2008 STPM Maths T Q&A

1 Actual 2008 STPM Mathematics Examination Paper

PAPER 1

1 The functions f and g are defined by 1 f : x → —, x ∈R \ {0}; x g : x → 2x – 1, x ∈R. Find f ° g and its domain. [4 marks]

3 (x – 2)2 5 2 2 Show that ∫ ——— dx = — + 4 ln�—� [4 marks] 2 x2 3 3

3 Using definitions, show that, for any sets A, B and C, A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C). [5 marks]

1 4 If z is a complex number such that |z| = 1, find the real part of ——–. [6 marks] 1 – z 1 5 The polynomial p(x) = 2x3 + 4x2 + —x – k has factor (x + 1). 2

(a) Find the value of k. [2 marks]

(b) Factorise p(x) completely. [4 marks]

sin x – cos x d 2y dy 6 If y = —————–, show that —— = 2y—–. [6 marks] sin x + cos x dx2 dx

1 0 0 7 Matrix A is given by A = � 1 –1 0 � . 1 –2 1

(a) Show that A2 = I, where I is the 3 × 3 identity matrix, and deduce A–1. [4 marks]

1 4 3(b) Find matrix B which satisfies BA = � 0 2 1 � . [4 marks]

–1 0 2

8 The lines y = 2x and y = x intersect the curve y2 + 7xy = 18 at points A and B respectively, where A and B lie in the first quadrant.(a) Find the coordinates of A and B. [4 marks]

(b) Calculate the perpendicular distance of A to OB, where O is the origin. [2 marks]

(c) Find the area of the OAB triangle. [3 marks]

9 Find the solution set of the inequality 4 3 �——� > 3 – —. [10 marks] x – 1 x x 10 Show that the gradient of the curve y = ——– is always decreasing. [3 marks] x2 – 1 Determine the coordinates of the point of inflexion of the curve, and state the intervals for

which the curve is concave upward. [5 marks]

Sketch the curve. [3 marks]

AAhead STPM Math V2 Act 08 4th.indd 1AAhead STPM Math V2 Act 08 4th.indd 1 07/01/2009 10:08:49 AM07/01/2009 10:08:49 AM


11 Sketch, on the same coordinate axes, the curves y = 6 – ex and y = 5e–x, and find the coordinates of the points of intersection. [7 marks]

Calculate the area of the region bounded by the curves. [4 marks]

Calculate the volume of the solid formed when the region is rotated through 2π radians about the x-axis. [5 marks]

12 At the beginning of this year, Mr. Liu and Miss Dora deposited RM10 000 and RM2000 respectively in a bank. They receive an interest of 4% per annum. Mr Liu does not make any additional deposit nor withdrawal, whereas, Miss Dora continues to deposit RM2000 at the beginning of each of the subsequent years without any withdrawal.(a) Calculate the total savings of Mr. Liu at the end of nth year. [3 marks]

(b) Calculate the total savings of Miss Dora at the end of nth year. [7 marks]

(c) Determine in which year the total savings of Miss Dora exceeds the total savings of Mr. Liu. [5 marks]

PAPER 2

1 Show that the substitution u = x2 + y transforms the differential equation dy (1 – x)—– + 2y + 2x = 0 dx into the differential equation du (1 – x) —– = –2u. [3 marks] dx

2 In triangle ABC, the point X divides BC internally in the ratio m : n, where m + n = 1. Express AX2 in terms of AB, BC, CA, m and n. [5 marks]

θ 2t 1 – t2

3 If t = tan —, show that sin θ = ——– and cos θ = ——–. [4 marks] 2 1 + t2 1 + t2

Hence, find the values of θ between 0° and 360° that satisfy the equation 10 sin θ – 5 cos θ = 2. [3 marks]

4 The diagram below shows the circumscribed circle of the triangle ABC.

The tangent to the circle at A meets the line BC extended to T. The angle bisector of the angle ATB cuts AC at P, AB at Q and the circle at R. Show that(a) triangles APT and BQT are similar, [4 marks]

(b) PT • BT = QT • AT, [2 marks]

(c) AP = AQ. [4 marks]

5 The position vectors of the points A, B and C, with respect to the origin O, are a, b and c respectively. The points L, M, P and Q are the midpoints of OA, BC, OB, and AC respectively.

