Add Math Sbp 07 Paper 1

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SULIT 3472/1 Matematik Tambahan Kertas 1 Mei 2007 2 hours

3472/1 Name : ………………..…………… Form : ………………………..……

SEKTOR SEKOLAH BERASRAMA PENUH KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERTENGAHAN TAHUN TINGKATAN 5 2007 MATEMATIK TAMBAHAN

Kertas 1 Dua jam JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU 1

This question paper consists of 25 questions.

2. Answer all questions. 3. Give only one answer for each question. 4. Write your answers clearly in the spaces provided in the question paper. 5. Show your working. It may help you to get marks. 6. If you wish to change your answer, cross out the work that you have done. Then write down the new answer. 7. The diagrams in the questions provided are not drawn to scale unless stated. 8. The marks allocated for each question and sub-part of a question are shown in brackets. 9. A list of formulae is provided on pages 2 to 3. 10. A booklet of four-figure mathematical tables is provided. . 11 You may use a non-programmable scientific calculator. 12 This question paper must be handed in at the end of the examination .

For examiner’s use only Total Marks Question Marks Obtained 1 2 2 4 3 2 4 3 5 3 6 2 7 3 8 3 9 3 10 4 11 4 12 4 13 4 14 3 15 3 16 2 17 4 18 4 19 4 20 4 21 3 22 3 23 3 24 3 25 3 TOTAL

80

Kertas soalan ini mengandungi 13 halaman bercetak

3472/1

[ Lihat sebelah SULIT

SULIT

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2

The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. ALGEBRA

b  b 2  4ac 2a

x

1 2

am  an = a m + n

3

am  an = a m - n

4

(am) n = a nm

5

loga mn = log am + loga n

6

loga

7

log a mn = n log a m

8

logab =

log c b log c a

9 Tn = a + (n-1)d

n [2a  ( n  1) d ] 2

10

Sn =

11

Tn = ar n-1

a (r n  1) a (1  r n ) , (r  1)  r 1 1 r a 13 S∞  , r <1 1 r 12

m = log am - loga n n

Sn =

CALCULUS 1

2

3

y = uv ,

dy dv du =u +v dx dx dx

du dv v −u u y = , dx , = dx 2 dx v dy v

4 Area under a curve b

∫y

=

dx or

a

b

=

∫ x dy a

5 Volume generated b

dy dy du = × dx du dx



2 = πy dx or a

b

=

∫ πx

2

dy

a

GEOMETRY

1 Distance =

( x1 − x 2 ) 2 + ( y1 − y 2 ) 2

2 Midpoint

y1 + y 2   x1 + x 2  , 2   2

(x , y) = 

3

r = x2 + y2

4

rˆ =

3472/2

5 A point dividing a segment of a line  nx1 + mx 2 ny1 + my 2  ,  ( x,y) =  m+n   m+n 6 Area of triangle = 1 ( x1 y 2 + x 2 y 3 + x3 y11 ) − ( x 2 y1 + x3 y 2 + x1 y 3 ) 2

xi + yj x2 + y 2

SULIT

SULIT

3

3472/1

STATISTICS

1

x =

2

x =

3

4

5

6

∑x N

7

∑ fx ∑f

8

σ =

σ=

∑ f ( x − x) ∑f

m =

1  2N−F L+ C  fm   

I=

∑x

∑ ( x − x )2 = N

N

2

=

2

9

_2

−x

∑ fx ∑f

2

−x

2

10

P(A ∪ B) = P(A)+P(B)- P(A ∩ B)

11

P (X = r) = nCr p r q n − r , p + q = 1

12

Mean µ = np

13

σ = npq x−µ z= σ

14

Q1 × 100 Q0

∑ w1 I1 ∑ w1 n! n Pr = (n − r )! n! n Cr = (n − r )!r!

I=

TRIGONOMETRY

1 Arc length, s = r θ

9 sin (A ± B) = sinA cosB ± cosA sinB 1 2 rθ 2

2 Area of sector , L = 3 sin 2A + cos 2A = 1

10 cos (A ± B) = cosA cosB sinA sinB 11 tan (A ± B) =

tan A ± tan B 1 tan A tan B

4 sec2A = 1 + tan2A 5 cosec2 A = 1 + cot2 A

12

a b c = = sin A sin B sin C

13

a2 = b2 + c2 - 2bc cosA

6 sin 2A = 2 sinA cosA 2

2

7 cos 2A = cos A – sin A = 2 cos2A - 1 = 1 - 2 sin2A 8 tan 2A =

3472/1

14 Area of triangle

=

1 absin C 2

2 tan A 1 − tan 2 A

[ Lihat sebelah SULIT

Answer all questions. 1. Diagram 1 shows the relation between two sets of numbers . 15 12 9 6 3 0

. . 1

. 2

. 3

DIAGRAM 1 State, (a)

the image of 1,

(b)

the type of relation [ 2 marks ]

1

Answer : (a) …………………….. (b) ……………………...

2

2. Given f : x →

5 , x ≠ 0 and g : x → 3 x + 6 . Find x

(a) g −1 ( x ) , (b) fg −1 (3) . [ 4 marks]

Answer : (a) …………………….. 2 (b) ……………………... 4

SULIT 3

3

3472/1

For examiner’s use only

1 and 4 . 3 Give your answer in the form ax2 + bx + c = 0, where a, b and c are constants. [2 marks ] Form the quadratic equation which has the roots −

Answer : (a) ...………………………......

