SMJK PHOR TAY Paper 1

ADDITIONAL MATHEMATICS FORM 4 FINAL REVISION 2013

Time: 2 hrs

This paper consists of 25 questions. Answer all questions. Write your answer clearly in the spaces provided in the question paper. Show your working. It may help you to get marks. If you wish to change your answer, erase the answer that you have done. Then write down the new answer. The diagrams in the questions provided are not drawn to scale unless stated. The marks allocated for each question are shown in brackets. This question paper must be handed in at the end of examination. 1

Given {(4, u), (4, t), (9, s), (9, t), (14, u), (14, v), (19, t)}. State the type of the relation. [2 marks] Answer:

2

Diagram 1 represents a function f: x → px2 + qx + 4.

Diagram 1 Find the (a) values of p and q, (b) the object which is mapped onto itself. [4 marks] Answer:

3

The functions of f and g are defined as f : x → −x − 2 and g : x → . Find the composite function of gf and fg. [3 marks] Answer:

4

It is given that −1 is one of the roots of the quadratic equation −10x2 − 6x + k = 0. Find the value of k. [2 marks] Answer:

5

Find the value of x in logx 512 = −.

1

6

The quadratic equation (−9n − 9)x2 − 2x + 1 = 0 has two different roots, n is a constant. Find the range of values of n. [2 marks] Answer:

7

Given y = + 9x2 − 7x, find . [3 marks] Answer:

8

Find the range of the values of x such that (6 − 7x)(x + 8) ≤ −127x + 216. [3 marks] Answer:

9

STU is a triangle with length of side TU = 12.3 cm, US = 26.2 cm and ∠T = 66°. Solve the triangle. [3 marks] Answer:

10

Express 5 logx a + 4 logx b − logx ab as a single logarithm. [2 marks] Answer:

11

The curve y = −2x2 − 53x + 62 has a maximum point at x = k, where k is a constant. Find the value k. [2 marks] Answer:

12

Given p = 2m and q = 2n, express each of the following in terms of m and/or n. (a) log2 (b) log16 p − log128 q [3 marks] Answer:

13

Find the point of intersection of the straight lines −7x + 2y = −4 and −9x − 8y = −9. [3 marks]

2

14

Find the equation of the straight line which passes through A(−7, 8) and is perpendicular to the straight line joining B(−5, −3) dan C(−9, 3). [3 marks] Answer:

15

Given the gradient of a straight line which pass through (0, 2) is −. Find the x-intercept of the straight line. [2 marks] Answer:

16

Diagram 1 shows the scores obtained by a group of students in a contest. 48, 50, 72, 64, 70, 58, 45 Diagram 1 Find the interquartile range for the data. [3 marks] Answer:

17

The functions f and g are defined as f : x → 5x + 7 and g : x → 7x − 1. Find gf−1(x). [3 marks] Answer:

18

19

Diagram 2 shows a sector OQR of a circle with centre O.

Diagram 2

3

Given that ∠QOR = 0.6798 radians and OP = PQ = OS = SR = 15 cm. Find (a) the length of arc QR, (b) the area of the shaded region. [3 marks] Answer:

20

Given y = 9x − 9, find using differentiatian by the first principle. [3 marks] Answer:

21

Diagram 4 shows a circle with centre O and radius 10 cm.

Diagram 4 Given that OR = RP and ∠ORQ = 90°, find (a) ∠ROQ in radians, (b) the area of the shaded region. [2 marks] Answer:

22

PQR is a triangle with length of side QR = 17.3 cm, RP = 27.2 cm and ∠Q = 125°. Solve the triangle. [3 marks] Answer:

23

ABC is a triangle with AB = 108 cm, BC = 8 cm and AC = 13 cm. Find the smallest angle in the triangle. [3 marks] Answer:

24

Table 3 shows the price index of two items for three years. Item

1990

1994

4

1994

R S

(1985 = 100) 150 176

(1985 = 100) c 184 Table 3

(1990 = 100) 71.3 d

Calculate the values of c and d. [3 marks] Answer:

25

Table 4 shows the price indices and weightages of four items in the year 1998 based on the year 1989. Item P Q R S

Price Index 92 113 155 j Table 4

Weightage 2 7 k 5

Given the price of item S is RM5.10 in the year 1989 and RM5.76 in the year 1998. The composite index for the 1998 is 120. Calculate (a) the value of j, (b) the value of k. [3 marks] Answer:

Many-to-many relation

2

(a) f(x) = px2 + qx + 4 f(−5) = −1 p(−5)2 + q(−5) + 4 = −1 25p − 5q + 4 = −1 25p − 5q = −5 −−−− E1 f(2) = 20 p(2)2 + q(2) + 4 = 20 4p + 2q + 4 = 20 4p + 2q = 16 −−−− E2 E1 × 2: 50p − 10q = −10 E2 × (−5): −20p − 10q = −80

