3

1

MODUL SUPER SCORE SBP 2014

KERTAS 1SET 4

NAMA MARKAHTARIKH

Answer all questions.Jawab semua soalan.

1. The relation between set P and set Q is defined by the set of ordered pairs {(1, a), (1, c), (3, a),(4, b),(4, d), (6, d)}. Hubungan antara set P dan set Q ditakrifkan oleh set pasangan bertertib {(1, a), (1, c), (3, a),(4, b), (4, d), (6, d)}. StateNyatakan (a) the image of 4,

imej bagi 4,(b) the type of the relation,

jenis hubungan tersebut,(c) the relation in the graph form.

hubungan dalam bentuk graf.[3 marks]

[3 markah]Answer/ Jawapan :(a)

(b)

(c)

1

For examiner’s

use only

2

3

2

3

3

MODUL SUPER SCORE SBP 2014

2. The following information refers to the function f and g.Maklumat berikut merujuk kepada fungsi f dan g.

f : x 4x – 6 g : x 2x – 3

Find gf–1(x).Cari gf–1(x).

[3 marks][3 markah]

Answer/ Jawapan :

3. Given the function h(x) = 4x + 5 and the composite function hg(x) = 8x + 7.Diberi fungsi h(x) = 4x + 5dan fungsi gubahan hg(x) = 8x + 7.FindCari(a) hg(–1)(b) g(x) [3 marks]

[3 markah]

Answer/ Jawapan :

2

For examiner’s

use only

3

4

MODUL SUPER SCORE SBP 2014

4. Find the roots for the quadratic equation x(x – 3) = 9 – x. Give your answer correct to three significant figures.Cari punca-punca bagi persamaan kuadratik x(x – 3) = 9 – x. Beri jawapan anda betul kepada tiga angka bererti.

[3 marks] [3 markah]

Answer/ Jawapan :

5. It is given that 4 and m + 1 are the roots of the quadratic equation x2 + (2 – n)x + 8 = 0, where m and n are constants. Find the value of m and n.Diberi bahawa 4 dan m + 1 ialah punca-punca bagi persamaan kuadratik x2 + (2 – n)x + 8 = 0, dengan keadaan m dan n ialah pemalar. Cari nilai m dan n.

[3 marks][3markah]

Answer/ Jawapan :

3

For examiner’s

use only

3

5

3

6

MODUL SUPER SCORE SBP 2014

6. The diagram shows the graph of a quadratic function y = –(x – 4)2 – 1.Rajah menunjukkan graf fungsi kuadratiky = –(x – 4)2 – 1.

FindCari(a) the value of k,

nilai k,(b) the equation of the axis of symmetry,

persamaan paksi simetri,(c)the coordinate of the maximum point.

koordinat titik maksimum.[3 marks]

[3 markah]Answer/ Jawapan :

7. The quadratic function f(x) = p(x + q)2 + r, where p, q and r are constants, has maximum value of 10. The equation of axis of symmetry is x = 2.Fungsi kuadratik f(x) = p(x + q)2 + r, dengan keadaan p, q dan r ialah pemalar, mempunyai nilai maksimum 10. Persamaan paksi simetri ialah x = 2.StateNyatakan(a) the range of value of p,

julat nilai p,

4

–

y

(k, –3)

0 x

For examiner’s

use only

3

7

4

8

4

9

MODUL SUPER SCORE SBP 2014

(b) the value of q,nilai q,

(c) the value of r.nilai r.

[3 marks]

[3 markah]Answer/ Jawapan :

8. Find the range of values of x for x2 – 2(2x + 1) 2x2 – x. Cari julat nilai-nilai x bagi x2 – 2(2x + 1) 2x2 – x.

[4 marks][4markah]

Answer/ Jawapan :

9. Solve the equation(27x)x ×

1

243x = 9.

Selesaikan persamaan (27x)x ×

1

243x = 9. [4 marks]

[4markah]Answer/ Jawapan :

5

For examiner’s

use only

3

10

4

11

MODUL SUPER SCORE SBP 2014

10. Solve the equation3x– 1 +3x + 2= 2.Selesaikan persamaan 3x – 1 + 3x + 2 = 2.

[3 marks][3markah]

Answer/ Jawapan :

11. Given that loga 2 = x and loga 3 = y, express loga( 89a )

in terms of x and y.