3 2

4

A quadratic equation 2x2 – x + p – 1 = 0 , has no roots. Find the range of values of p . [3 marks ]

4 Answer : .........…………………

5

2 Given that the roots of quadratic equation 2 x   h  1 x  k  0 are -3 and 6. Find (a) the value of h, (b) the value of k .

3

[1 marks] [2 marks]

Answer : (a) …………………….. For examiner’s use only

(b) ................................... 6 3472/1

It is given that the quadratic function f ( x)  2[ x  3  5] . 2

[ Lihat sebelah SULIT

5 3

(a) (b)

Write the equation of the axis of symmetry, State the coordinates of the minimum point. [2 marks]

Answer : (a) ……........................

6

(b)....……………….... 2

7

Find the range of the values of x for x(2x+5) ≥ 12. [3 marks]

7

Answer :

………………………...

3

8

Solve the equation

4 x 1  16 x 1 . 2

[3 marks]

8 3

Answer : .................................

9

Solve the equation log 5 (2 x  3)  1  log5 ( x  1)

For examiner’s use only

SULIT

3

3472/1 [3 marks]

9 Answer : ……..……...……….....

10

3

Given that log 3 2  m and log 3 5  n , express log 9 20 in terms of m and n. [4 marks]

10 Answer : ……………...……….....

4

11 The 5th term of an arithmetic progression is 45 and the 7th term is 5. Find a)

the first term and the common difference

[2 marks]

b)

the sum of the first six terms

[2

marks]

Answer: a)…...…………..…....... b) ....................................

3472/1

[ Lihat sebelah SULIT

11 4

12.

The nth term of a geometric progression can be determined by using the formula Tn = 2 3 - 2 n . Calculate a)

the common ratio of the progression.

[2 marks]

b)

the sum to infinity.

[2 marks]

Answer :a)……………...……….....

12

b) ......................................

4

13 Diagram 3 shows a linear graph of xy against

1 x

xy

 (2 , 3) 1 x

O

 (1 , −2) DIAGRAM 3 (a) Express y in terms of x. [ 3 marks ] (b) Find the value of y when x = 4 [ 1 marks ]

13 Answer : (a)

…………………….

4 (b) ……………………..

14

The points P ( 2k , k), Q ( h , t ) and R ( 2h , 3t) are on a straight line. Q divides PR internally in the ratio 2:3 . Express h in terms of t

For examiner’s use only

SULIT

3

3472/1 [ 3 marks ] .

14 Answer : h = ………..……….......... 15

3

Given point R (-2,0 ) and point S ( 2,3 ) . Point P moves such that PR : PS = 3: 2. Find the equation of the locus of P. [3 marks]

15 Answer : ........................................... 16 Diagram 3 shows a triangle OAB such that  = a  = b and 2  =  OA OB AP AB

3

A P  B

DIAGRAM 3

Find  in terms of a and b OP

For examiner’s use only

O [2 marks] 16

Answer : …...…………..……..…... 17.

3472/1

Diagram 4 shows vectors OR , OS

and OT drawn on a Cartesian plane. [ Lihat sebelah SULIT

2

y

T(8, 5) 

R(1, 2) 

S(3, 1) x

O DIAGRAM 4 Find the value of m and n such that OR  m OS  n OT .

[4 marks]

17 Answer : …...…………..……..…... 4

18.

Table 2 shows the number of story books read by a group of students in a certain school. Number of story books read Number of students

0 7

1 9

2 3

3 x

TABLE 2 (a) State the largest possible value of x given that the mode is 1. (b) State the largest possible value of x given that the median is 1. (c) Calculate the value of x given that the mean is 1. [2marks]

[1 mark] [1 mark]

Answer : a)…...…………..……..…...

18

b)…………………………

4

c)………………………… 19

Diagram 5 shows a circle with centre O. The length of the radius is 2.5 cm and the

For examiner’s use only

SULIT

3

3472/1

area of sector AOB is 6.25 cm2. B

O



A DIAGRAM 5 Calculate (a) the value of θ (b) the perimeter of the sector AOB.

[2 marks] [2 marks]

19 Answer : (a)…...…………..…….. 4

(b)................................ 20

Solve the equation cos 2x - cos x = 0 for 0o  x  360o .

[4 marks]

20 Answer : …...…………..……..…... For examiner’s use only

21 3472/1

It is given that sin A = p , 0o < A < 90o [ Lihat sebelah SULIT

4

Find (a) cos ( 90o – A ) (b) sin 2A in terms of p [3 marks]

21

Answer : 3

(a) …...…………..……..…... (b) ........................................

22 The curve y = x2 – 3x + 2 has a gradient of 2 at point P ( t , 5 ). Find (a) the value of t. (b) the equation of the normal at point P. [3 marks]

Answer : …..

22 3

(a)…...…………..…

(b) .................................. 23. Given that y  2 x ( x  5) . Find the rate of change of y at (2,1) when the rate of change of x is 3 units per second [3 marks]

23 3

Answer : …..…...

…...…………..…

SULIT

3

3472/1

For examiner’s use only

0

24 Given that ∫ (2 x + 3)dx = − 4 , where k is a constant. Find the possible value of k. k

[3 marks]

Answer : …...…………..……..…...

24 3

25. Given that y =

3x + 1 dy = 4 k ( x ) with k(x) is a function in terms of x. and 2 dx x 1

Find the value of

∫ k ( x) dx .

−1

[3 marks]

Answer : …...…………..……..…...

25

END OF THE QUESTION PAPER

3

3472/1

[ Lihat sebelah SULIT

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