Đề Thi toán HKII Lớp 11

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N TP HC K 2 Nm hc 2013-2014 Mn TON Lp 11

Thi gian lm bi 90 pht

I. Phn chung cho c hai ban Bi 1. Tm cc gii hn sau:

1) x

x x

x

2

1

2lim

1

2) x

x x4lim 2 3 12

+ 3)x

x

x3

7 1lim

3+

4) x

x

x23

1 2lim

9+

Bi 2. 1) Xt tnh lin tc ca hm s sau trn tp xc nh ca n:

x xkhi xf x x

x khi x

2 5 63( ) 3

2 1 3

+ >

=

+

2) Chng minh rng phng trnh sau c t nht hai nghim : x x x3 22 5 1 0 + + = . Bi 3. 1) Tm o hm ca cc hm s sau: a) y x x2 1= + b) y

x 23

(2 5)=

+

2) Cho hm s xyx

11

=

+ .

a) Vit phng trnh tip tuyn ca th hm s ti im c honh x = 2. b) Vit phng trnh tip tuyn ca th hm s bit tip tuyn song song vi d: xy 2

2

= .

Bi 4. Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA vung gc vi y, SA = a 2 . 1) Chng minh rng cc mt bn hnh chp l nhng tam gic vung. 2) Chng minh rng: (SAC) (SBD) . 3) Tnh gc gia SC v mp (SAB) . 4) Tnh gc gia hai mt phng (SBD) v (ABCD) .

II . Phn t chn. 1 . Theo chng trnh chun.

Bi 5a. Tnh x

x

x x

3

22

8lim

11 18+

+ +.

Bi 6a. Cho y x x x3 21 2 6 83

= . Gii bt phng trnh y / 0 .

2. Theo chng trnh nng cao.

Bi 5b. Tnh x

x x

x x21

2 1lim

12 11

+.

Bi 6b. Cho x xyx

2 3 31

+=

. Gii bt phng trnh y / 0> .

--------------------Ht------------------- H v tn th sinh: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . SBD :. . . . . . . . . .

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I . Phn chung cho c hai ban. Bi 1. Tm cc gii hn sau:

1) x

x x x

x

2 1 3lim

2 7 +

+ 2)

xx x3lim ( 2 5 1)

+ + 3)

x

x

x5

2 11lim

5+

4) x

x

x x

3

20

1 1lim

+

+.

Bi 2 .

1) Cho hm s f(x) = x

khi xf x xm khi x

3 11( ) 1

2 1 1

=

+ =

. Xc nh m hm s lin tc trn R..

2) Chng minh rng phng trnh: m x x2 5(1 ) 3 1 0 = lun c nghim vi mi m. Bi 3. 1) Tm o hm ca cc hm s:

a) x xyx

2

22 2

1

+=

b) y x1 2 tan= + .

2) Cho hm s y x x4 2 3= + (C). Vit phng trnh tip tuyn ca (C): a) Ti im c tung bng 3 . b) Vung gc vi d: x y2 3 0+ = . Bi 4. Cho t din OABC c OA, OB, OC, i mt vung gc v OA = OB = OC = a, I l trung im BC

1) Chng minh rng: (OAI) (ABC). 2) Chng minh rng: BC (AOI). 3) Tnh gc gia AB v mt phng (AOI). 4) Tnh gc gia cc ng thng AI v OB . II . Phn t chn. 1 . Theo chng trnh chun .

Bi 5a. Tnh nn n n2 2 21 2 1

lim( .... )1 1 1

+ + ++ + +

.

Bi 6a. Cho y x xsin2 2 cos= . Gii phng trnh y / = 0 . 2 . Theo chng trnh nng cao . Bi 5b. Cho y x x22= . Chng minh rng: y y3 / /. 1 0+ = .

Bi 6b . Cho f( x ) = f x xxx3

64 60( ) 3 16= + . Gii phng trnh f x( ) 0 = .