C

T

P

A

QR

B



1 1(a) Show that the position vector of any point on the line LM is —a + —λ(b + c – a) for 2 2 some scalar λ, and express the position vector of any point on the line PQ in terms of

a, b and c. [6 marks]

(b) Find the position vector of the point of intersection of the line LM and the line PQ. [4 marks]

6 A 50 litre tank is initially filled with 10 litres of brine solution containing 20 kg of salt. Starting from time t = 0, distilled water is poured into the tank at a constant rate of 4 litres per minute. At the same time, the mixture leaves the tank at a constant rate of k litres per minute, where k > 0. The time taken for overflow to occur is 20 minutes.(a) Let Q be the amount of salt in the tank at time t minutes. Show that the rate of change of

Q is given by

dQ Q k —– = – ——————. dt 10 + (4 – k)t

Hence, express Q in terms of t. [7 marks]

(b) Show that k = 4, and calculate the amount of salt in the tank at the instant overflow occurs. [6 marks]

(c) Sketch the graph of Q against t for 0 ≤ t ≤ 20. [2 marks]

7 There are 12 towels, two of which are red. If five towels are chosen at random, find the probability that at least one is red. [4 marks]

1 8 The random variable X has a binomial distribution with parameters n = 500 and p = —. 2 Using a suitable approximate distribution, find P(|X – E(X)| ≤ 25). [6 marks]

9 In a basket of mangoes and papayas, 70% of mangoes and 60% of papayas are ripe. If 40% of the fruits in the basket are mangoes,(a) find the percentage of the fruits which are ripe, [3 marks]

(b) find the percentage of the ripe fruits which are mangoes. [4 marks]

10 A sample of 100 fuses, nominally rated at 13 amperes, are tested by passing increasing electric current through them. The current at which they blow are recorded and the following cumulative frequency table is obtained.

Current (amperes) Cumalative frequency

< 10 0

< 11 8

< 12 30

< 13 63

< 14 88

< 15 97

< 16 99

< 17 100

Calculate the estimates of the mean, median and mode. Comment on the distribution. [8 marks]



SUGGESTED ANSWERS

PAPER 1 1 1. f : x → —, x ∈R \ {0} x g : x → 2x – 1, x ∈R f ° g = fg(x) = f(2x – 1) 1 1 = ———, x ≠ — 2x – 1 2 1 The domain of f ° g is �x : x ∈R, x ≠ —�. 2 3 (x – 2)2

2. ∫ ——— dx 2 x2

3 x2 – 4x + 4 = ∫ �————–—� dx 2 x2

3 4 = ∫ �1 – — + 4x–2� dx 2 x x –1 3

= �x – 4 ln |x| + 4�——�� –1 2

4 3

= �x – 4 ln |x| – —� x 2

4 4 = 3 – 4 ln 3 – — – �2 – 4 ln 2 – —� 3 2 5 = — + 4 ln 2 – 4 ln 3 3 5 = — + 4(ln 2 – ln 3) 3 5 2 = — + 4 ln �—� [shown] 3 3

3. By using basic defi nitions of sets, let x ∈A ∩ (B ∪ C), then, x ∈A and x ∈ (B ∪ C) x ∈A and (x ∈B or x ∈C) (x ∈A and x ∈B) or (x ∈A and x ∈C) (x ∈A ∩ B) or (x ∈A ∩ C) x ∈ (A ∩ B) ∪ (A ∩ C) Thus, all elements in A ∩ (B ∪ C) are also

found in (A ∩ B) ∪ (A ∩ C). Hence, A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C)

[shown]

4. Let z = x + yi. |z| = 1

x2 + y2 = 1 x2 + y2 = 1 … (1) 1 1 ——– = ————— 1 – z 1 – (x + yi) 1 = ————— 1 – x – yi 1 – x + yi = —————————— (1 – x – yi)(1 – x + yi) 1 – x + yi = —————— (1 – x)2 + y2

1 – x + yi = ——————— 1 – 2x + x2 + y2

1 – x + yi = ————— 1 – 2x + 1

1 – x + yi = ————– 2 – 2x

A \ B means A – B or A ∩ B'.

11 The continuous random variable X has probability density function

0, x < 0 5

f(x) =

⎫⎪⎪⎬⎪⎪⎭

— – x, 0 ≤ x < 1, 4 1 ——, x ≥ 1. 4x2

(a) Find the cululative distribution function of X. [7 marks]

(b) Calculate the probability that at least one of two independent observed values of X is greater than three. [4 marks]

12 A car rental shop has four cars to be rented out on a daily basis at RM50.00 per car. The average daily demand for cars is four.(a) Find the probability that, on a particular day,

(i) no cars are requested, [2 marks]

(ii) at least four requests for cars are received. [2 marks]

(b) Calculate the expected daily income received from the rentals. [5 marks]

(c) If the shop wishes to have one more car, the additional cost incurred is RM20.00 per day. Determine whether the shop should buy another car for rental. [5 marks]

From (1)



1 – x + yi = ————— 2(1 – x) 1 – x y = ———— + ————i 2(1 – x) 2(1 – x) 1 y = — + ————i 2 2(1 – x) 1 1 Hence, the real part of ——– is —. 1 – z 2

1 5. (a) p(x) = 2x3 + 4x2 + —x – k 2 Since (x + 1) is a factor of p(x), then p(–1) = 0 1 2(–1)3 + 4(–1)2 + —(–1) – k = 0 2 1 –2 + 4 – — – k = 0 2 3 — – k = 0 2 3 k = — 2