Diberi loga 2 = x dan loga 3 = y, nyatakan loga( 8

9a )dalam sebutan x dan

y.[4 marks]

[4 markah]Answer/ Jawapan :

12. Solve the equation 2 + log4 x = log22x.

6

For examiner’s

use only

4

12

MODUL SUPER SCORE SBP 2014

Selesaikan persamaan 2 + log4 x = log2 2x.[4 markah]

[4 marks]Answer/ Jawapan :

13. In a geometric progression, the first term is

12 and the fourth term is

− 427 .

Dalam suatu janjang geometri, sebutan pertama ialah

12 dan sebutan

keempat ialah − 4

27 .CalculateHitung

(a) the common ratio,nisbah sepunya,

(b) the sum to infinity of the geometric progression.hasil tambah hingga sebutan ketakterhinggaan bagi janjang geometri itu.

[3 marks][3 markah]

Answer/ Jawapan :

7

For examiner’s

use only

3

13

4

14

MODUL SUPER SCORE SBP 2014

14. The 3rd term of a geometric progression is 18. The sum of the 3rd term and 4th term is 27.Sebutan ke-3 suatu janjang geometri ialah 18. Hasil tambah sebutan ke-3 dan sebutan ke-4ialah 27FindCari (a) the first term and the common ratio of the progression,

sebutan pertama dan nisbah sepunya janjang itu,(b) the sum to infinity of the progression.

hasil tambah hingga sebutan ketakterhinggaan janjang itu.[4 marks]

[4 markah]Answer/ Jawapan :

15. The nth term of a geometric progression, Tn, is given by Tn = 2( 2

3 )n−2

, find

Sebutan ke-n bagi suatu janjang geometri, Tn, diberi oleh Tn = 2( 2

3 )n−2

, cari (a) the first term and the common ratio,

sebutan pertama dan nisbah sepunya.

(b) the sum to infinity of the progression.hasil tambah hingga sebutan ketakterhinggaan bagi janjang itu.

[ 3 marks][3 markah]

Answer/ Jawapan :

8

For examiner’s

use only

3

15

3

16

MODUL SUPER SCORE SBP 2014

16. The diagram shows two vectors, OA and OB.

Rajah menunjukkan dua vektor, OA dan OB.

ExpressUngkapkan

(a) AO in the form (xy) ,

OA dalam bentuk (xy) ,

(b) AB in the form xi + yj.

AB dalam bentuk xi + yj.

[3 marks][3markah]

Answer/ Jawapan :

17. Given p = ( 5−12)

and q = (m+1

3 ), find

Diberi p = ( 5−12)

dan q = (m+1

3 ), cari

(a) | p |,

(b) the value of m such that p + q is parallel to y-axis.

nilai m dengan keadaan p + q adalah selari dengan paksi y.[4 marks]

9

For examiner’s

use only

y

x

A(4, 10)

B(2, –6 )

O

~ ~

~ ~

~

4

17

b~a ~

2

18

MODUL SUPER SCORE SBP 2014

[4 markah]Answer / Jawapan :

18. The diagram shows the vectors OA, OB and OP drawn on a grid of equal squares with sides of 1 unit.Rajah menunjukkan vektor OA, OB dan OP dilukis pada grid segiempat sama yang sama besar bersisi 1 unit.

DetermineTentukan

(a) | OP |,

(b) OP in terms of a and b.

OP dalam sebutan a and b.

[2 marks][2 markah]

Answer/ Jawapan :

10

A B P

~ ~

~ ~

O

MODUL SUPER SCORE SBP 2014

Jawapan/Answer :

No

AnswerNo

Answer

1

(a) b and d(b) many to many(c)

16 (a)

(−4−10)

(b) –2i – 16j

2x2

17

(a) 13(b) –6

3

(a) –1

(b) 2x +

12

18

(a) √20(b) 2b – a

4 4.16 and –2.165 m = 1 and n = 8

6(a) 8(b) x = 4(c) (4, –1)

7(a) p < 0(b) q = –2(c) r = 10

8 –2 x –1

9x = 2,

−13

10

x = –1.4022

11

3x – 2y – 1

12

x = 0 and x = 4

13

(a) −2

3

(b)

310

11

1 3 4 6Set P

Set Q

a

b

c

d

MODUL SUPER SCORE SBP 2014

14 (a) a = 72, r =

12

(b) 144

15 (a) a = 3, r =

23

(b) 9

12