KSHS v CC BI TON LIN QUAN

Qua cc k thi tuyn sinh i Hc(T nm 2002 n 2010)

Phn I: Tip Tuyn.

Bi 1.(D-02)

Cho hn s : y =(2m 1)xm2

x 1 (1) (m l tham s).1. Kho st s bin thin v v th (C) ca hm s (1) ng vi m= 1.2. Tnh din tch hnh phng gii hn bi ng cong (C) v hai trc ta .3. Tm m th hm s (1) tip xc vi ng thng y = x.Bi 2.(D-05)

Gi (Cm) l th hm s y =1

3x3 m

2x2 +

1

3(*) (m l tham s).

1. Kho st s bin thin v v th ca hm s (*) ng vi m = 2.2. Gi M l im thuc (Cm) c honh bng 1 . Tm m tip tuyn ca (Cm) ti imM song song vi ng thng 5x y = 0.Bi 3.(D-07)

Cho hm s y =2x

x + 1.

1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm ta im M thuc (C), bit tip tuyn ca (C) ti M ct hai trc Ox, Oy ti

A, B v tam gic OAB c din tch bng1

4.

Bi 4.(D-10)

Cho hm s y = x4 x2 + 6.1. Kho st s bin thin v v th (C) ca hm s cho.2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn vung gc vi ng thng

y =1

6x 1.

Bi 5.(B-04)

Cho hm s y =1

3x3 2x2 + 3x (1) c th (C).

1. Kho st hm s (1).2. Vit phng trnh tip tuyn ca (C) ti im un v chng minh rng l tip tuynca (C) c h s gc nh nht.Bi 6.(B-06)

Cho hm s y =x2 + x 1x + 2

.

1. Kho st s bin thin v v th (C) ca hm s cho.2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn vung gc vi tim cnxin ca (C).

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Bi 7.(B-08)Cho hm s y = 4x3 6x2 + 1 (1).

1. Kho st s bin thin v v th ca hm s (1).2. Vit phng trnh tip tuyn ca th hm s (1), bit rng tip tuyn i qua imM(1;9).Bi 8.(A-09)

Cho hm s y =x + 2

2x + 3(1).

1. Kho st s bin thin v v th ca hm s (1).2. Vit phng trnh tip tuyn ca th hm s (1), bit tip tuyn ct trc honh,trc tung ln lt ti hai im phn bit A, B v tam gic OAB cn ti gc ta O.Phn II: Cc Tr.

Bi 1.(B-02)Cho hn s : y = mx4 + (m2 9)x2 + 10 (1) (m l tham s).

1. Kho st s bin thin v v th ca hm s (1) ng vi m= 1.2. Tm m hm s (1) c ba im cc tr.Bi 2.(B-05)

Gi (Cm) l th ca hm s y =x2 + (m + 1)x + m + 1

x + 1(*) (m l tham s).

1. Kho st v v th hm s (*) khi m= 1.2. Chng minh rng vi m bt k, th (Cm) lun lun c im cc i, im cc tiu vkhong cch gia hai im bng

20.

Bi 3.(B-07)Cho hm s: y = x3 + 3x2 + 3(m2 1)x 3m2 1 (1), m l tham s.

1. Kho st s bin thin v v th hm s (1) khi m= 1.2. Tm m hm s (1) c cc i, cc tiu v cc im cc tr ca th hm s (1) cchu gc ta O.Bi 4.(A-02)

Cho hm s: y = x3 + 3mx2 + 3(1m2)x + m3 m2 (1) (m l tham s).1. Kho st s bin thin v v th ca hn s (1) khi m = 1.2. Tm k phng trnh: x3 + 3x2 + k3 3k2 = 0 c ba nghim phn bit.3. Vit phng trnh ng thng i qua hai im cc tr ca th hm s (1).Bi 5.(A-05)

Gi(Cm) l th ca hm s y = mx +1

x(*) (m l tham s).

1. Kho st s bin thin v v th ca hm s (*) khi m =1

4.

2. Tm m hm s (*) c cc tr v khong cch t im cc tiu ca (Cm) n tim cn

xin ca (Cm) bng12.

Bi 6.(A-07)

Cho hm s y =x2 + 2(m + 1)x + m2 + 4m

x + 2(1), m l tham s.

1. Kho st s bin thin v v th ca hm s (1) khi m = 1.2. Tm m hm s (1) c cc i v cc tiu, ng thi cc im cc tr ca th cngvi gc ta O to thnh mt tam gic vung ti O.

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Phn III: Tng Giao Th.