1 3(b) Therefore, p(x) = 2x3 + 4x2 + —x – — 2 2 3 2x2 + 2x – — 2 1 3 x + 1 2x3 + 4x2 + —x – — 2 2 2x3 + 2x2

1 2x2 + — x 2 2x2 + 2x 3 3 – —x – — 2 2 3 3 – —x – — 2 2

0 3 Hence, p(x) = (x + 1)�2x2 + 2x – —� 2 4x2 + 4x – 3 = (x + 1)�——————� 2 1 = — (x + 1)(2x + 3)(2x – 1) 2

sin x – cos x 6. y = —————— sin x + cos x (sin x + cos x)y = sin x – cos x dy (sin x + cos x)—– + dx y(cos x – sin x) = cos x + sin x dy (sin x + cos x)�—– – 1� + y(cos x – sin x) = 0 dx

d2y dy (sin x + cos x) —— + �—– – 1�(cos x – sin x) dx2 dx dy + y(–sin x – cos x) + (cos x – sin x) —— = 0 dx

d2y (sin x + cos x)—— dx2

dy dy + �—– – 1 + —–�(cos x – sin x) dx dx + y(–sin x – cos x) = 0 d2y (sin x + cos x) —— dx2

dy + �2—– – 1�(cos x – sin x) dx + y(–sin x – cos x) = 0 d2y dy (sin x + cos x) —— = �2—– – 1�(sin x – cos x) + dx2 dx y(sin x + cos x) d2y dy sin x – cos x —— = �2—– – 1��——–—––—� dx2 dx sin x + cos x sin x + cos x + y�——————�

sin x + cos x d2y dy —— = �2—– – 1� y + y dx2 dx d2y dy —— = 2y —– [shown] dx2 dx

7. (a) A2 = AA 1 0 0 1 0 0 = � 1 –1 0 � � 1 –1 0 � 1 –2 1 1 –2 1

1 0 0 = � 0 1 0 � 0 0 1 = I [shown] A2 = I AA = I

AA–1 = I By comparison, A–1 = A 1 0 0 = � 1 –1 0 � 1 –2 1

1 4 3(b) BA = � 0 2 1 � –1 0 2

1 4 3 BAA–1 = � 0 2 1 � A–1

–1 0 2

1 4 3 BI = � 0 2 1 � A–1

–1 0 2

1 4 3 1 0 0 B = � 0 2 1 � � 1 –1 0 � –1 0 2 1 –2 1

8 –10 3 = � 3 –4 1 � 1 –4 2

8. (a) y = 2x ............. (1) y = x ............. (2) y2 + 7xy = 18 ............. (3)



Substituting (1) into (3), (2x)2 + 7x(2x) = 18 4x2 + 14x2 = 18 18x2 = 18 x2 = 1 x = ±1 x = –1 is not accepted because point A lies

in the fi rst quadrant. Thus, x = 1. When x = 1, y = 2(1) = 2 Hence, the coordinates of point A are

(1, 2). Substituting (2) into (3), x2 + 7x2 = 18 8x2 = 18 9 x2 = — 4 3 x2 = ± — 2 3 x = – — is not accepted because point B 2 lies in the fi rst quadrant. 3 Thus, x = —. 2 3 3 When x = —, y = — 2 2 Hence, the coordinates of point B are 3 3 �—, —�. 2 2

3 — – 0 y – 0 2(b) The equation of OB is ——— = ———– x – 0 3 — – 0 2 y = x x – y = 0 The perpendicular distance from A(1, 2) to

OB

|1 – 2| = ————— 12 + (–1)2

1 = —– 2

2 = —– units 2 3 0 1 — 0 1 2(c) Area of ∆OAB = — � � 2 3 0 2 — 0 2

1 3 = —�— – 3� 2 2

3 = �– —� 4

3 = — units2

4

4 9. y = �———� x – 1

4 ——–, x > 1 x – 1 y =

⎫⎪⎬⎪⎭

4 – �——–�, x < 1 x – 1

As y → ∞, x – 1 → 0 x → 1 Thus, x = 1 is the asymptote. As x → ±∞, y → 0. 4 The graph of y = �——–� is as shown. x – 1

3 y = 3 – — x As y → ∞, x → 0 Thus, x = 0 (the y-axis) is the asymptote. As x → ±∞, y → 3. 3 The graph of y = 3 – — is as shown. x The x-coordinate of point A is obtained by

solving the following equations simultaneously. 4 y = ——– ... (1) x – 1 3 y = 3 – — ... (2) x 4 3 ——— = 3 – — x – 1 x 4 3x – 3 ——– = ——— x – 1 x (3x – 3)(x – 1) = 4x 3x2 – 6x + 3 – 4x = 0 3x2 – 10x + 3 = 0 (3x – 1)(x – 3) = 0 1 x = — or 3 3 1 x = — is not accepted 3 Thus, x = 3 The solution set for which

