NAM HOC 2010-20ll

TRTIONG THPT r,A

On

rHr CnUyfN

DE LOp

rr r,AN 1

y6n: Torln- ruiii O
hoi gian: 180'(khdng

Ciu I. Cho b6t phucng trinh: J. *2 - J
"
1. Ciai b6t phuong trinh vcri m:-2

I , r* -rr[;J

kC thcri gian giao
+

*

2. fimm AC U6t phucmg trinh da cho c6 nghi6m.
Ciu

,

II.

:

1.

Giai phucrng trinh:

2.

ciaihephucrngoi"h, {tl2

3.

Tfnh gioi han

4.

Tinh gi6 tri biCu thuc:

4cosr.ror[1-r).^"[I*r)-+"o.zx+3cosx-4

[3 / \.3 )
x+3v)(x-1)=14

=

0I
;

lx' + x+3y =9

sau:

)
x'
+ sin(x - 1) - 1
A = lim
x-+l 4lx -I

A=

4Cloo+8c,fu +l2Cfoo+...+200c,1jj.

Ciu III.
Trong mat phang voi h0 toa dQ Oxy cho Elip (E) c6 ti6u diOm thri nhdt
qua di6m

CAu

t[t'+)

"*

xic itlnh chu vi

h F1(3;0) vd di

cria hinh chfr nh6t co s6 cria

Elip{E).
i
t

IV.

.

Cho hinh ch6p S.ABC diy ld tam giric rl6u
dey (ABC). Gqi I td trung di6m cira BC.

c$h

alt, SA: 2avdSA vu6ng g6c voi m{t

1. chrmg minh rnng mat phang (sAD vu6ng g6c v6i mat phrng (sBC).
2- Gqi G ld trgng tAm cria tam giiic d6u ABC. Tfnh khoang c6ch tir G dtin mat ph6ng (SBC).
CAU

V
Cho a,b,c > 0: abc =

l.

Chring minh rdng:

lll

a+b+l - b-+c+l -.;;=

'


N6i du
Gi6i bdt phucrng trinh
Ei6u kiQn:
t = J a a2

-

-r >

vli m:-2

z

J. -zUatphucrng

t'-t+m<0(2)

Khi m:-2, ta c6: t2 -t -2

<0

..l;.2-J.-2<2e x>2

trinh c6 dpng:

e -l
trinh c6 nghiQm x,>
Tim m

OC U6t

Z

phuong trinh d6 cho c6 nghiQm.

Bdt phuong trinh dd cho c6 nghiQm khi b
th6a mdn: 0 Xdt hem sO f(t)=t2 -t,v t e(0;2)
t

0.25

I

0

2

2

(0

'1
,4

0.25

4 cos3 x

-

3

cos x

- 4(2

cosz -r

-

1) + 3 cos

V4y phuong trinh c6 nghiQm

"

=

x

- 4 = 0 <> 4 cosz x(cosx-2)={ <+ cos*=0

I + fttr

I

x'-x=7
, = 14 <+
^ f(f - x)(zx +3y)
e1
Lt'-" +2x+3y =)

2x+3y

=)

x'-x=2
2x+3y =7

Vfly h0 phucmg trinh c6 nghi6m:
(*;

v)= (-r; :), (2

rr,(*P, =#),

[

-tP!P)


Ciu Y

NQi dung

Di6m

4
Ta c6:(1+x)'oo = C$o +C,toox+ C?rr*'

(t-r)'o

= c$o -c]oox+ cr'oo*'

1.0

+...+C]jfx'oo (l)

0r25

-clor*t +...+cfifx'oo (2)

Ley Q)+Q) ta dusc:
(t + r)'oo + (1 - r)'oo :2clro +2clorx2 +2cl*xa +... + 2clff

0r25

xr00

L6y dao hdm hai v6 theo 6n x ta dugc
1 00 (1 * r)nn
- 1 00 (1 - r)nn = 4Clrox+ 8C$ox3 +... + 200C]ffx"
Thay

:)
3

x:l

vdo

A =100.2ee = 4Cr'ro+ 8C,fo +... + 200clff

0,25

A=lim x2+sin(x-1)-1

;;i

-4 =

A=

lim
x-+1

lr

1li

x2

f

*t

+sin(x-1)-l
?,1; _

1

= lim

("'-t)G[? +G +t)

iJ

+ lim

1ll, -r)Gl7 +.8 + 1)
r1[7
^ r-=
(*' -r)(tr[P+J7+1) rtth', ,,^(llxt +Jx +l)sin(x-l)
x_+r

x-1

x->l

G +r)6,[7
.4 = lim

III

0,25

+

x-+

rE

+ 1)

x-->l

I

1.0

0.25

*Ji+ryffi-1y

x-l

0.5

^ t----.-

+ lim

(11

x' + Jx + l)sin(x - I)

x-I

-9

Gia sft FrC3;0) ,vi 2 tiou di6m cria Elip il6i ximg nhau qua g5c at d0 ncn ticu"d
cdn l4i c6 tga d0 Fz(3;0). Theo dinh nghia cria Elip tac6: MF, + MF2:2a:10;h6n a:S
Ta c6: b2: t --c2 : 14,viiyb:4.
Chu vi cua hinh cht nh{t co sd cria Elip: C:2(2a+2b):f6
.

0.25
1.0
0.5

0.5

IV
l.Ta c6 BC r SA, BC l_ AI n6n BC r (SAI)
vay (sAI) r (SBC).
2.Ha AH vudng g6c vdi SI suy ra AH .J- (SBC)
Ta c6: d(G;(SBC)) =1 af er(sBc)) =?

33

111146
AH2 SAz'

AI2

YQy d(G;(sBQ)

ou

2.0
0.5
0.5

0.5

g .,l,r,Ia = _Lt
-- -:---;-r -----;
go, n n"'

4a2'

=la
5

5

0.5

IV

1.0
Ta c6:
a

* b = (.,6. V6XJT - V.b

. {6t) = V* (Ji * .,6)

0.25

+ a+b+r > 16(16.Vb)+r = r6b(16.Vo)*iffi = #o(16*Vb *16)

-

.l_{6f
u*b+l - V*(S*V6 *#) - {;*Vu *i6

Tucrnstutac6:
!:' !s vv'

1

b+c+1

0.25

=_8.._
-=4
{E($*Vu.S) - 1,1;*Vu *i6

rlvb
c+a+l=6=6;6m

0.25

Cgtre v6 vdi vti ta suy ra dugc dpcm.
Ddu bdng s6y ra khi a:b:c:l

0.2s

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