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Bi 1. Tnh cc gii hn sau:

1) x

x x x3 2lim ( 1)

+ + 2) x

x

x1

3 2lim

1+

+ 3)

x

x

x2

2 2lim

7 3+

+

4) x

x x x

x x x

3 2

3 23

2 5 2 3lim4 13 4 3

+ 5) lim

n n

n n

4 5

2 3.5

+

Bi 2. Cho hm s: x

khi x >2 xf x

ax khi x 2

3 3 2 22( )14

+

= +

. Xc nh a hm s lin tc ti im x = 2.

Bi 3. Chng minh rng phng trnh x x x5 43 5 2 0 + = c t nht ba nghim phn bit trong khong (2; 5).

Bi 4. Tm o hm cc hm s sau:

1) xyx x25 3

1

=

+ + 2) y x x x2( 1) 1= + + + 3) y x1 2 tan= + 4) y xsin(sin )=

Bi 5. Cho hnh chp S.ABC c ABC vung ti A, gc B = 600 , AB = a; hai mt bn (SAB) v (SBC) vung gc vi y; SB = a. H BH SA (H SA); BK SC (K SC).

1) Chng minh: SB (ABC) 2) Chng minh: mp(BHK) SC. 3) Chng minh: BHK vung . 4) Tnh cosin ca gc to bi SA v (BHK).

Bi 6. Cho hm s x xf xx

2 3 2( )

1 +

=

+ (1). Vit phng trnh tip tuyn ca th hm s (1), bit tip

tuyn song song vi ng thng d: y x5 2= .

Bi 7. Cho hm s y x2cos 2= . 1) Tnh y y, . 2) Tnh gi tr ca biu thc: A y y y16 16 8 = + + .


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Bi 1. Tnh cc gii hn sau:

1) x xx

3 2lim ( 5 2 3) +

2) x

x

x1

3 2lim

1++

+ 3)

x

x

x2

2lim

7 3

+

4) x

x

x

3

0

( 3) 27lim

+ 5)

n n

n n

3 4 1lim

2.4 2

+

+

Bi 2. Cho hm s: x

khi xf x x

ax khi x

11( ) 1

3 1

>=

. Xc nh a hm s lin tc ti im x = 1.

Bi 3. Chng minh rng phng trnh sau c t nht mt nghim m: x x3 1000 0,1 0+ + =

Bi 4. Tm o hm cc hm s sau:

1) x xyx

22 6 52 4

+=

+ 2) x xy

x

2 2 32 1

+=

+ 3) x xy

x x

sin cossin cos

+=

4) y xsin(cos )=

Bi 5. Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA (ABCD) v SA = 2a. 1) Chng minh SAC SBD( ) ( ) ; SCD SAD( ) ( ) 2) Tnh gc gia SD v (ABCD); SB v (SAD) ; SB v (SAC). 3) Tnh d(A, (SCD)); d(B,(SAC))

Bi 6. Vit phng trnh tip tuyn ca th hm s y x x3 23 2= + : 1) Ti im M ( 1; 2) 2) Vung gc vi ng thng d: y x1 2

9= + .

Bi 7. Cho hm s: x xy2 2 22

+ += . Chng minh rng: y y y 22 . 1 = .

Ht H v tn th sinh: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . SBD :. . . . . . . . . .

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A. PHN CHUNG: Bi 1: Tm cc gii hn sau:

a) n nn

3

32 2 3

lim1 4

+

b) x

x

x21

3 2lim

1+

Bi 2: Xt tnh lin tc ca hm s sau trn tp xc nh ca n:

x x khi xf x x

khi x

2 3 22( ) 2

3 2

+ + = +

=

Bi 3: Tnh o hm ca cc hm s sau: a) y x x x2sin cos tan= + b) y xsin(3 1)= + c) y xcos(2 1)= + d) y x1 2 tan4= +

Bi 4: Cho hnh chp S. ABCD c y ABCD l hnh thoi cnh a, BAD 060= v SA = SB = SD = a. a) Chng minh (SAC) vung gc vi (ABCD). b) Chng minh tam gic SAC vung. c) Tnh khong cch t S n (ABCD).