4 3 �——–� > 3 – — is given by the part of the x – 1 x 4 graph where the curve y = �——–� x – 1 3 is above the curve y = 3 – —, that is, x {x : 0 < x < 1 or 1 < x < 3}.

x

y

31

34

4y = ——— x – 1

3y = 3 – —– x

3y = 3 – —– x

4y = – �———� x – 1

O

A



x10. y = ——— x2 – 1 dy (x2 – 1)(1) – x(2x) —– = ———————— dx (x2 – 1)2

–x2 – 1 = ———— (x2 – 1)2

–(x2 + 1) = ———— (x2 – 1)2

dy Since —– < 0 for all real values of x, then dx the gradient of the curve is always decreasing. [shown]

d2y –(x2 – 1)2(2x) + (x2 + 1)(2)(x2 – 1)(2x) —— = ———————————————–– dx2 (x2 – 1)4

–2x(x2 – 1)[x2 – 1 – 2(x2 + 1)] = ————————————— (x2 – 1)4

–2x (–x2 – 3) = —————— (x2 – 1)3

2x(x2 + 3) = ————— (x2 – 1)3

d 2y When —— = 0, dx2

2x(x2 + 3) ————– = 0 (x2 – 1)3

x = 0 0 When x = 0, y = ——— = 0 02 – 1

(x2 – 1)3(6x2 + 6) – d3y (2x)(x2 + 3)(3)(x2 – 1)2(2x) —— = ———————————— dx3 (x2 – 1)6

6(x2 – 1)3(x2 + 1) – (12x2)(x2 + 3)(x2 – 1)2

= ———————————————— (x2 – 1)6

6(x2 – 1)2[(x2 – 1)(x2 + 1) – (2x2)(x2 + 3)] = ———————————————— (x2 – 1)6

6(x2 – 1)2(x4 – 1 – 2x4 – 6x2) = ———————————— (x2 – 1)6

6(–x4 – 1 – 6x2) = ——————— (x2 – 1)4

d3y 6[–04 – 1 – 6(0)2] When x = 0, —— = ———————— = –6 dx3 (02 – 1)4

(that is ≠ 0)

d 2y d3y Since —— = 0 and —— ≠ 0 dx2 dx3

when x = 0, then (0, 0) is the point of infl exion. When the curve concaves upwards. d 2y —— > 0 dx2

2x(x2 + 3) ————— > 0 (x2 – 1)3

2x(x2 + 3) ——————— > 0 [(x + 1)(x – 1)]3

2x(x2 + 3) ——————— > 0 (x + 1)3(x – 1)3

Hence, the intervals for which the curve concaves upwards are –1 < x < 0 or x > 1

and the intervals for which the curve concaves downwards are x < –1 or 0 < x < 1.

x The curve y = ——— is as shown below. x2 – 1

11. y = 6 – ex

On the x-axis, y = 0 6 – ex = 0 ex = 6 x = ln 6 Thus, the curve y = 6 – ex intersects the x-axis

at (ln 6, 0). On the y-axis, x = 0 y = 6 – e0

y = 5 Thus, the curve y = 6 – ex intersects the y-axis

at (0, 5). As x → ∞, y → –∞ As x → –∞, y → 6

y = 5e–x

On the y-axis, x = 0 y = 5(e0) y = 5 Therefore, the curve y = 5e–x intersects the

y-axis at (0, 5). As x → ∞, y → 0. As x → –∞, y → ∞ The curve y = 6 – ex and y = 5e–x are as

shown.

x

y

1–1 O

+ + + +

– – – +

– – + +

– + + +

– + – +–1 0 1x

(x + 1)3 > 0

x > 0

(x – 1)3 > 0

x2 + 3 > 0

x

y

ln 6

(ln 5, 1)y = 5e–x

y = 6 – ex

65

O



y = 6 – ex ... (1) y = 5e–x ... (2) Substituting (1) into (2), 6 – ex = 5e–x

6ex – (ex)2 = 5 Letting ex = p, 6p – p2 = 5 p2 – 6p + 5 = 0 (p – 1)(p – 5) = 0 p = 1 or 5 When p = 1, ex = 1 x = ln 1 x = 0 When x = 0, y = 6 – e0 = 5 When p = 5, ex = 5 x = ln 5 When x = ln 5, y = 6 – eln 5 = 6 – 5 = 1 Hence, the points of intersection are (0, 5) and

