MathsH2

(1)

JC-2 Examination Papers

2013

Maths

College

H2

Yishun Junior College

P1

P2

Victoria Junior College

P1

P2

Tampines Junior College

P1

P2

Temasek Junior College

P1

P2

Serangoon Junior College

P1

P2

St Andrew’s Junior College

P1

P2

River Valley High School

P1

P2

Raffles Institution

P1

P2

Pioneer Junior College

P1

P2

Nanyang Junior College

P1

P2

National Junior College

P1

P2

Meridian Junior College

P1

P2

Millennia Institute

P1

P2

Jurong Junior College

P1

P2

Innova Junior College

P1

P2

HWA Chong Institute

P1

P2

Dunman High School

P1

P2

Catholic Junior College

P1

P2

Anderson Junior College

P1

P2

(2)

This question paper consists of 4 printed pages.

YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE

Y

ISHUN

J

UNIOR

C

OLLEGE

2013 JC2 PRELIMINARY EXAMINATION

MATHEMATICS

9740/01

Higher 2

Paper 1

20 AUGUST 2013

TUESDAY 0800h – 1100h Additional materials : Answer paper Graph paper List of Formulae (MF15)

TIME

3 hours

READ THESE INSTRUCTIONS FIRST

Write your name and CTG in the spaces provided on all the work you hand in. Write in dark blue or black pen on both sides of the paper.

You may use a soft pencil for any diagrams or graphs.

Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.

You are expected to use a graphic calculator.

Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise.

Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands.

You are reminded of the need for clear presentation in your answers.

At the end of the examination, write down the question number of the questions attempted, model of calculator used on the spaces provided on the cover page. Tie your cover page on top of the answer scripts before submission.

(3)

Yishun Junior College  2013 JC2 Preliminary Exam  H2 Mathematics 9740/01

Page

2

of 4

1 Use the method of mathematical induction to prove that for all n ,









1 sin 2 1 sin cos 2 , 0 2sin n r n r               



. [6] Hence find 2

 

1 cos n r r 



. [2]

2 (i) Using the method of differences, show that







1 1 1 3 2 3, n r B C A r r n n        



where A, B, and C are constants to be determined. [4] (ii) Explain why the series







1 1 1 3 r r r    



converges, and write down its value. [2] (iii) Hence show that





1 2 1 5 2 2 1 r r    



. [2]

3 Four housewives bought three different types of meat from the same meat seller. The mass of meat and the total each housewife paid are shown in the table below.

Mrs Lee Mrs Nasri Mrs Vicnesh Mrs Parker

Chicken (kg) 0.3 0.5 0.6 0.8

Mutton (kg) 0.4 0 0.5 a

Duck (kg) 0.6 0.4 0 1

Total amount ($) 10.7 6.5 7 18

Find the value of a. [4]

4 (i) Without using a calculator, solve the inequality ₂5 1 1 6

x

x x



   . [3]

(ii) Hence solve

 2 1 5ln ln ln 6 1 x 0 x x      . [3]

5 (a) (i) Find the value of A such that





2 2 d 2 1 1 x x x e A x C e e      _  _      , where C is an arbitrary constant. [2]

(ii) The region under the curve

2 2 1 x x e y e   , 0 x ln 2 is rotated 2 radians

about the x-axis to form a solid. Using the result in (i), find the exact

volume of the solid. [4]

(b) A curve has parametric equations 4sin , cot

x  y , where 0   .

(i) Sketch the curve. [2]

(ii) Find the exact area of the region bounded by the curve, the 2 axes and the

(4)

Page

3

of 4

6 The points A, B and C have position vectors i k ,  4i j 6k, 7i xj 4k, where 0

x .

(i) Evaluate x, given that BC 17. [2] The point P lies on AC such that BP is perpendicular to AC. Find

(ii) the coordinates of P, [5]

(iii) the coordinates of B , the reflection of B in the line AC, ' [2] (iv) the exact area of the quadrilateral ABCB . ' [3] 7 The equation of a curve is y2xy 1.

(i) Find the equations of all tangents to the curve that are parallel to the y-axis. [4] (ii) State and justify whether the curve has any stationary points. [2] (iii) Find the area of the region bounded by the axes, and the normal to the curve at

the point where the y-coordinate is 2. [5]

8 (i) Expand 4 1

x x



 in ascending powers of x, up to and including the term in

2

x .

State the set of values of x for which the expansion is valid. [5] (ii) By substituting 1

10

x  , obtain an estimate for 390 , leaving your answer as a

fraction. [2]

9 (i) An open container is made in the form of an upright prism with an equilateral triangular base of side x, and height h.

Each unit of area of the base costs a and each unit of area of the sides costs b, where a and b are constants. The total cost of the container is a fixed amount c. Prove that the capacity, V, of the container is



2



3

4 3

48bx c ax . [3]

(ii) Hence show that V is maximum when the cost of the base is 3

c

. [4]

10 (i) Given that w 2, Re

 

w 0, arg _* 3 w w   _{ }     , prove that w 3 i . [3] (ii) Hence find the equation of the locus z a r, where a r,  , which contains

the points representing i and w. [2]

(a) Find the exact maximum and minimum possible values of z3i . [3] (b) Find the range of arg 3 7

2 z        . [2] x h

(5)

Page

4

of 4

11 (a) Find the equation of the curve in the form yf

 

x , given that it passes through the points

 

1, 0 ,



1, 2



, and

2 4 2 3 d 6 2 d y x x x   . [4]

(b) Perry is jogging in a park. The distance covered, x kilometres, varies with time, t hours. As he jogs, the difference between his initial speed and speed at time t, is proportional to the distance he has covered. Given that he starts his jog with a speed of 5 km/h, and his speed is 4 km/h when he has jogged 2 km, find the exact time he takes to complete a 5 km jog. [6]

12 The curve C has equation

2 2 x y x   .

