SMJK PHOR TAY Revision Add Maths P2 Form 4 2013 Paper 2 1

Time: 2 hours 30 minutes

The functions f and g are defined as f : x → , x ≠ −and g : x → 8x. Find (a) f−1(x) [2 marks] −1

(b) f g(x) [2 marks] −1

(c) gf (x) [2 marks] 2

The functions f and g are defined as f : x → , x ≠ and g : x → −2x. Find (a) f−1(x) [2 marks] −1

(b) f g(x) [2 marks] −1

(c) gf (x) [2 marks] 3

The quadratic equation 5x2 + mx + n = 0 has roots −9 and 3. Find (a) the values of m and n [3 marks] 2

(b) the range of the values of 5x + mx + n = 7s such that s does not have real roots [4 marks] 4

The quadratic equation −3x2 + hx + k = 0 has roots 6 and −5. Find (a) the values of h and k [3 marks] (b) the range of the values of −3x + hx + k = −3s such that s does not have real roots [4 marks] 2

5

Diagram 1 shows a cuboid where its height is longer than the sides of the base.

Diagram 1 Given the total length of the sides of the cuboid is 84 cm and the total surface area of the cuboid is 144 cm2, find the volume of cuboid. [5 marks]

6

Table 1 shows the prices and the price indices for the year 1996 based on the year 1993 of four components, A, B, C and D, used to make a type of toy. Diagram 2 shows a pie chart which represents the relative quantity of the components used. Price per unit (RM) Component Year Year 1993 1996 A 8.70 15.57 B 5.00 10.00 C x 5.32 D 2.20 3.98 Table 1

Price index 179 w 140 181

Diagram 2 (a) Find the values of (i) w (ii) x [4 marks] (b) Calculate the composite price index for the production cost of the toy in the year 1996 based on the year 1993. [2 marks] (c) The composite price index for the production cost of the toy increases by 50% from the year 1996 to the year 2005. Calculate (i) the composite price index for the production cost of the toy in the year 2005 based on the year 1993, (ii) the price of a box of the toy in the year 2005 if its price in the year 1993 is RM28.30. [4 marks]

7

Solution by scale drawing is not accepted. Diagram 3 shows a trapezium OABC. The line OA is perpendicular to the line AB which intersects with y-axis at the point D. It is given that the equation of OA is y = −x and the equation of AB is 8y = ex − 113.

Diagram 3 (a) Find (i) the value of e, (ii) the coordinates of A. [4 marks] (b) Given AD : DB = 7 : 9, find (i) the coordinates of D. (ii) the equation of the straight line BC. [4 marks] (c) A point J moves such that JA = JB. Find the equation of the locus of J. [2 marks] 8

Solution by scale drawing is not accepted. Diagram 4 shows a triangle OAB. Point C(−6, 3) lies on the straight line AB. The coordinates of point B is (−, j).

Diagram 4 (a) Find the equation of the locus of a moving point P such that its distance from C is always 2units. [2 marks] (b) Given that point A and point B lie on the locus of P. Calculate (i) the value of j, (ii) the coordinates of A. [6 marks] (c) Hence, find the area, in unit2, of triangle OAB. [2 marks]

9

Table 1 shows the marks obtained by 122 students in a test. Marks 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69

Number x 34 y 25 3 Table 1

Given that the median mark is 43.5, find the values of x and y. Hence, state the modal class. [6 marks] 10

Table 2 shows the frequency distribution of the ages of residents in an apartment. Age (years) 1−5 6 − 10 11 − 15 16 − 20 21 − 25 Table 2

Number 5 33 40 16 3

Calculate the variance and the standard deviation of this distribution. [6 marks] 11

Diagram 5 shows two sectors OAB and OCD of two cocentric circles with centre O. AOD and BOC are straight lines.

Diagram 5 Given that ∠AOB = 3 radian, OB = p cm, OD = (p + 6) cm and the perimeter of the figure is 100 cm. Find (a) the value of p, [6 marks] (b) the difference between the area of sector OAB and the area of sector OCD. [4 marks] 12

Diagram 6 shows two circles. The larger circle has centre X and radius 16 cm. The smaller circle has centre Y and radius 13 cm. The circles touch at point C. The straight line AB is a

common tangent to the circles at point A and point B.