(ln 5, 1). Area of the shaded region ln 5

= ∫ [(6 – ex) – 5e–x] dx 0

5 ln 5 = �6x – ex – —— e–x� (–1) 0

5 ln 5

= �6x – ex + —–� ex 0

5 5 = 6 ln 5 – eln 5 + —— – �0 – e0 + —–� eln 5 e0

5 = 6 ln 5 – 5 + — – (–1 + 5) 5 = 6 ln 5 – 5 + 1 + 1 – 5 = (6 ln 5 – 8) units2

Volume of the solid generated

ln 5

= π∫ [(6 – ex)2 – (5e–x)2] dx 0

ln 5

= π∫ [36 – 12ex + e2x – 25e–2x] dx 0

1 25 ln 5

= π �36x – 12ex + —e2x – —— e–2x� 2 (–2) 0

1 25 ln 5

= π �36x – 12ex + — e2x + ——–� 2 2e2x 0

1 25 = π�36 ln 5 – 12eln 5 + —e2 ln 5 + ——–– 2 2e2 ln 5

1 25 – �0 – 12e0 + —e0 + ——�� 2 2e0

1 25 = π[36 ln 5 – 12(5) + —(25) + ——— 2 2(25) 1 25 – �–12 + — + —–�� 2 2 = π(36 ln 5 – 48) = 12(3 ln 5 – 4)π units3

12. (a) For Mr. Liu: At the end of the 1st year, the total

savings, U1 = 1.04 × 10 000 At the end of the 2nd year, the total

savings, U2 = 1.04(1.04 × 10 000) = 1.042 × 10 000 At the end of the 3rd year, the total

savings, U3 = 1.04(1.042 × 10 000) = 1.043 × 10 000 � At the end of the nth year, the total

savings, Un = 1.04n × 10 000 = RM10 000(1.04n)(b) For Miss Dora: At the end of the 1st year, the total

savings, U1 = 1.04 × 2000 At the beginning of the 2nd year, the total

savings = (1.04 × 2000) + 2000 At the end of the 2nd year, the total

savings, U2 = 1.04[(1.04 × 2000) + 2000] = 1.042 × 2000 + 1.04 × 2000 At the beginning of the 3rd year, the total

savings = (1.042 × 2000 + 1.04 × 2000) + 2000

At the end of the 3rd year, the total savings,

U3 = 1.04[1.042 × 2000 + 1.04 × 2000 + 2000]

= 1.043 × 2000 + 1.042 × 2000 + 1.04 × 2000 � At the end of the nth year, the total

savings, Un = 1.04n × 2000 + ... + 1.042 × 2000

+ 1.04 × 2000 = 2000(1.04n + ... + 1.042 + 1.04) 1.04 (1.04n – 1) = 2000�——–——–——� 1.04 – 1 = 52 000(1.04n – 1)(c) When the total savings of Miss Dora

exceeds the total savings of Mr. Liu, 52 000(1.04n – 1) > 10 000 (1.04n) 5.2(1.04n – 1) > 1.04n

5.2(1.04n) – 5.2 > 1.04n

(5.2 – 1)(1.04n) > 5.2 4.2 (1.04n) > 5.2 5.2 1.04n > —— 4.2 5.2 n ln 1.04 > ln �——� 4.2 5.2 ln �——� 4.2 n > ————— ln 1.04 n > 5.45 Smallest integer value of n = 6 Hence, the total savings of Miss Dora

exceeds the total savings of Mr. Liu at the end of the 6th year.



Paper 2

1. u = x2 + y

du dy —– = 2x + —– dx dx

dy du —– = —– – 2x dx dx dy (1 – x)—– + 2y + 2x = 0 dx

du (1 – x)�—– – 2x� + 2(u – x2) + 2x = 0 dx

du (1 – x)—– – 2x(1 – x) + 2u – 2x2 + 2x = 0 dx

du (1 – x)—– – 2x + 2x2 + 2u – 2x2 + 2x = 0 dx

du (1 – x)—– + 2u = 0 dx

du (1 – x)—– = –2u dx [shown]

m 2. BX = �———�BC m + n

m BX = �—� BC 1 BX = mBC

In ∆ABC, by using the cosine rule,

AB2 + BC 2 – CA2

cos ∠ABC = ——————— 2(AB)(BC) In ∆ABX, by using the cosine rule, AX2 = AB2 + BX2 – 2(AB)(BX) cos ∠ABC = AB2 + (mBC)2

AB2 + BC 2 – CA2

– 2(AB)(mBC)�———–————� 2(AB)(BC) = AB2 + m2BC 2 – m(AB2 + BC 2 – CA2) = AB2 + m2BC 2 – mAB2 – mBC 2 + mCA2

= (1 – m)AB2 + (m2 – m)BC 2 + mCA2

= (1 – m)AB2 + m(m – 1)BC 2 + mCA2

= nAB2 + m(–n)BC 2 + mCA2

= nAB2 – mnBC 2 + mCA2

θ 2 tan — 2 3. tan θ = —————— θ 1 – tan2 — 2

2t tan θ = ——— 1 – t2

AB = AC2 + BC2

= (1 – t2)2 + (2t)2

= 1 – 2t2 + t4 + 4t2

= 1 + 2t2 + t4

= (1 + t2)2

= 1 + t2

A

Cnm XB

Based on ∆ABC, 2t sin θ = ——— [shown] and 1 + t2

1 – t2 cos θ = ——— [shown] 1 + t2

10 sin θ – 5 cos θ = 2 2t 1 – t2

10�——–� – 5�———� = 2 1 + t2 1 + t2

10(2t) – 5(1 – t2) = 2(1 + t2) 20t – 5 + 5t2 = 2 + 2t2

3t2 + 20t – 7 = 0 (3t – 1)(t + 7) = 0 1 t = — or t = –7 3 1 When t = —, 3 θ 1 tan — = — 2 3 θ — = 18.43° 2 θ = 36.9° [correct to one decimal place] When t = –7, θ tan — = –7 2 θ — = 98.13° 2 θ = 196.3° [correct to one decimal place]

4.