(i) Find the equation(s) of the asymptote(s) of C. [1] (ii) Sketch the curve C, labelling the equation(s) of its asymptote(s) and coordinates

of any axial intercepts and turning points. [2] (iii) Hence find the range of values of k for which the equation 2



2



4

x k x  has no

real roots. [2]

(6)

Page 1 of 11

YISHUN JUNIOR COLLEGE

Mathematics Department

SOLUTION

Examination : JC2 Preliminary Examination Date : 20/08/2013

Subject : JC2 H2 Maths Paper No. : 1

Qn Solution

1

(i) Let P be the statement “n









1 sin 2 1 sin cos 2 2sin n r n r           



”, n_  1: n





 

1 1 LHS cos 2 cos 2 r r   





 









 

sin 3 RHS 2sin 1 1 2cos 3 sin 3 2 2 2sin 2c sin 2sin cos 2 os 2 sin                                LHS RHS , P1 is true.

Assume P is true for some _k k_ _.

i.e.









1 sin 2 1 sin cos 2 2sin k r k r           



To show Pk1 is true. i.e. 1









1 sin 2 3 sin cos 2 2sin k r k r            



















_

_

























1 1 1 1 LHS cos 2 cos 2 cos 2 2 sin 2 1 sin cos 2 2 2sin

sin 2 1 sin 2cos 2 2 sin

2sin

sin 2 1 sin sin 2 3 sin

2sin sin 2 2 1 3 sin RHS 2sin k r k r k r r k k k k k k k k k P                            _  _         _  _                                       



is true Hence by induction, Pn is true for all n 

(7)

Page 2 of 11 Qn Solution (ii)

 





2 1 sin 2 1 sin 2cos 1 2sin n r n r               



 





2 1 sin 2 1 sin 2 cos 2sin n r n r n            



 





2 1 sin 2 1 sin 2 cos 2sin 2 sin n r n r   n           



 









2 1 sin 2 1 2 1 sin cos 4sin n r n r  n           



2 (i)







1 1 3 1 3 a b r r r r By cover-up method, 1 2 a , 1 2 b 







1 1 1 1 1 1 1 3 2 1 3 1 1 1 2 2 4 n n r r r n r r    _  _   _   _  



1 1 3 5      1 4  1 6  1 5  1 7  1 1 n   1 1 n   1 n  1 2 1 1 n n     1 3 n   _  









5 1 1 12 2 n 2 2 n 3      Hence 5 12 A , 1 2 B C   (ii) As n , 1 2 0 n  , 1n30 Hence

_

_

_

1 1 1 3 r r r    



converges and

_

_

_

1 5 1 1 1 3 2 r r r    



(iii)











2 ₂ 2 2 4 4 1 3 4 3 r r r r r r r          Hence



r2

 

2  r 1



r 3 0



(8)

Page 3 of 11 Qn Solution



 

2





1 1 1 3 2 r r r   





2







1 1 1 1 5 1 3 12 2 r r r r r        



3 Let x, y, z be cost (per kg) of chicken, mutton, and duck respectively. 0.3 0.4 0.6 10.7 0.5 0.4 6.5 0.6 0.5 7 x y z x z x y      _ _   _ _  From GC, x5, y8, z10

     

0.8 5 a 8 1 10 18  Hence a0.5 4 (i) 2 5 1 ₁ 6 x x x    





2 2 6 5 6 0 1 x x x x x        2 2 0 4 5 6 x x x x     









31



52



0 x x x x      Hence x 3,   1 x 2, or x5. (ii)  2 0

_{ }

2 1 5ln 5ln 1 1 ln ln 6 ln ln 6 1 x x x x x x           2 5 1 ₁ 6 y y y     , where ylnx Hence lnx 3,  1 lnx2, or lnx5. i.e. _{0 x e}_{ } 3_,_e1_{ }_{x e}2_{, or}_{x e}_ 5_. 5 (a) (i)













2 ₂ 2 2 ₂ 2 2 2 2 d d 1 ₁ 1 2 1 d 2 1 2 1 x x x _x x x x e _x e _x e _e e e x C e      _               _  



x ve  _ve 3 1 ve  _ve ve 2 5

(9)

Page 4 of 11 Qn Solution (ii) Volume = ln 2 ₂ 2 2 0 d 1 x x e _x e  _ _      





4 2 ln 2 0 2 1 d x x e e x       









2 2 2 2 2 2 1 1 2 2 1 x x x x x e u e v e u e v e          Volume





ln 2 ln 2 2 2 2 2 0 0 d 1 2 1 x x x x e e _x e e     _   _     _ _      









ln 2 2 0 4 1 _{1 ln} ₁ 2 4 1 2 1 1 2 e x       __  _ _  _ _       





1 2 1 ln5 ln2 4 5 2    _    _   1 5 3_ln 2 2 20 _  _    (b) (i)

