Modul3 Jpnt

KEM SPM TERBILANG TERENGGANU 2008 (MODUL 3)

ADDITIONAL MATHEMATICS PAPER 2

 MODULE 3  PAPER 2 (3472/2)  TRIGONOMETRIC FUNCTIONS 1.

(a) Prove that sec2 x + cosec2 x = sec2 x cosec2 x. (b) (i)

3

Sketch the graph of y = 2 cos

2

x for 0  x  2.

(ii) Find the equation of the suitable straight line to solve the equation cos

3 2

x=

3 4

x – 1.

Hence, by using the same axes, sketch the straight line obtained and state the number of solutions for the equation cos

2.

3 2

x=

3 4

x – 1 for 0  x  2.

(a) Prove that cosec 2A + cot 2A = cot A (b) Solve the following equations for 0  x  180. (i) 2 sin 2x = tan x. (ii) 3 sec2 x = 5(1 + tan x). (c) Using the same axes, sketch the graph of y = | 2 cos 2x + 1 | and y = Hence, state the number of solutions for the equation | 2 cos 2x + 1 | =

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x

for 0  x  .

 x



.

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 3)

3.

ADDITIONAL MATHEMATICS PAPER 2

(a) Show that 2 cos( + 45) cos(  45) = cos 2 . (b) Solve the following equations for 0  x  360. (i)

2 sin2 x + sin 2x = 2

(ii) sin (60 – x) – sin (60 + x) =

3 2

.

(c) Using the same axes, sketch the graph of the curve y = 2 sin x and the straight line of 3 1 x  y = 1  . Hence, state the number of solutions for the equation  for 0  x  2 2  2  x  4 sin x = 1. 2

4.

(a) Given that tan (A + B) = k and tan B =

2

.

2k  1

. k +2 (ii) Hence, find an expression for tan (A – B) in terms of k. 1 (iii) Hence, if tan (A – B) = , where A is an acute angle, find the value of A, without using 3 a calculator. (i)

Prove that tan A =

1

(b) Solve each of the following equations for 0  X  360. (i) 8 sin X + 3 sec X = 0 (ii) 2 tan X = 3 tan (45  X)

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 3)

5.

(a) Show that (cot x + cosec x)2 =

1  cos x 1  cos x

ADDITIONAL MATHEMATICS PAPER 2

.

(b) Solve the following equations for 0  x  360. (i) (ii) .

6.

2 cot2 x + 11 = 9 cosec x. 1 2

tan x

+ cosec2 x – 3 = 0.

(c) Sketch the graph of y = 3 sin 2x for 0 ≤ x ≤ . Hence, using the same axes, draw a suitable straight line to find the number of solutions to the equation 2x – 3 sin 2x = 0 for 0 ≤ x ≤ .

(a) Prove that cos 3A = 4 kos3 A  3 cos A. Hence, find the acute angle that satisfies the equation 2 cos 3A  8 cos3 A + 5 = 0. (b) Solve the following equations for 0  x  360. (i) 3 cos 2x – sin x = 1 (ii) 3 tan 2x tan x = 6 (c) Find the values of A and B that satisfy the equation sin (2A + B) = 08 and sin (2A – B) = 04 such that 0 ≤ 2A + B ≤ 90 and 0 ≤ 2A – B ≤ 90.

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 3)

7.

(a) Prove that

sin  1  cos 

=

1  cos  sin 

ADDITIONAL MATHEMATICS PAPER 2

.

(b) Given that g = cos 25 and h = sin 35, express in terms of g and h : (i) cos 60 (ii) cos 50 (iii) cos 12·5

 x for 0  x  3. 2  1 From your graph, find the approximate values of x that satisfy the equation cos x + = 0 2 4 in the range of 0  x  3.

(c) Using the scale of 2 cm to 05 unit on both axes, draw a graph of y = 2 cos

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 3)

ADDITIONAL MATHEMATICS PAPER 2

 PROBABILITY DISTRIBUTIONS 8.

(a) In a survey carried out in a college, it was found that one out of four students owned a car. (i) If five students are randomly selected from the college, find the probability that at least one student owned a car. (ii) If the college has 800 students, find the mean and the variance of the number of students that owned a car. (b) In a survey, it was found that the weight of a student in an institute has a normal distribution with mean 60 kg and standard deviation 4 kg. (i) If a student is randomly selected from the institute, find the probability that the student weigh less than 62 kg. (ii) Find the percentage of the students who weigh between 55 kg and 63 kg.