B. PHN T CHN: 1. Theo chng trnh chun Bi 5a: Cho hm s y f x x x3( ) 2 6 1= = + (1) a) Tnh f '( 5) . b) Vit phng trnh tip tuyn ca th hm s (1) ti im Mo(0; 1) c) Chng minh phng trnh f x( ) 0= c t nht mt nghim nm trong khong (1; 1).

2. Theo chng trnh Nng cao

Bi 5b: Cho x xf x x xsin3 cos3( ) cos 3 sin3 3

= + +

.

Gii phng trnh f x'( ) 0= .

Bi 6b: Cho hm s f x x x3( ) 2 2 3= + (C). a) Vit phng trnh tip tuyn ca (C) bit tip tuyn song song vi ng thng d: y x22 2011= +

b) Vit phng trnh tip tuyn ca (C) bit tip tuyn vung gc ng thng : y x1 20114

= +


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A. PHN CHUNG Cu 1: Tm cc gii hn sau:

a) x xx x

23 4 1lim1 1

+

b) x

x x

2 9lim3 3

+ c) x

x x

2lim2 7 3

+ d) x x

x x

2 2 3lim2 1

+

+

Cu 2: Cho hm s x x

khi xf x x

m khi x

2 2 2( ) 2

2

=

=

.

a) Xt tnh lin tc ca hm s khi m = 3 b) Vi gi tr no ca m th f(x) lin tc ti x = 2 ?

Cu 3: Chng minh rng phng trnh x x x5 43 5 2 0 + = c t nht ba nghim phn bit trong khong (2; 5)

Cu 4: Tnh o hm ca cc hm s sau:

b) y x x2 3( 1)( 2)= + c) yx2 21

( 1)=

+ d) y x x2 2= + e) xy

x

42

22 1

3

+=

B.PHN T CHN: 1. Theo chng trnh chun Cu 5a: Cho tam gic ABC vung cn ti B, AB = BC= a 2 , I l trung im cnh AC, AM l ng

cao ca SAB. Trn ng thng Ix vung gc vi mp(ABC) ti I, ly im S sao cho IS = a. a) Chng minh AC SB, SB (AMC). b) Xc nh gc gia ng thng SB v mp(ABC). c) Xc nh gc gia ng thng SC v mp(AMC).

2. Theo chng trnh nng cao Cu 5b: Cho hnh chp u S.ABCD c cnh y bng a v cnh bn bng 2a. Gi O l tm ca y

ABCD. a) Chng minh rng (SAC) (SBD), (SBD) (ABCD). b) Tnh khong cch t im S n mp(ABCD) v t im O n mp(SBC). c) Dng ng vung gc chung v tnh khong cch gia hai ng thng cho nhau BD v SC.


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I. PHN BT BUC: Cu 1: Tnh cc gii hn sau:

a) ( )x

x x2lim 5+

+ b)x

x

x23

3lim

9+

Cu 2 (1 im): Cho hm s x

khi xx xf x

A khi x

22 1 1

22 3 1( )12

+ + +=

=

Xt tnh lin tc ca hm s ti x 12

=

Cu 3 (1 im): Chng minh rng phng trnh sau c t nht mt nghim trn [0; 1]: x x3 5 3 0+ = . Cu 4 (1,5 im): Tnh o hm ca cc hm s sau:

a) y x x( 1)(2 3)= + b) xy 21 cos2

= +

Cu 5 (2,5 im) : Cho hnh chp S.ABCD c y ABCD l hnh thoi tm O cnh a, BAD 060= , ng cao SO = a.