(ii) _cot d _cosec2 d y  y     1 / 4 0 / 2 y y          





















0 4 2 2 4 2 1 2 4 Area d 4sin cosec d 4 cosec d 4 ln cosec cot 4 ln 2 1 ln 1 0 4ln 2 1 x y                    _ _       



 0 y x  









2 2 2 4 2 4 4 4 4 2 2 2 Area 2 2 dx 2 2 d cos 2 2 4 d sin 2 2 4 cosec d 2 2 4 ln cosec cos 1 2 2 4 l cot 4cos sin n 2 1 2 4 n t l co 2 1 y                                _  _     _   _           

(10)

Page 5 of 11 Qn Solution 6 (i) 7 4 3 1 1 4 6 2 BC x x             _{    }   _      _       



  

2 2 2 3  x 1  2  17





2 1 4 x  1 2 x   3 x or x 1 (rej. since x0) (ii) 7 1 6 2 3 0 3 3 1 4 1 3 1 AC                 _{     }   _{ }                 1 2 : 0 1 , 1 1 AC l           _{ } _{ }          r

Since P lies on AC,

1 2 0 1 , for some 1 1 OP           _{ } _{ }          . 3 1 2 4 2 0 1 1 1 1 6 1 5 BP                     _{ } _{    }  _                    BP perpendicular to AC 2 2 1 0 2 1 3 1 5          _{     }                1 2 5 0 1 2 1 1 2 3 OP             _{ } _{   }             Thus P(5, 2, 3) (iii) _' 2 OB OB OP  5 4 6 ' 2 2 2 1 3 3 6 0 OB OP OB                _{     }              Thus B’ (6, 3, 0) (iv) Since ABCB' is a kite

(AC cuts BB' at the midpoint perpendicularly), Area of quadrilateral ABCB'

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Page 6 of 11 Qn Solution





2 Area of triangle 2 3 6 3 2 4 2 3 3 2 1 3 7 2 3 2 1 1 3 16 49 1 3 6 1 2 6 AC ABC BC                        _ _{  }  _ _{  }  _ _ _    _    _                    7 (i) 2 ₁ y xy  d d 2 0 d d y y y y x x x   _  _  



2



d d y y x y x   d d 2 y y x  y x

Tangent parallel to y-axis implies 2y x 0 2 x y 2 1 2 2 x _x x   _  _{ }         2 1 ₁ 4x    2 ₄ x  2 x 

Hence equation of tangents are x2 and x 2 (ii) d d 2 y y x  y x Hence d 0 d y x  implies y0. But _y2__xy_{ }₁_. Thus y 0 LHS 0 RHS 

Hence there are no stationary points. (iii) 2 22 x 1 5 2 x 5 2 x , y2

 

d 2 4 5 d _{2 2} 3 2 y x    

Hence gradient of normal is 3 4

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Page 7 of 11 Qn Solution 3 5 2 4 2 y   _x _   Equation of normal: 3 31 4 8 y  x 31 3.875 8 0 y x    31 5.16 6 0 y  x 

Hence area of region





1 31 31 2 6 8 961 10.0 3 s.f. 96       _ _ _   8 (i) ₁4 _xx  





1 2 ₁ 4 1 1 4 x _x     _  _   





2 2 1 1 1 2 2 2 1 1 2 4 2 4 x x _{x x}  _      _{ } _ __{ }      _{ } _{ }              





2 2 2 1 1 8 128 x x _{x x}    _    _      2 2 2 2 1 8 128 8 x x _x x _x    _      _   2 9 143 2 4x 64 x     Valid for: 1 4 x _ _and _x _₁_. Hence



x : 1  x 1



(ii) 2 9 1 143 1 2 4 1 1 4 0 64 1 1 1 0 0 0 1 1       _ _ _       1 11503 6400 0 39 11 10  126533 390 6400  9 (i) ₂ 2 Area of b C ase 1 si ost of base n 60 2 3 4 a x ax a     

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Page 8 of 11

Qn Solution

Area of s Cost of sides ides

3 3 b x b xh bh      Hence 3 2 ₃ 4 ax  bhx c , i.e. 2 3 4 3 c ax h bx  





2 2 2 heigh Area o t 3 4 f base 3 4 3 3 ₄ ₃ 48 V c ax bx x c ax b x        (ii)



3



3 4 3 48 V cx ax b  



2



d _{3 4 3 3} d 48 V _c _ax x  b  2 4 d ₀ _{3 3} d c ax V x   

Hence cost of base 3 2

4 3 c ax  





2 2 d _{3 6 3} ₀ d 48 V _ax x  b   for x0

Hence V is maximum when cost of base is 3 c 10 (i) arg _ww* ₃   _{ }      

 

* arg arg 3 w  w    2arg 3 w    arg 6 w 

Hence 2 cos isin

6 6 3 1 2 i 2 2 3 i w _    _      _  _    

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Page 9 of 11

Qn Solution

(ii) Let A, B, C represent a, w, i respectively on an Argand diagram.

Then AB AC r  , thus A lies on perpendicular bisector of BC.