(a) ∠BTQ = ∠ATP [RQPT is the angle bisector of ∠ATB]

∠PAT = ∠QBT [Alternate segment theorem]

APT Hence, ∆—— are similar [shown]. BQT APT(b) Since ∆—— are similar, then BQT AT PT —– = —– BT QT PT • BT = QT • AT [shown] APT(c) Since ∆—— are similar, BQT

then ∠APT = ∠BQT. Let ∠APT = ∠BQT = θ. ∠APQ = 180° – θ [Angles on a straight

line] ∠AQP = 180° – θ [Angles on a straight

line] Since ∠APQ = ∠AQP, then ∆APQ is an isosceles triangle where

AP = AQ [shown].

2t

CA

B

θ

1 + t2

1 – t2

T

B

R

Q

A

CP

θ

θ



5.

(a) Let X be a point on the line LM. → → → OX = OL + LX → → = OL + λ LM, where LX = λ LM → → → → = OL + λ(LO + OB + BM)

→ → → 1 → = OL + λ�LO + OB + —BC� 2

→ → → 1 → → = OL + λ�LO + OB + —(BO + OC)� 2

1 1 1 = —a + λ�– —a + b + —(–b + c)� 2 2 2

1 1 1 1 = —a + λ �– —a + b – —b + —c� 2 2 2 2

1 1 1 1 = —a + λ�– —a + —b + —c� 2 2 2 2

1 1 = —a + —λ(–a + b + c) 2 2 1 1 = —a + —λ(b + c – a) [shown] 2 2 Hence, the position vector of any point on 1 1 the line LM is —a + —λ(b + c – a). [shown] 2 2 Let Y be a point on the line PQ. → → → OY = OP + PY → → = OP + µPQ, where PY = µPQ → → → → = OP + µ(PO + OA + AQ)

→ → → 1 → = OP + µ�PO + OA + —AC� 2

→ → → 1 → → = OP + µ[PO + OA + —(AO + OC)] 2 1 1 1 = —b + µ�– —b + a + —(–a + c)� 2 2 2

1 1 1 1 = —b + µ�– —b + a – —a + —c� 2 2 2 2

1 1 1 1 = —b + µ�– —b + —a + —c� 2 2 2 2 1 1 = —b + —µ(–b + a + c) 2 2 1 1 = —b + —µ(a + c – b) 2 2 Hence, the position vector of any point on 1 1 the line PQ is —b + —µ(a + c – b). 2 2

→ → →(b) YX = OX – OY 1 1 = —a + —λ(b + c – a) 2 2 1 1 – �—b + —µ(a + c – b)� 2 2 1 1 1 1 1 = —a + —λb + —λc – —λa – —b 2 2 2 2 2 1 1 1 – —µa – —µc + —µb 2 2 2 1 1 1 1 1 = �— – —λ – —µ�a + �—λ – — 2 2 2 2 2 1 1 1 + —µ�b + �—λ – —µ�c 2 2 2 Let T be the point of interesection of the

line LM and the line PQ. → At point T, YX = 0. 1 1 1 1 1 1 �— – —λ – —µ�a + �—λ – — + —µ�b 2 2 2 2 2 2 1 1 + �—λ – —µ� c = 0 2 2 1 1 1 — – —λ – —µ = 0 2 2 2 1 – λ – µ = 0 λ + µ = 1 ...(1) 1 1 1 —λ – — + —µ = 0 2 2 2 λ – 1 + µ = 0 λ + µ = 1 1 1 —λ – —µ = 0 2 2 λ – µ = 0 ...(2) (1) + (2): 2λ = 1 1 λ = — 2 1 From (1): — + µ = 1

2 1 µ = — 2 1 → Substituting µ = — into OY, 2

→ 1 1 1 OT = —b + —�—�(a + c – b) 2 2 2 1 1 1 1 = —b + —a + —c – —b 2 4 4 4 1 1 1 = —a + —b + —c 4 4 4 1 = —(a + b + c)

4 Hence, the position vector of the point of

intersection of the line LM and the line 1 PQ is —(a + b + c).