a) Gi K l hnh chiu ca O ln BC. Chng minh rng: BC (SOK) b) Tnh gc gia SK v mp(ABCD). c) Tnh khong cch gia AD v SB. II. PHN T CHN 1. Theo chng trnh chun Cu 6a (1,5 im): Cho hm s: y x x32 7 1= + (C). a) Vit phng trnh tip tuyn ca th (C) ti im c honh x = 2. b) Vit phng trnh tip tuyn ca th (C) c h s gc k = 1. Cu 7a (1,5 im): Cho hnh chp tam gic S.ABC c y ABC l tam gic u, SA (ABC), SA= a. M

l mt im trn cnh AB, ACM = , h SH CM. a) Tm qu tch im H khi M di ng trn on AB. b) H AK SH. Tnh SK v AH theo a v . 2. Theo chng trnh nng cao

Cu 6b (1,5 im): Cho cc th (P): xy x2

12

= + v (C): x xy x2 3

12 6

= + .

a) Chng minh rng (P) tip xc vi (C). b) Vit phng trnh tip tuyn chung ca (P) v (C) ti tip im. Cu 7b (1,5 im): Cho hnh chp S.ABCD c y ABCD l hnh vung tm O, cnh a; SA = SB = SC

= SD = 52a

. Gi I v J ln lt l trung im BC v AD.

a) Chng minh rng: SO (ABCD). b) Chng minh rng: (SIJ) (ABCD). Xc nh gc gia (SIJ) v (SBC). c) Tnh khong cch t O n (SBC).


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A. PHN BT BUC Cu 1: Tnh cc gii hn sau:

a)x

x x

x x

2

22 3 4

lim4 2 1+

+

+ + b)

x

x x

x

2

21

3 2lim

1 +

c) x

x

x

2

2

5 3lim

2+

Cu 2: Cho hm s x khi xf xax khi x21 1

( )4 1 +

= >

. nh a hm s lin tc ti x = 1

Cu 3: Chng minh rng phng trnh x x32 6 1 0 + = c 3 nghim trn [2; 2].

Cu 4: Tnh o hm ca cc hm s sau:

a) xyx

3 52 1

+=

+ b) y x xsin .cos3=

Cu 5: Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, hai mt bn (SAB), (SBC) vung gc vi y, SB = a.

a) Gi I l trung im SC. Chng minh rng: (BID) (SCD). b) Chng minh rng cc mt bn ca hnh chp l cc tam gic vung. c) Tnh gc ca mp(SAD) v mp(SCD). B. PHN T CHN 1. Theo chng trnh chun

Cu 6a: Cho hyperbol (H): yx

1= . Vit phng trnh tip tuyn ca (H):

a) Ti im c honh x0 1= .

b) Tip tuyn song song vi ng thng d: y x14

= .

Cu 7a: Cho lng tr tam gic ABC.ABC. Gi I, J, K ln lt l trng tm ca cc tam gic ABC, ABC, ACC. Chng minh rng:

a) (IJK) // (BBCC) b) (AJK) // (AIB) 2. Theo chng trnh nng cao Cu 6b: Gii v bin lun phng trnh f x( ) 0 = , bit f x x m x mx( ) sin2 2(1 2 )cos 2= + .

Cu 7b: Cho hnh chp S.ABCD c y ABCD l hnh thang vung, AD // BC, AB = a, BC = a, ADC 045= . Hai mt bn SAB, SAD cng vung gc vi y, SA = a 2 .

a) Tnh gc gia BC v mp(SAB) b) Tnh gc gia mp(SBC) v mp(ABCD) c) Tnh khong cch gia AD v SC


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Bi 1: 1) Tnh cc gii hn sau: a) + +

+

4

2

2 2lim

1n n

n b)

3

2

8lim

2xx

x c)

+

+

+13 2

lim1x

x

x.

2) Cho y f x x x3 2( ) 3 2= = + . Chng minh rng phng trnh f(x) = 0 c 3 nghim phn bit.

3) Cho x x

khi xf x x

a x khi x

2 22( ) 2

5 3 2

=

=

. Tm a hm s lin tc ti x = 2.