Bi 1.(D-03)

1. Kho st s bin thin v v th ca hm s y =x2 2x + 4

x 2 (1).2. Tm m ng thng dm: y = mx + 2 2m ct th hm s (1) ti hai im phnbit.Bi 2.(D-06)

Cho hn s : y = x3 3x + 2.1. Kho st s bin thin v v th (C) ca hm s cho.2. Gi d l ng thng i qua im A(3; 20) v c h s gc l m. Tm m ng thngd ct th (C) ti 3 im phn bit.Bi 3.(D-08)

Cho hn s : y = x3 3x2 + 4 (1).1. Kho st s bin thin v v th ca hm s (1).2. Chng minh rng mi ng thng i qua im I(1; 2) vi h s gc k (k> 3) u ct th ca hm s (1) ti ba im phn bit I, A, B ng thi I l trung im ca on thngAB.Bi 4.(D-09)

I. Cho hn s y = x4 (3m + 2)x2 + 3m c th l (Cm), m l tham s.1. Kho st s bin thin v v th ca hm s cho khi m= 0.2. Tm m ng thng y = 1 ct th (Cm) ti 4 im phn bit u c honh nhhn 2.II. Tm cc gi tr ca tham s m ng thng y = 2x + m ct th hm sy =

x2 + x 1x

ti hai im phn bit A, B sao cho trung im ca on thng AB thuc

trc tung.Bi 5.(B-09)I. Cho hm s y = 2x4 4x2 (1).1. Kho st s bin thin v v th ca hm s (1).2. Vi gi tr no ca m, phng trnh x2|x2 2| = m c ng 6 nghim thc phn bit?II. Tm cc gi tr ca tham s m ng thng y = x+m ct th hm s y = x

2 1x

ti hai im phn bit A, B sao cho AB= 4.Bi 6.(B-10)

Cho hm s y =2x + 1

x + 1.

1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm m ng thng y = 2x + m ct th (C) ti hai im phn bit A, B sao chotam gic OAB c din tch bng

3 (O l gc ta ).

Bi 7.(A-03)

Cho hm s y =mx2 + x + m

x 1 (1) (ml tham s).1. Kho st s bin thin v v th hm s (1) khi m = 1.2. Tm m th hm s (1) ct trc honh ti hai im phn bit v hai im chonh dng.

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Bi 8.(A-04)

Cho hm s y =x2 + 3x 3

2(x 1) (1).1. Kho st hm s (1).2. Tm m ng thng y = m ct th hm s (1) ti hai im A, B sao cho AB= 1.Bi 9.(A-06)1. Kho st s bin thin v v th hm s y = 2x3 9x2 + 12x 4.2. Tm m phng trnh sau c 6 nghim phn bit: 2|x3| 9x2 + 12|x| = m.Bi 10.(A-10)

Cho hm s y = x3 2x2 + (1m)x + m (1), m l tham s thc.1. Kho st s bin thin v v th ca hm s (1) khi m = 1.2. Tm m th ca hm s (1) ct trc honh ti 3 im phn bit c honh x1, x2, x3tha mn iu kin x21 + x

22 + x

23 < 4.

Phn IV: Bi Ton Khc.Bi 1.(D-04)

Cho hn s : y = x3 3mx2 + 9x + 1 (1) (m l tham s).1. Kho st hm s (1) ng vi m = 2.2. Tm m im un ca th hm s (1) thuc ng thng y = x + 1.Bi 2.(B-03)

Cho hn s : y = x3 3x2 + m (1) (m l tham s).1. Tm m th hm s (1) c hai im phn bit i xng vi nhau qua gc ta .2. Kho st s bin thin v v th ca hm s (1) ng vi m= 2.Bi 3.(A-08)

Cho hm s y =mx2 + (3m2 2)x 2

x + 3m(1), vi m l tham s thc.

1. Kho st s bin thin v v th ca hm s (1) khi m = 1.2. Tm cc gi tr ca tham s m gc gia hai ng tim cn ca th hm s (1) bng45o.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

p s:

Phn I: 1(1 + 4 ln 43;m 6= 1). 2(m = 4). 3(M(1

2;2);M(1; 1)). 4(y = 6x + 10).

5(y = x + 83). 6(y = x + 22 5; y = x 22 5). 7(y = 24x + 15; y = 15

4x 21

4).

8(y = x 2).Phn II:1(m < 3 or 0 < m < 3). 2(M(2;m 3);N(0;m + 1)). 3(m = 1

2).

4(1 < k < 3 k 6= 0, k 6= 2; y = 2xm2 + m). 5(m = 1). 6(m = 4 26).Phn III: 1(m > 1). 2(m > 15

4,m 6= 24). 4(I(1

3< m < 1); II(m = 1)).

5(I(0 < m < 1); II(m = 26)). 6(m = 2). 7(12< m < 0). 8(m = 1

5

2).

9(4 < m < 5). 10(14< m < 1,m 6= 0).

Phn IV: 1(m = 0 orm = 2). 2(m > 0). 3(m = 1).

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