   

Im w Im i 1 . Thus perpendicular bisector of BC is vertical. Hence 3 2 a Thus 2 3 i 2 3 ₁ 3 ₁ 2 4 7 2 r              (ii) (a) Maximum





2 2 3 7 3i 3 2 2 39 7 4 2 1 39 7 2 z  _ _        3 Im A Re B C A Im Re

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Page 10 of 11 Qn Solution Minimum





2 2 3 7 3i 3 2 2 39 7 4 2 1 39 7 2 z  _ _        (b) 1 2 s n 6 i      Hence arg ₁ 3 7 2 6 z 6             11 (a) 2 4 3 2 3 d 6 _{2 6 2} d y x _x _x x x      2 2 d ₃ d y _x _x _C x     3 1 _{, ,} y x x_ _  _Cx D C D_ _

 

0 1 1 2 - - - (1) 1, 0 C D C D         





2 1 1 4 - - - ( 2 ) , 2 1 C D C D            Hence C 3, D1. Thus _{y x x}_ 3_ 1__{3 1}_x_ (b) d 5 dx xt   , initial d 5 d x t  Hence 5 d , 0 dx kx kt    2 x , d 4 d x t  . Thus 4 5 k

 

2 . 1 2 k  Hence d 5 1 d 2 x _x t   d 10 d 2 x x t   7 3 ₇ 2  Im 3 2 Re 7 2

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Page 11 of 11 Qn Solution d 2 d 10 t x x 2 d 10 2ln 10 t x x x C          0 x , t0. Thus 0 2ln10 C 2ln10 C Hence t2ln10 2ln 10 x 2ln10 2ln 5 10 2ln 5 2ln 2 5 x t      12 (i) 2 ₄ 2 2 2 x y x x x       Asymptotes:y x 2, x2 (ii) (iii) _x2 __{k x}



2_4







2 2 2 x _{k x} x  

For all k , y k x



2



cuts



2, 0



. No real roots, hence 2

2

x y

x



 does not intersect y k x



2



.

Thus 0 k 1. x   (4, 8) 0 2 x 2 y x  y x

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This question paper consists of 4 printed pages.

YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YISHUN JUNIOR COLLEGE YSIHUN JUNIOR COLLEGE

Y

ISHUN

J

UNIOR

C

OLLEGE

2013 JC2 PRELIMINARY EXAMINATION

MATHEMATICS

9740/02

Higher 2

Paper 2

21 AUGUST 2013

WEDNESDAY 1400h – 1700h Additional materials: Answer paper List of Formulae (MF15)

TIME

3 hours

READ THESE INSTRUCTIONS FIRST

Write your name and CTG in the spaces provided on all the work you hand in. Write in dark blue or black pen on both sides of the paper.

You may use a soft pencil for any diagrams or graphs.

Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions.

Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise.

Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands.

At the end of the examination, write down the question number of the questions attempted, model of calculator used on the spaces provided on the cover page. Tie your cover page on top of the answer scripts before submission.

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Page

2

of 4

Section A: Pure Mathematics [40 marks] 1 The functions f and g are defined by





2 1 2 ln 1 , , 1, 4 6 , , 2. f : g : x x x x x x x x x        

(i) Explain why the function 1

g exists and express 1

g in a similar form. [4] (ii) Show that the composite function fg exists. Define fg and find its range. [4] (iii) State a sequence of transformations which transform the graph of





ln 1

y  x to the graph of yf

 

x . [3] 2 The plane ₁ contains the point (1, , 1) and the line l with equation ₁ y1,x 1 z.

(i) Find the equation of the plane ₁ in scalar product form. [3] (ii) Find the Cartesian equation of the plane ₂ that also contains the line l but is ₁

perpendicular to the plane ₁. [2]

(iii) The line l with equation 2 r 



5 

 

i 7 2

 

j 3 a



k, intersects 1

 at 30. Find the possible value(s) of a in exact form. [3]

3 (i) Find the series expansion of e4xsinx, up to and including the term in x . 4 [3] (ii) Using your answer in (i), find the series expansion of e4xcosx , up to and

including the term in x . 3 [2]

4 (i) Let f

 

z z3 z2 bz4,b . Given f 1+i 3





0, obtain all the roots of

 

f z 0. [3]

(ii) Hence find the exact roots of the equation 4w3bw2  w 1 0. [2] (iii) For the value of z found in (i) for which Im

 

z 0, find the smallest positive

integer n such that zn . State the value of z when n takes this value. n [4] 5 Andy and Bob played a computer game last year.

(a) Andy gained 450 points on the first day he played. On each subsequent day, he gained 13 points less than what he gained on the previous day. He played daily up to and including the day he gained less than 55 points. Find the total points Andy

gained. [4]

(b) Bob gained 400 points on the first day he played. On each subsequent day, he gained 0.9 times of the points he gained on the previous day. Find the minimum number of days for Bob to gain at least 95% of the theoretical maximum points

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Page

3

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Section B: Statistics [60 marks]

6 Betty has one black, one yellow, two red, two green and three blue beads. The nine beads are arranged randomly in a line. Find the probability that

(i) the two red beads are next to each other, [2] (ii) all the red and green beads are separated, [2] (iii) either the two red beads are next to each other or the two green beads are next to

each other or both. [3]

The nine beads are now threaded randomly on a ring.