4 6. (a) Let V be the volume of solution in the

tank at time t minutes.

y

O

L

A

Q

CMB

P

X

Y T

a

c

b

x



Rate of change of volume of solution

Change in volume of solution = —————————————– Change in time

dV δV —– = —– dt δt

dV δV = —– × δt dt = (4 – k)(t – 0) = (4 – k)t

Hence, the volume of solution in the tank at time t minutes

= Initial volume of solution + δV = 10 + (4 – k)t

It is given that the amount of salt in the tank at time t minutes is Q.

Rate of change of amount of salt (kg/min) Amount of salt at time t minutes (kg)

× Rate of change of volume of mixture leaving the tank (l/min)

= ————————————————– Volume of solution at time t minutes (l)

dQ Q × (– k) Hence, —— = —————— dt 10 + (4 – k)t

Q k = – —————— [shown] 10 + (4 – k)t

Q dQ t k dt ∫ —– = –∫ —————— 20 Q 0 10 + (4 – k)t

Q dQ t dt ∫ —– = – k∫ —————— 20 Q 0 10 + (4 – k)t

Q dQ – k t (4 – k) dt ∫ —– = ——– ∫ —————— 20 Q 4 – k 0 10 + (4 – k)t

Q k t [ln |Q|] = �——–ln �10 + (4 – k)t �� 20 k – 4 0

k ln |Q| – ln 20 = ——— [ln |10 + (4 – k)t| k – 4 – ln 10]

Q k 10 + (4 – k)t ln�—–� = ———�ln�———–——�� 20 k – 4 10

k Q 10 + (4 – k)t ——

ln�—–� = ln��———–—–—� k – 4 � 20 10

k Q 10 + (4 – k)t ——

—– = �——————� k – 4

20 10

k 4 – k ——

Q = 20�1 + �———� t� k – 4

10

dV(b) —– = 4 – k dt 50 20

∫ dV = ∫ (4 – k) dt 10 0

50 20

[V] = �(4 – k)t� 10 0

50 – 10 = (4 – k)(20 – 0) 2 = 4 – k k = 2 k = 4 [shown] When k = 4, 4 4 – 4 ———

Q = 20�1 + �——–—�t� 4 – 4

10

1 = 20�1 + —t�

–1

5 1 = 20�——–—� 1 1 + —t 5 5 = 20�—–—� 5 + t 100 = ——— 5 + t 100 When t = 20, Q = ——— = 4 5 + 20 Hence, at the instant overfl ow occurs, the

amount of salt is 4 kg.(c)

7. P(at least one red towel is chosen)= 1 – P(all the fi ve towels chosen are not red) 10 9 8 7 6= 1 – �—– × —– × —– × — × —� 12 11 10 9 8 7= 1 – —– 22

15 = —– 22

Alternative method P(at least one red towel is chosen) = 1 – P(all the fi ve towels chosen are not red)

2C0 × 10C5 = 1 – ————— 12C5

Rate of volume of distilled water pouring into the tank –Rate of volume of mixture leaving the tank

t

Q

O

20

4

20

Take NoteTake Note

This event does not follow a binomial distribution because the towels are chosen without replacement.



252 = 1 – ——– 792 15 = —– 22

8. X ~ B(n, p) 1 X ~ B�500, —� 2 1 µ = E(X) = np = 500 × — = 250 2 1 1 σ 2 = npq = 500 × — × — = 125 2 2 1 Since n > 50, p = — and np > 5, then the 2 normal approximation is used. X ~ N(250, 125) approximately P(|X – E(X)| ≤ 25) = P(|X – 250| ≤ 25) = P(–25 ≤ X – 250 ≤ 25) = P(225 ≤ X ≤ 275) = P(225 – 0.5 < X < 275 + 0.5) [Taking

continuity correction] = P(224.5 < X < 275.5) 224.5 – 250 X – 250 275.5 – 250 = P �————— < ———–– < ———–––—� 125 125 125 = P(–2.281 < Z < 2.281) = 1 – 0.0113 – 0.0113 = 0.9774

9. M – Event that a mango is chosen P – Event that a papaya is chosen R – Event that a ripe fruit is chosen

R

—R

0.7

0.3

M

R

—R

0.6

0.4

P

0.4

0.6

Outcomes

MR

—MR

PR

—PR

(a) P(fruits are ripe) = P(MR) + P(PR) = (0.4 × 0.7) + (0.6 × 0.6) = 0.64 Hence, the percentage of fruits which are

ripe = 0.64 × 100 = 64%(b) P(M / R) P(M ∩ R) = ———— P(R)

0.4 × 0.7 = ———— 0.64 = 0.4375 Hence, the percentage of ripe fruits which

are mangoes = 0.4375 × 100 = 43.75% = 43.8% [ correct to three signifi cant

fi gures]10.