Bi 2: Cho y x2 1= . Gii bt phng trnh: y y x2. 2 1 < .

Bi 3: Cho t din OABC c OA = OB = OC = a, AOB AOC BOC0 060 , 90= = = . a) Chng minh rng ABC l tam gic vung. b) Chng minh OA vung gc BC. c) Gi I, J l trung im OA v BC. Chng minh IJ l on vung gc chung OA v BC.

Bi 4: Cho y f x x x3 2( ) 3 2= = + . Vit phng trnh tip tuyn ca th hm s f(x) bit tip tuyn song song vi d: y = 9x + 2011.

Bi 5: Cho xf xx

2 1( )

= . Tnh nf x( ) ( ) , vi n 2.


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A. PHN BT BUC: Cu 1: Tnh cc gii hn sau:

a) x

x

x x23

3lim

2 3+

+ b)

x

x

x

3

0

( 1) 1lim

+ c)

x

x

x

2

2

5 3lim

2+

+

Cu 2: a) Chng minh rng phng trnh sau c t nht 2 nghim: x x32 10 7 0 =

b) Xt tnh lin tc ca hm s x

xf x x

x

3, 1( ) 1

2 , 1

+

=

=

trn tp xc nh .

Cu 3: a) Vit phng trnh tip tuyn ca thi hm s y x3= ti im c honh x0 1= . b) Tnh o hm ca cc hm s sau: y x x y x x x x2 21 (2 )cos 2 sin = + = + Cu 4: Cho hnh chp S.ABCD c SA (ABCD) v ABCD l hnh thang vung ti A, B . AB = BC = a, ADC SA a045 , 2= = .

a) Chng minh cc mt bn ca hnh chp l cc tam gic vung. b) Tnh gc gia (SBC) v (ABCD). c) Tnh khong cch gia AD v SC. B. PHN T CHN: 1. Theo chng trnh chun

Cu 5a: a) Tnh x xx22

1 1lim

24+

b) Cho hm s f xx

8( ) = . Chng minh: f f( 2) (2) =

Cu 6a: Cho y x x3 23 2= + . Gii bt phng trnh: y 3 < . Cu 7a: Cho hnh hp ABCD.EFGH c AB a AD b AE c, ,= = =

. Gi I l trung im ca on BG. Hy

biu th vect AI

qua ba vect a b c, ,

.

2. Theo chng trnh nng cao Cu 5b: a) Tnh gn ng gi tr ca 4,04 b) Tnh vi phn ca hm s y x x2.cot=

Cu 6b: Tnh x

x x

x

2

3

3 1lim

3+ +

Cu 7b 3: Cho t din u cnh a. Tnh khong cch gia hai cnh i ca t din .


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II. Phn bt buc Cu 1: 1) Tnh cc gii hn sau: a)

x

x

x x21 2

lim2 3+

+ b)

x

x x x

x x

3 2

32

3 9 2lim

6+

c) ( )x

x x x2lim 3

+ +

2) Chng minh phng trnh x x3 3 1 0 + = c 3 nghim phn bit .

Cu 2: 1) Tnh o hm ca cc hm s sau: a) ( )y x x

x

23 1

= +

b) y x xsin= + c) x xyx

2 21

=

2) Tnh o hm cp hai ca hm s = tany x 3) Tnh vi phn ca ham s y = sinx.cosx Cu 3: Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, ( )SA ABCD v = 6SA a . 1) Chng minh : BD SC SBD SAC, ( ) ( ) . 2) Tnh khong cch t A n mt phng (SBD). 3) Tnh gc gia SC v (ABCD)

II. Phn t chn 1. Theo chng trnh chun Cu 4a: Vit phng trnh tip tuyn ca th hm s = 1y x

x ti giao im ca n vi trc honh .

Cu 5a: Cho hm s = + +360 64

( ) 3 5f x xx x

. Gii phng trnh f x( ) 0 = .