(iv) Find the probability that the two red beads are next to each other and the two

green beads are next to each other. [2]

7 I have a choice of two routes to get to school. The probability that I am punctual for school is 43

50 and the probability that I choose the first route is 3

5 . If I get to school punctually, the probability that the first route is chosen is 27

43. Find the probability that I get to school punctually if

(i) I choose the first route, [2]

(ii) I choose the second route. [2]

8 In a factory, a machine fills and seals tin cans of milk automatically. It is given that 5 % of the cans are dented.

(i) Given a random sample of 10 cans, find the probability that more than two cans

are dented. [2]

(ii) Given 30 random samples of 100 cans, find the probability that the mean number of dented cans per sample does not exceed four. [3] (iii) Find the least number of cans that must be taken such that the probability for

more than one can to be dented exceeds 0.95. [3] It is found that 99.7 % of the cans are not rusty.

(iv) Given a random sample of 1000 cans, use a suitable approximation to find the probability that at most 998 cans are not rusty. [3]

9 The random variable X has the distribution N



 , 2



. It is known that





P X 10 0.1 and P 9



X 10



0.2. Calculate the values of  and  . [4] Another random variable Y has the distribution N 4.8, 1.3



2



. Given that 10 and

1.5



 , find P



X 2Y



. [3]

10 A journalist intends to conduct a survey to find out the perception of different age groups of citizens on the effectiveness and relevance of the education system. Describe how a quota sample of size 50 might be carried out in this context. Explain a disadvantage of quota sampling in the context of your answer. [3] State, with a reason, whether stratified sampling is realistic to be used in this case. [1]

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Page

4

of 4

11 At a car rental firm, there are 5 cars and 3 vans available each day. The demand for cars follows a Poisson distribution with a mean of 2 on a weekday and a mean of 5 per day on weekends. The demand for vans follows a Poisson distribution with a mean of 2 on any day of the week. The demand for cars and the demand for vans on any day are independent.

(i) Find the probability that, on a particular weekday, the demand for cars is 5. [1] (ii) Find the least number of cars the rental firm should have to meet the demand on a

particular weekday with at least a probability of 0.9. [2] (iii) Using a suitable approximation, find the probability that, out of 104 Saturdays,

there are at least 50 days in which the total demand for vehicles exceeds 6 each

day. [4]

(iv) Find the probability that, on a Wednesday, 5 cars are hired out given that at most 6 vehicles were hired out on that day. [4]

12 The college wishes to find out the number of hours spent per week, on average, by a typical student on social media. A survey was conducted on a random sample of 80 students and the time spent per week, x hours, was recorded and summarised by



x18



208



,





x18



2 8967.

(i) Find, correct to 1 decimal place, the unbiased estimates of the population mean

and variance. [2]

(ii) The Head of Department of Student Well-being claims that the population mean weekly time spent on social media exceeds ₀, where ₀ is a constant.

Use the earlier sample to determine, correct to 1 decimal place, the least value of 0

 in order for the claim not to be valid at 5% level of significance. [4]

13 An experiment was conducted to investigate how the mass of a drug in a human body varies with time, measured from when the drug is given. The results are summarised in the following table.

t (minutes) 10 15 20 30 40 55 70 90

x (milligrams) 1.36 0.93 0.72 0.50 0.39 0.31 0.26 0.21

(i) Draw a scatter diagram for the data. [2] (ii) It is desirable to predict the mass of the drug in the long run. Explain why, in this

context, neither a linear nor a quadratic model is likely to be appropriate. [2] (iii) It is decided that a model of the form t x a







b will be used. Calculate the

least squares estimates of a and b and estimate the mass of the drug after two

hours. [3]

(iv) Give an interpretation, in context, of the value of a. [1]

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Page 1 of 9

YISHUN JUNIOR COLLEGE

Mathematics Department

SOLUTION

Examination : JC2 Preliminary Examination Date : 21/08/2013

Subject : JC2 H2 Maths Paper No. : 2

Qn Solution

1

(i) Graph of yg

 

x ,x2

Any horizontal line yk k,  cuts the graph of yg

 

x ,x2 at most once. Hence g

is one-one. Thus _g1_exists. Let _y__x2_₄_x_₆





2 2 2 y x  2 2 x   y 2 2 x  y (rej. y2 since x2) 1 ₂ _2, ₂ g : x  x x (ii) R_g 

_

2, 

_{ }

1, 

_

D_f Hence fg exists.



2



fg( ) 1 2lnx   x 4x5 , x2





fg ,1 R  

(iii) Translation of 2 units in the positive x-direction Scaling parallel to the y-axis by a factor of 2, followed by a translation of 1 unit in the positive y-direction.