9.

(a) 70% of the candidates in a certain school pass their examination. If 10 candidates are randomly selected, find the probability that at least 8 candidates pass their examination. (b) The height, in cm, for the students in a certain school has normal distribution with mean 160 cm and variance 225 cm2. Find (i) the standard score for the height of 180 cm, (ii) the height of the students correspondence to the standard score 08, (iii) the probability that the height of a student that was randomly selected lies between 160 cm and 180 cm.

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 3)

ADDITIONAL MATHEMATICS PAPER 2

10. (a) In a survey carried out in a certain school, it was found that one out five students can speak in English fluently. (i)

If 8 students was randomly selected, find the probability that 3 students can speak in English fluently. (ii) Calculate the mean and the standard deviation of the number of students that can speak in English fluently, given that the enrolment of the school is 500 students. (b) In a trial examination for the English subject, the marks obtained by the students has a normal distribution with mean 61 marks and standard deviation of 12 marks. Find the probability that a student who sat for the examination will score (i) between 51 marks and 70 marks, (ii) more than 76 marks. (iii) If 65% of the students fail the English subject, find the minimum marks to pass the subject.

11. (a) Five participants for the National Service Program was selected for undergoing a fitness test. The probability that the participants pass the test is 08. Find the probability that (i) none of the participants fail the test, (ii) at least 4 participants pass the test. (b) The weight, in kg, of the students in a girls boarding school is normally distributed with mean 40 kg and variance 25 kg2. (i) Find the standard score for the weight of 47 kg. (ii) Find the number of students that weigh between 36 kg and 47 kg if the school has 350 students. (iii) Given that 60% of the girls weigh more than k kg, find the value of k.

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 3)

ADDITIONAL MATHEMATICS PAPER 2

12. (a) During the arching practice, Din hits the target 60% out of the trials. Find the number of arching such that the probability that at least one of the arch hits the target is greater than 09. (b) In a binomial distribution, the mean of the number of success is 40 and the standard deviation is 3 2 . Find the probability of success and the number of trials.

(c) The weight of a group of 400 students is normally distributed with mean 240 N and standard deviation 20 N. Find (i) the probability that a student that was randomly selected from the group weighs between 220 N and 280 N. (ii) the percentage of the number of students that weigh more than 300 N.

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 3)

ADDITIONAL MATHEMATICS PAPER 2

ANSWERS MODULE 3 1.

(b)

(i)

(ii)

y 2



0

2

3

x2 2 No. of solutions = 3

y=

x

-2

2.

(b)

(i)

0, 60, 120, 180

(ii)

6343, 16157

(c) 3 (ii)

1 0

3.

(b)

(i)

(c)

90

No. of solutions = 4

180 180

45, 90, 225, 270

(ii)

240, 300

y 2



 2

0

3 2

x No. of solutions = 2

2

4.

5.

(a)

(ii)

3k  4 4k  3

(b)

(i)

1143,

(b)

(i)

1947, 4181, 13819, 16053

(iii) 45 15571

(ii)

2657, 10843, 20657, 28843

(ii)

45, 135, 225, 315

y

(c)

1

0

y=

2x



 4

 2

3 4



x

, no. of solutions = 2

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 3)

6.

(a) (b) (c)

3357 (i) 30, 150, 22182, 31818 A = 1918 , B = 1477

(b)

(i)

(c)

x = 115 , 285

8.

(a) (b)

(i) (i)

9.

(a) (b)

10.

11.

7.

12.

g

2

1 h  h 1 g

2

(ii)

3526, 14474, 21526, 32474

(ii)

2g2 – 1

(ii) (ii)

x = 200,  2 = 150 6678 %

03828 (i) 1333

(ii)

148

(a) (b)

(i) (i)

01468 05710

(ii) (ii)

x = 100,  = 89443 01057 (iii) 4283

(a) (b)

(i) (i)

03277 14

(ii) (ii)

07373 248

(a)

3

07627 06915

(b)

73

ADDITIONAL MATHEMATICS PAPER 2

(iii)

1 g 2

(iii) 04087

(c)

(iii) 3874 (i) 08185

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(ii)

0135%

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