Cu 6a: Cho hnh lp phng ABCD.EFGH c cnh bng a . Tnh .AB EG .

2. Theo chng trnh nng cao Cu 4b: Tnh vi phn v o hm cp hai ca hm s y x xsin2 .cos2= .

Cu 5b: Cho = + 3 2

23 2x x

y x . Vi gi tr no ca x th y x( ) 2 = .

Cu 6b: Cho hnh lp phng ABCD.ABCD c cnh bng a. Xc nh ng vung gc chung v tnh khong cch ca hai ng thng cho nhau BD v BC.


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Bi 1: Tnh cc gii hn sau:

a) n n

n

1

13 4

lim4 3

+

+ b)

x

x

x23

1 2lim

9+

Bi 2: Chng minh phng trnh x x3 3 1 0 + = c 3 nghim thuc ( )2;2 .

Bi 3: Chng minh hm s sau khng c o hm ti x 3=

x khi xf x x

khi x =

2 93( ) 3

1 3

= +

Bi 4: Tnh o hm cc hm s sau: a) y x x x2(2 1) 2= + b) y x x2.cos=

Bi 5: Cho hm s xyx

11

+=

c th (H). a) Vit phng trnh tip tuyn ca (H) ti A(2; 3). b) Vit phng trnh tip tuyn ca (H) bit tip tuyn song song vi ng thng y x1 5

8= + .

Bi 6: Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA = a, SA vung gc vi (ABCD). Gi I, K l hnh chiu vung gc ca A ln SB, SD.

a) Chng minh cc mt bn hnh chp l cc tam gic vung. b) Chng minh: (SAC) vung gc (AIK). c) Tnh gc gia SC v (SAB). d) Tnh khong cch t A n (SBD).


HUYNH CH HAORectangle

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a) x

x x

x

2

21

2 3 5lim

1+

b) x

x x

x

3

1

1lim

1++ +

Bi 2: Chng minh rng phng trnh x mx x m3 22 0 + = lun c nghim vi mi m.

Bi 3: Tm a hm s lin tc ti x = 1.

x x x khi x 1f x x a

x a khi x = 1

3 2 2 2( ) 3

3

+

= + +

Bi 4: Tnh o hm ca cc hm s:

a) y xx x x2 42 3 1

3 1= + + + b) x xyx x

cossin

= +

Bi 5: Cho ng cong (C): y x x3 23 2= + . Vit phng trnh tip tuyn ca (C): a) Ti im c honh bng 2. b) Bit tip tuyn vung gc ng thng y x1 1

3= + .

Bi 6: Cho hnh chp S.ABCD c y ABCD l hnh thoi tm O cnh a, aOB 33

= , SO ABCD( ) ,

SB a= . a) Chng minh: SAC vung v SC vung gc vi BD. b) Chng minh: SAD SAB SCB SCD( ) ( ), ( ) ( ). c) Tnh khong cch gia SA v BD.


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a) ( )x

x x x2lim 3 2

+ b) ( )x

x x x2lim 4 1 2+

+ +

Bi 2: Chng minh rng phng trnh x x32 10 7 0 = c t nht hai nghim.

Bi 3: Tm m hm s sau lin tc ti x = 1

x khi xf x x

mx khi x

2 11( ) 1

2 1

<

= + +

Bi 4: Tnh o hm ca cc hm s sau:

a) xyx

3 2

2 5

=

+ b) y x x x2( 3 1).sin= +

Bi 5: Vit phng trnh tip tuyn ca th hm s yx

1= :

a) Ti im c tung bng 12

.

b) Bit tip tuyn song song vi ng thng y x4 3= + .

Bi 6: Cho t din S.ABC c ABC u cnh a, SA ABC SA a3( ),2

= . Gi I l trung im BC.

a) Chng minh: (SBC) vung gc (SAI). b) Tnh khong cch t A n (SBC). c) Tnh gc gia (SBC) v (ABC).


Đề Thi toán HKII Lớp 11

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