OR

Translation of 0.5 unit in the positive y-direction, followed by scaling parallel to the y-axis by a factor of 2. 2 (i) 1 0 : 1 , 1 1 1 0 l       _    _       _{ }     r y x (2, 2)

(22)

Page 2 of 9 Qn Solution 1 0 1 3 1 2 1 1 2           _ _            _ _        1 2 2 1 1 1 0 2 2            _ _       _         1 2 1 2 : 1 3 1 3 2 1 2              _{     }       _       r (ii) 2 1 2 1 1 0 4 1 1           _ _            _        2 1 0 1 : 4 1 4 3 1 1 1              _ _{   }  _             r 2:x 4y z 3      (Cartesian equation) (iii) 2 2 1 1 2 2 sin 30 4 1 4 1 4 a a       _                    2 1 2 2 _{9 5} a a   2 3 5a 4a



2



2 9 5a 16a 2 7a 45 45 7 a  3 (i)

_{   }

  



2 3 4 4 2 3 3 2 3 3 sin 4 4 1 1 4 2! 3! 3! 32 1 1 4 8 1 4 3 47 ₁₀ 6 6 4 x x x x x e x x x x x x x x x x x x  _ _      _   _  _ _        _   _   _ _         

(ii) _{4 sin}4 4 _cos _{1 8} 47 2 ₄₀ 3 2 x x e x e x  x x  x 4 2 3 2 3 2 3 47 4 4 6 15 26 47 cos 1 8 1 4 40 3 2 2 x e x x x x x x x_ x x x _ _   _           

(23)

Page 3 of 9

Qn Solution

4

(i) By conjugate root theorem, _Hence

_

_{1 i 3}

_

_

_{1 i 3}z 1 i 3

_

is also a root. 2 ₂ ₄

z  z  z  z is a factor of f z

 

.







2 ₂ ₄



3 2 ₄

AzB z  z z  z bz

By comparison of coefficients, A1,B1.

Hence z 1 is also a root. (ii) ₄_w3__bw2_{  }_w _{1 0} 2 3 1 1 1 4 b 0 w w w        _{   } _{ }        Hence 1 1 _, 1 _{, or} 1 1 1 i 3 1 i 3 1 i 3 1 i 3 1, , or 4 4 w z          (iii) i 3 1 i 3 2 z e      , 3 n n k k z   _ _     Hence n 3k

Thus least n3 and z3 23ei  8

5

(a) ₁₃an 55 450 13_n_₄₀₈



n 1 55



n31.4 3 s.f.





Hence Andy stopped playing on day 32. Total points









32 32 2 450 13 32 1 3 7952 S   _   _  (b) Sn 0.95S





400 1 0.9 ₄₀₀ 0.95 1 0.9 1 0.9 n     1 0.9 n 0.95 0.9n 0.05





ln 0.05 28.4 3 s.f. ln 0.9 n 

Minimum number of days = 29 days. 6

(i) Number of arrangements without restrictions = _2!2!3!9! Required probability = 8! 9! 2

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Page 4 of 9

Qn Solution

(ii) No. of ways to ‘slot’ in ‘red beads’ and ‘green beads’ = 6

4 _2!2!4! C Required probability = 6 4 5! 4! 9! 5 3! C 2!2! 2!2!3! 42 

(iii) P (2 ‘red beads together’ and 2 ‘green beads together’) = 7! 9! 1 3! 2!2!3! 18 





 





Required probability = P A B P A P B P A B      = 2 2 1 7 9 9 18 18  

(iv) Number of arrangements without restrictions =

_

8!

_

2 2!2!3! Required probability = 6! 8! 1 2(3!)2! 2(2!2!3!)2! 14  7 (i)













punctual | 1st route punctual and 1st route

1st route 43 27 50 43 3 5 9 10 P P P     (ii)













punctual | 2nd route punctual and 2nd route

2nd route 43 16 50 43 2 5 4 5 P P P     Punctual 1 st_route 2nd_route 27/43 43/50 Late 7/50 1 st_route 2nd_route 16/43

(25)

Page 5 of 9

Qn Solution

8

(i) Let X be the r.v. “number of tin cans that are dented out of 10 cans”. X ~ 10,0.05B







2 1





2 0.0115



P X   P X  

(ii) Method 1

Let Y be the r.v. “number of tin cans that are dented out of 100 cans”. Y ~ 100,0.05B





E(Y) = 5 and Var(Y) = 4.75 By CLT, ~ 5, 4.75 30 Y N_ _   approximately P Y



4 0.00598



 Method 2

Let T be the r.v. “number of tin cans that are dented out of 3000 cans”.





~ 3000,0.05

T B

Required probability = P T



120 0.00551





(iii) Let W be the r.v. “number of tin cans that are dented out of n cans”. W~B n



,0.05





1 0.95



P W   Method 1



1 0.05



P W   When n92, P W



 1 0.0521



When n93, P W



 1 0.04998



Hence least n = 93 Method 2









1P W0 P W  1 0.95











1 0.05 0.95 n 0.05 0.95 n 0 n     Using GC, n > 92.98857. Hence least n = 93

(iv) Let V be the r.v. “number of cans that are rusty out of 1000 cans”. V ~ 1000,0.003B





Since n = 1000 is large, p = 0.003 is small and np = 3 < 5

(26)

Page 6 of 9 Qn Solution 9 P X



10



0.1 10 _0.1 P Z     _ _     10  1.281551567 (equation 1)



9



0.7 P X   9 _0.7 P Z     _ _     9  0.5244005101 (equation 2) Solving, we have 8.31, 1.32





2 ~ 0.4, 9.01 X  Y N



2 0 0.553



P X Y  

10 Quota sampling might be carried out by

- dividing the range of ages into several groups (strata such as 13-16, 17-19, 20-22, 23-30, 31-50 etc)

- assigning a quota for each age group with a total of 50

- by waiting outside a library and picking citizens at the journalist’s discretion until all the quotas are met.