Current (C amperes)

Mid-point (x)

Cumulative Frequency

Frequency(f)

fx

10 ≤ C < 11 10.5 8 8 84.0

11 ≤ C < 12 11.5 30 22 253.0

12 ≤ C < 13 12.5 63 33 412.5

13 ≤ C < 14 13.5 88 25 337.5

14 ≤ C < 15 14.5 97 9 130.5

15 ≤ C < 16 15.5 99 2 31.0

16 ≤ C < 17 16.5 100 1 16.5

Σf = 100 Σfx = 1265

– Σfx 1265Mean, x = —— = ——— = 12.65 Σf 100The median class is 12 ≤ C < 13. N — – F 2Median, M = Lm + �——–—� c fm

100 —— – 30 2 = 12 + �—————�(13 – 12) 33 = 12.61The modal class is 12 ≤ C < 13. d1Mode, mo = Lmo

+ �——–—�c d1 + d2

11 = 12 + �———� (1) 11 + 8 = 12.58Since mean > median > mode, the distribution is positively skewed.

11. 0, x < 0, 5

f(x) =

⎫⎪⎬⎪⎭

— – x, 0 ≤ x < 1, 4 1 ——, x ≥ 1. 4x2

x

(a) For x < 0, F(x) = ∫ f(x) dx –∞ x

= ∫ 0 dx –∞

= 0 x

For 0 ≤ x < 1, F(x) = ∫ f(x) dx –∞ 0 x

= ∫ f(x) dx + ∫ f(x) dx –∞ 0

x 5 = F(0) + ∫ �— – x� dx 0 4

0.0113

–2.281 2.281

0.0113

z



5 x2 x

= 0 + �—x – —� 4 2 0

5 x2

= —x – — 4 2 x

For x ≥ 1, F(x) = ∫ f(x) dx –∞

1 x

= ∫ f(x) dx + ∫ f(x) dx –∞ 1

x 1 = F(1) + ∫ —— dx 1 4x2

x 1 = F(1) + ∫ —x–2 dx 1 4

5 12 1 x–1 x

= —(1) – — + �—�—–�� 4 2 4 –1 1

3 1 x

= — – �——� 4 4x 1

3 1 1 = — – �—– – ——� 4 4x 4(1)

1 = 1 – —– 4x

0, x < 0, 5 x2

F(x) =

⎫⎪⎬⎪⎭

—x – —–, 0 ≤ x < 1, 4 2 1 1 – —–, x ≥ 1. 4x

3

(b) P(X ≤ 3) = ∫ f (x) dx –∞

= F(3) 1 = 1 – —— 4(3) 11 = —– 12 P(at least one of two independent

observed of X is greater than 3) = 1– P(both the independent observed

values of X is less than or equal to 3)

= 1 – P(X ≤ 3) • P(X ≤ 3) 11 11 = 1 – �—–��—–� 12 12 23 = —— 144

12. Let X represent the number of cars requested. X ~ Po (4) in a day

e–440

(a) (i) P(X = 0) = ——– = 0.0183 0! [correct to three signifi cant fi gures] (ii) P(X ≥ 4) = 1 – P(X = 0) – P(X = 1) – P(X = 2)

– P(X = 3)

e–440 e–441 e–442 e–443

= 1 – ——– – ——– – ——– – ——– 0! 1! 2! 3!

= 1 – 0.0183 – 0.0733 – 0.1465 – 0.1954

= 0.567 [correct to three signifi cant fi gures]

(b)

x 0 1 2 3 ≥4

P(X = x) 0.0183 0.0733 0.1465 0.1954 0.5665

xP(X = x) 0 0.0733 0.2930 0.5862 2.2660

E(X) = ΣxP(X = x) = 0 + 0.0733 + 0.2930 + 0.5862

+ 2.2660 = 3.2185 Hence, the expected daily income from

the rentals of cars = 3.2185 × 50 = RM160.93 e–444

(c) P(X = 4) = ——— = 0.1954 4! P(X ≥ 5) = 1 – P(X = 0) – P(X = 1) – P(X = 2)

– P(X = 3) – P(X = 4) = 1 – 0.0183 – 0.0733 – 0.1465 – 0.1954

– 0.1954 = 0.3711

x 0 1 2 3 4 ≥5

P(X = x) 0.0183 0.0733 0.1465 0.1954 0.1954 0.3711

xP(X = x) 0 0.0733 0.2930 0.5862 0.7816 1.8555

E(X) = ΣxP(X = x) = 0 + 0.0733 + 0.2930 + 0.5862 +

0.7816 + 1.8555 = 3.5896 Hence, the expected daily income from

the rentals of cars = 3.5896 × 50 = RM179.48 Additional daily income = RM179.48 – RM160.93 = RM18.55 Since the additional daily income

(RM18.55) is less than the additional cost incurred per day (RM20.00), then the shop should not buy another car for rental.

Even if the number of cars requested is more than 4, the number of cars rented out is still 4 because there are only 4 cars available to be rented out.

Even if the number of cars requested is more than 5, the number of cars rented out is still 5 because there are only 5 cars available to be rented out.


2008 STPM Maths T Q&A

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