−−−− E3 −−−− E4

5

E3 − E4: 70p = 70 p=1 Substitute p = 1 into E1. 25(1) − 5q = −5 −5q = −5 − 25 = −30 q=6 (b) f(x) = x2 + 6x + 4 x2 + 6x + 4 = x x2 + 5x + 4 = 0 (x + 1)(x + 4) = 0 x = −1, −4 3

Given f(x) = −x − 2 and g(x) = . gf(x) = g(f(x)) = g(−x − 2) = fg(x) = f(g(x)) =f =−−2 = −x − 2

4

−10(−1)2 − 6(−1) + k = 0 −10 + 6 + k = 0 k=4

5

logx 512 = − 512 = x x = 512 =2 =4

6

(−9n − 9)x2 − 2x + 1 = 0 The equation has two different roots b2 − 4ac > 0 (−2)2 − 4(−9n − 9)(1) > 0 4 + 36n + 36 > 0 36n > −40 n>−

7

y = 9x−3 + 9x2 − 7x = −27x−4 + 18x − 7 = 108x−5 + 18 = + 18

8

(6 − 7x)(x + 8) ≤ −127x + 216 −7x2 − 50x + 48 ≤ −127x + 216 −7x2 + 77x − 168 ≤ 0 −7(x2 − 11x + 24) ≤ 0 −7(x − 3)(x − 8) ≤ 0

6

The range of values of x is x ≤ 3 or x ≥ 8 9

Using sine rule, = sin S = = 0.4289 ∠S = 25.4° ∠U = 180° − 66° − 25.4° = 88.6° Using sine rule, = u= = 28.67 cm 10

5 logx a + 4 logx b − logx ab = logx a5 + logx b4 − logx ab = logx = logx a4b3

11

y = −2x2 − 53x + 62 = −4x − 53 When y is maximum, = 0 −4x − 53 = 0 x=− ∴k=−

12

(a) log2 = log2 p7 − log2 128 − log2 q5 = 7 log2 p − 7 log2 2 − 5 log2 q = 7m − 7 − 5n (b) log16 p − log128 q =− =− =−

13

−7x + 2y = −4 −−−− (1) −9x − 8y = −9 −−−− (2)

7

From (1), x = y + −−−− (3) Substitute (3) into (2), −9(y + ) − 8y = −9 −y − − 8y = −9 −y = − y= Substitute y = into (1), −7x + 2() = −4 −7x = − x= Intersection point = (, ) 14

Gradient of BC = =− (−)m2 = −1 m2 = Equation: y − 8 = (x + 7) 3y − 24 = 2x + 14 2x − 3y + 38 = 0

15

Gradient = − −= − x−intercept = −2 × −10 = 20

16

Rearrange the data, 45 48 50 58 64 70 72 First quartile = 48 Third quartile = 70 Interquartile range = 70 − 48 = 22

17

Let f−1(x) = y So f(y) = x 5y + 7 = x y= Therefore f−1(x) = gf−1(x) = g(f−1(x)) = g() = 7() − 1 =

18

8

= 105.88° 19

(a) Length of arc QR = 30 × 0.6798 = 20.394 cm (b) Area of sector OQR = (30)2(0.6798) = 305.91 cm2 Area of triangle OPS = ab sin C = × 15 × 15 × sin 0.6798 = 70.722 cm2 Area of the shaded region = 305.91 − 70.722 = 235.188 cm2

20

Step 1: Let δx be a small increment in x and δy be the corresponding small increment in y. Step 2: y = 9x − 9 ----- (1) y + δy = 9(x + δx) − 9 y + δy = 9x + 9δx − 9 ----- (2) Step 3: (2) − (1), δy = 9δx Step 4: =9

21

(a) ∠ROQ = cos−1() = 60° = 60 × = 1.0473 rad. (b) Area of sector POQ = r2θ = × (10)2 × 1.0473 = 52.365 cm2 Area of triangle OQR = × 5 × 10 × sin 60° = 21.652 cm2 Area of the shaded region = 52.365 − 21.652 = 30.713 cm2

22

Using sine rule,

9

= sin P = = 0.521 ∠P = 31.39° ∠R = 180° − 125° − 31.39° = 23.61° Using sine rule, = r= = 13.3 cm 23

Using sine rule, = sin C = = 0.5853 C = 35.82° A = 180° − 108° − 35.82° = 36.18° Area of ΔABC = bc sin A = × 13 × 8 sin 36.18° = 30.7 cm2 24

Item R: Given × 100 = 150 × 100 = 71.3 Therefore, c = × 100 = × × 100 = × × 100 = 107 Item S: Given × 100 = 176 × 100 = 184 Therefore, d = × 100 = × × 100 = × × 100 = 104.5

25

(a) Price index,

10

I = × 100 j = × 100 = 113 (b) Composite index, = = = Given = 120 = 120 1540 + 155k = 1680 + 120k 35k = 140 k=4

11

Add maths final revision 1 f4 2013