One disadvantage is that the journalist may likely have collected a biased sample of citizens due to his/her non random selection process. For example, he/she might choose people that he/she thought are more approachable.

Stratified sampling is not realistic in this context because it is difficult to obtain the sampling frame.

11

(i) Let C be the r.v. “demand for cars on a weekday”. C~ Po 2

 



5 0.0360894089 0.0361



P C  

(ii) P C



n



0.9

Using GC and listing all probabilities,



3 0.85712 0.9



P C  



4 0.94735 0.9



P C  

Least number of cars required = 4

(iii) Let A be the r.v. “demand for vehicles on a Saturday”. A~ Po 7

 



6 1





6



P A  P A

0.5502889

Let W be the r.v “number of Saturdays (out of 104) in which the total demand for vehicles exceeds 6 each day”.

W ~ 104,0.5502889B





Since n = 104 (large) and np = 57.2300456 > 5 and

nq =46.7699544 > 5,





~ 57.2300456, 25.73698676

W N approximately



50





49.5 0.936



(27)

Page 7 of 9

Qn Solution

(iv) Let V be the r.v. “demand for vans on a weekday”. V ~ Po 2

 









5 cars hired| 6 vehicles hired 5 cars hired and 6 vehicles hired

( 6 vehicles hired) P P P    









5 cars hired and 1 van hired













( 0) 0 ( 1) 0 ( 2) 4 ( 3) 3 P P V P C P V C P V P C P V P C             



 











5 1









( 0) 0 ( 1) 0 ( 2) 4 ( 3) 3 P C P V P V P C P V C P V P C P V P C              



0.052653 0.4060058





0.9395530118  0.0228  12

(i) Unbiased estimate of population mean

_

18

_

18 20.6 80 x x 



  









2 2 2 1 ₁₈ 18 42131 _106.7 79 80 395 x s x  _            



(ii) H0:  0 1: 0 H   Under H0, ~

 

0, 1 X Z N s n    approximately by CLT, where 2 42131 20.6, , 80 395 x  s  n ,   ₀

At 5% level of significance and one tail test, critical value = 1.645 In order not to reject ,

Test value < 1.645

  

0 20.6 _1.645 41231 395 80   _ 0 18.7006   Least 0 18.8. (1 d.p.)

(28)

Page 8 of 9

Qn Solution

13 (i)

(ii) The mass of the drug could not continue to decrease linearly to a negative value. Thus, a linear model is not appropriate.

A quadratic model would eventually lead to an increase in the mass of the drug.

However, the mass of the drug is likely to plateau off or stay constant after many hours. Thus a quadratic model would also not be appropriate.

(iii) t x a







b 1 x a b t         Using GC,a0.07177273 0.0718 and b12.8879265 12.9 When t = 120 mins, We have x0.1791721175 0.179 .

(iv) The amount of drugs that remains in the body after a long period of time will approximate to a.

(10 , 1.36)

(90 , 0.21)

x

(29)

1

VICTORIA JUNIOR COLLEGE Preliminary Examination

MATHEMATICS

(Higher 2)

9740/01

September 2013 3 hours

Paper 1

READ THESE INSTRUCTIONS FIRST

Write your name and CT group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs.

Do not use staples, paper clips, highlighters, glue or correction fluid.

Answer all the questions.

Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands.

At the end of the examination, fasten all your work securely together.

The number of marks is given in brackets [ ] at the end of each question or part question.

This document consists of 5 printed pages

(30)

2 1 Without using a calculator, solve the inequality

2 15 11 _1. 6 9 x x x  _   [4]

Deduce the range of values of x that satisfies

 

15 11ln2 1. ln 6ln 9 x x x     [2] 2 It is given that y 1 ln(1 ). x

(i) By differentiating successively, find the Maclaurin’s series for y, up to and

including the term in _x2_. _[3]

(ii) By substituting 1 , 4 x approximateln 4 5    

 , leaving your answer in exact form. [3]

3 At the beginning of January 2010, Robert borrowed $200 000 from a bank that charges 0.4% interest at the end of every month. Robert pays back $1500 at the beginning of every month, starting from February 2010. Show that the amount Robert still owes the bank at the end of n months is 376500 175000(1.004) . n [3]

How much did he pay on his last instalment? [3]

4 A sequence of positive real numbers x1, x2, x3, ... satisfies the recurrence relation

1 4 4

n n n

x _ x   x

for n1.

Given that x1 = 9, find the values of x2, x3 and x4. Hence, conjecture an expression for

xn in terms of n. [2]

Prove your conjecture using mathematical induction. [4]

5 Referred to the origin O, the points A, B and C have position vectors 6 4 3i j k,

3 3i j kand 3i j 4k respectively.

(i) Find the size of angle ABC. [3]

(ii) If P is a point on the line AB such that CP 83, find the possible coordinates

(31)

3 6

The diagram shows the graph of yf ( )x which has a minimum point at (0, 2) and

asymptotes x2 andy5.