Diagram 6 [Use π = 3.142] Given that ∠AXC = θ radians, calculate (a) the value of θ, [2 marks] (b) the length of the minor arc BC, [3 marks] (c) the area of the shaded region. [5 marks] 13

Table 3 shows the prices of four ingredients in the making a type of cake. Ingredient W X Y Z

Price per kilogram (RM) Year 2000 Year 2008 5.90 p 4.50 8.37 7.30 9.86 q r Table 3

(a) The price index of ingredient W in the year 2008 based on the year 2000 is 163. Calculate the value of p [2 marks] (b) The price index of ingredient Z in the year 2008 based on the year 2000 is 177. The price per kilogram of ingredient Z in the year 2008 is RM7.39 more than its price in the year 2000. Calculate the values of q and r. [3 marks] (c) The composite price index for the cost of making the cake in the year 2008 based on the year 2000 is 167. Calculate (i) the price of a cake in the year 2000 if its price in the year 2008 is RM23.70, (ii) the value of s if the quatities of ingredients W, X, Y and Z are used in the ratio of 2 : 4 : s : 6. [5 marks] 14

Diagram 7 shows a quadrilateral EFGH with acute angle ∠FEH.

Diagram 7 Given that the area of ΔEFH is 38.53 cm2, calculate (a) ∠FEH, [3 marks] (b) the length of FH. [2 marks] (c) ∠FHG, [3 marks] (d) the area of quadrilateral EFGH. [2 marks] 15

Diagram 8 shows the monthly expenditure on the items P, Q, R and S. Table 3 shows the prices, the price indices for the year 1997 based on the year 1987 and weightages for the items.

Diagram 8 Item P Q R S

Price per unit (RM) Price index Year 1987 Year 1997 3.40 6.77 199 9.60 b 171 c 2.28 134 5.50 6.49 a Table 3

(a) Find the values of (i) a (ii) b (iii) c [4 marks]

(b) Calculate the composite price index for the items in the year 1997 based on the year 1987. [2 marks] (c) The total expenditure for the items in the year 1987 is RM1370. Calculate the total expenditure for the items in the year 1997. [2 marks] (d) The cost of the items increases by 9% from the year 1997 to the year 2001. Find the composite price index for the year 2001 based on the year 1987. [2 marks]

Answer: 1

(a) Let f−1(x) = y So f(y) = x =x 6 = x(5y + 3) 6 = 5xy + 3x 6 − 3x = 5xy y= Therefore f−1(x) = , x ≠ 0 (b) f−1g(x) = f−1(g(x)) = f−1(8x) = =,x≠0 −1 (c) gf (x) = g(f−1(x)) = g() = 8() =,x≠0

2

(a) Let f−1(x) = y So f(y) = x =x y + 4 = x(6y − 7) y + 4 = 6xy − 7x y − 6xy = −7x − 4 y(1 − 6x) = −7x − 4 y= Therefore, f−1(x) = , x ≠ (b) f−1g(x) = f−1(g(x)) = f−1(−2x) = =,x≠− (c) gf−1(x) = g(f−1(x)) = g() = −2() =,x≠

3

(a) (x − a)(x − b) = 0 (x + 9)(x − 3) = 0 x2 + 6x − 27 = 0 5x2 + 30x − 135 = 0 Therefore,

m = 30 and n = −135 (b) 5x2 + mx + n = 7s 5x2 + 30x − 135 − 7s = 0 The equation does not have real roots b2 − 4ac < 0 (30)2 − 4(5)(−135 − 7s) < 0 900 + 2700 + 140s < 0 140s < −3600 s<− 4

(a) (x − a)(x − b) = 0 (x − 6)(x + 5) = 0 x2 − x − 30 = 0 −3x2 + 3x + 90 = 0 Therefore, h = 3 and k = 90 (b) −3x2 + hx + k = −3s −3x2 + 3x + 90 + 3s = 0 The equation does not have real roots b2 − 4ac < 0 (3)2 − 4(−3)(90 + 3s) < 0 9 + 1080 + 36s < 0 36s < −1089 s<−

5

Total length of the sides = 8x + 4y = 84 y = 21 − 2x −−−− (1) Total surface area = 2x2 + 4xy = 144 x2 + 2xy = 72 −−−− (2) Substitute (1) into (2), x2 + 2x(21 − 2x) = 72 x2 + 42x − 4x2 = 72 −3x2 + 42x − 72 = 0 3x2 − 42x + 72 = 0 (x − 12)(x − 2) = 0 x = 12 or x = 2 Substitute x = 12 into (1), y = 21 − 2(12) = −3 Substitute x = 2 into (1), y = 21 − 2(2) = 17 17 is longer. Therefore, the height is 17 cm. The length of the sides of the base is 2 cm. Volume of cuboid = 2 × 2 × 17 = 68 cm3

6

(a) (i) Price index I = × 100 w = × 100 = 200 (ii) 140 = × 100 x = 3.80 (b) = = = 179 (c) (i) I2005 = × 179 = 268.5 (ii) P2005 = × 28.3 = RM75.99