On separate diagrams, sketch the graphs of

(i) 1 , f ( ) y x  [3] (ii) y2 f ( ),x [3] (iii) yf '( ).x [2]

7 The variables x and y are connected by the differential equation d 1 3 .

d 2 2 2

y y

x x   x

By using the substitution u y x, show that d 1 3 .

d 2 2

u

x

x   x [3]

Sketch, for x0, three typical members of the family of solution curves of the differential equation d 1 3 .

d 2 2 2

y y

x x   x [5]

8 (i) Show that









2 _{3 1 1} 1 . 2 ! ! 2 ! r r r r r   _ _   [2]

(ii) Hence find





2 1 3 1 2 ! n r r r r    



. (There is no need to express your answer as a single

algebraic fraction.) [3]

(iii) Using the result in part (ii) and the standard series expansion for e ,x find the exact value of 2

_

_

1 3 3. 2 ! r r r r     



[4] [Turn over x 3 2 y = 5 0 x = 2 y y = f(x)

(32)

4

9 The complex numbers w and z satisfy the relations w 6 8i   w 8 6i and

8 6i 5

z   respectively.

(i) Illustrate both of these relations on a single Argand diagram. [4] (ii) Find, in either order,

(a) the least value of z w , and

(b) the value of z in the form xiy when z w attains its least value,

leaving your answers in exact form. [5]

(iii) Find the range of values for  such that the system of equations 8 6i 5,

z  





arg z4i 

has more than one solution for z. [4]

10 (a) The curve C has parametric equations 3 cos

x t, y3sint, for . 2 t 2

 

  

(i) Sketch C, indicating clearly the exact coordinates of the axial intercepts. [2] (ii) Find the exact value of the area of the region bounded by C and the y – axis.

[5]

(b) (i) Differentiate 1 x 2 with respect to x. [2] (ii) Show that

1 2 1 0 1 cos d 1 4 2 2 x x   _ _{ }



. [3]

(iii) The diagram below shows the curve with equation y cos1x. The

region bounded by the curve, the line 2

y  and the y-axis is denoted by

R.

Find the exact volume of revolution when R is rotated completely about the

x-axis. [4] y 1 cos y  x x 2 y  R

(33)

5 11 The line l1 has equation 3 2

2 1

x y z

k

 

  ; the line l2 passes through the points with

coordinates ( 1, 4, 2 )   k and ( 3,1, k), where k is a constant.

(i) Determine a cartesian equation of l2. [2]

(ii) Show that there are no real values of k for which l₁ and l₂ are perpendicular to

each other. [2]

(iii) The plane p has equation x4y z a, where a is a constant. If l₁ and p have

no common points, what can be said about the values of k and a? [4] For the value of k found in part (iii) and for a2, find the shortest distance between l₁

and p. [3]

It is given instead that k3 and that l₁ and l₂ intersect.  is the plane containing l₁

and l₂. By first finding a vector perpendicular to  or otherwise, find an equation of  in the form r u vw, such that v is perpendicular to w. [4]

(34)

Solutions to VJC Prelim 2013 Paper 1 Q Solutions 1)























2 2 2 2 2 2 2 2 15 11 ₁ 6 9 15 11 6 9 0 6 9 5 _{6 0} 3 5 _{6 0} 3 6 1 0 3 x x x x x x x x x x x x x x x x x                   _      6 or 1 3 or 3 x x x       Replace x with ln x 6 3 3 ln 6 or 1 ln 3 or ln 3 0 e or e e or e x x x x  x x           2i)

_

_









2 2 2 2 2 1 ln 1 1 ln 1 d 1 2 d 1 d d 1 2 2 d d 1 y x y x y y x x y y y x x x          _{ } _{ }      When x = 0, 2 2 2 d 1 1, , d 2 d 1 _{1 2} 1 3 d 2 2 4 y y x y x    _{ }   _  _{ } _      2 2 1 3 1 2 4 2! 1 3 1 2 8 x y x x x      _ _       ii) When 1 4 x , 2 1 1 1 3 1 1 ln 1 1 4 2 4 8 4 5 141 1 ln 4 128 4 19881 1 ln 5 16384 4 3497 ln 5 16384        _  _  _{ } _{ }           _{ }      _{ }           6  1 3    

(35)

Q Solutions

3) n Amt owed (beginning) Amt owed (end)

1 200 000 _{1.004 200 000}

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2 _{1.004 200 000 1500}

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

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2 1.004 200 000 1500 1.004  3

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2 1.004 200 000 1500 1.004 1500  

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3 2 1.004 200 000 1500 1.004 1.004   n  

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1

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1.004 200 000 1500 1.004 1.004 n n   

Amount owed after n months

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1 1 1.004 200 000 1500 1.004 1.004 1.004 1 1.004 1.004 200 000 1500 1 1.004 1.004 200 000 375000 1.004 1.004 376500 175000 1.004 n n n n n n n               

At the last instalment,

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376500 175000 1.004 1500 15 1.004 7 15 ln 1.004 ln 7 15 ln 7 ln 1.004 190.92 n n n n n            

His last instalment is on the 191st_month.

Amount he owed at end of 191st_month

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191

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$ 376500 175000 1.004 $1373.97    4)

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1 2 3 4 2 9 9 4 4 9 25 25 4 4 25 49 49 4 4 49 81 2 1 n x x x x x n               

Let Pn be the statement

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2 2 1 , n x  n n . When n = 1, LHS u1 9 RHS

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2 2 1 1 9    = LHS 1 P  is true