7

(a) (i) mOA = − mAB = 8y = ex − 113 mAB = = e=7 (ii) y = −x −−−− (1) y = x − −−−− (2) Substitute (1) into (2), −x = x − −x = − x=7 Substitute x = 7 into (1), y = −(7) y = −8 Therefore, A = (7, −8) (b) (i) From 8y = 7x − 113, D = (0, −14) (0, −14) = (, ) =0 x = −9 =− y = −22 Therefore, B = (−9, −22) (ii) Equation of BC: y = −x + c −22 = −(−9) + c c=− y = −x − (c) Equation of locus J: Given JA = JB, = x2 − 14x + 49 + y2 + 16y + 64 = x2 + 18x + 81 + y2 + 44y + 484

0x2 − 32x − 28y − 452 = 0 8

(a) Given PC = Let P = (x, y), = Squaring both sides, (x + 6)2 + (y − 3)2 = x2 + 12x + 36 + y2 − 6y + 9 = 4x2 + 48x + 144 + 4y2 − 24y + 36 = 25 4x2 + 4y2 + 48x − 24y + 155 = 0 (b) (i) Given B(−, j) lies on the locus of P, 4(−)2 + 4j2 + 48(−) − 24j + 155 = 0 225 + 4j2 − 360 − 24j + 155 = 0 4j2 − 24j + 20 = 0 j2 − 6j + 5 = 0 (j − 5)(j − 1) = 0 ∴ j = 5 or j = 1 Based on the diagram, j = 1 (ii) C is the midpoint of AB. Let A = (x, y), = −6 x − = −12 x=− =3 y+1=6 y=5 The coordinates of A are (−4, 5) (c) Area of ΔOAB =|| =|−+| = |−33| = 16unit2

9

Marks Number 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69

Cumulative frequency

x

x

34

x + 34

y

x + y + 34

25

x + y + 59

3

x + y + 62

x + y + 62 = 122 x + y = 60 y = 60 − x

43.5 = 39.5 + []10 4 = ()10 240 − 4x = 270 − 10x 6x = 30 x=5 y = 60 − 5 = 55 Modal class = 40 − 49 10

Midpoint, fx f(x − )2 x 1−5 5 3 15 397.83 6− 33 8 264 507.09 10 11 − 40 13 520 46.66 15 16 − 16 18 288 591.46 20 21 − 3 23 69 368.3 25 ∑f ∑fx = ∑f(x − )2 = = 1156 1911.34 97 Mean, = = 11.92 Variance, σ2 = = 19.7 Standard deviation, σ = = 4.44

11

(a) Length of arc AB = (p)(3) = 3p Length of arc CD = (p + 6)(3) = Perimeter of the figure, 3p + + 2p + 2(p + 6) = 100 = 100 = 100 10p + 30 = 100 p=7 (b) Area of sector OAB = × 72 × 3 = 73.5 cm2 Area of sector OCD = × 132 × 3 = 253.5 cm2 Difference in area = 253.5 − 73.5 = 180 cm2

Age

f

12

(a) θ = cos−1 () = 84.06° = 84.06 × = 1.4673 rad. (b) ∠BYC = 180° − θ = 3.142 − 1.4673 = 1.6747 rad. Length of minor arc BC = 13 × 1.6747 = 21.771 cm (c) AB = = 28.844 cm Area of trapezium ABYX = × (16 + 13) × 28.844 = 418.238 cm2 Area of sector AXC = × 162 × 1.4673 = 187.814 cm2 Area of sector BYC = × 132 × 1.6747 = 141.512 cm2 Area of the shaded region = 418.238 − 187.814 − 141.512 = 88.912 cm2

13

(a) Price index, I = × 100 163 = × 100 p = 9.62 (b) Given 177 = × 100 r = q + 7.39 177 = × 100 177q = 100q + 739 q = 9.60 r = 9.6 + 7.39 = 16.99 (c) (i) 167 = × 100 P2000 = RM14.19 (ii) Given = 167 167 = 167 = s=4

14

(a) Area of ΔEFH = × 11.9 × 7.5 × sin ∠FEH 38.53 = × 11.9 × 7.5 × sin ∠FEH sin ∠FEH = 0.8634 ∠FEH = 59.7° (b) FH2 = 11.92 + 7.52 − 2 × 11.9 × 7.5 × cos 59.7° = 107.8018 FH = 10.38 cm (c) = sin ∠FGH = = 0.9647

∠FGH = 74.72° ∠FHG = 180° − 74.72° = 63.28° (d) Area of ΔFGH = × 10.38 × 7.2 × sin 63.28° = 33.38 cm2 Area of quadrilateral EFGH= 38.53 + 33.38 = 71.91 cm2 15

(a) (i) Price index I = × 100 a = × 100 = 118 (ii) 171 = × 100 b = 16.42 (iii) 134 = × 100 c = 1.70 (b) = = = 151 (c) × 100 = 151 P1997 = = RM2068.70 (d) × 100 = × × 100 = × × 100 = 164.59

Read more

Similar to

Popular now

Just for you