Amk Prelim 2009 Am1

ANG MO KIO SECONDARY SCHOOL PRELIMINARY EXAMINATION 2009 SECONDARY FOUR EXPRESS/FIVE NORMAL ACADEMIC

ADDITIONAL MATHEMATICS Monday

14 September 2009

4038/01 2 hours

Name of Setter: Mdm Karen Teng Additional materials: Answer paper Graph paper

READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen. 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. Write your answers on the separate Answer Paper provided. 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. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. 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. The total number of marks for this paper is 80.

This question paper consists of 5 printed pages. [Turn over

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax 2  bx  c  0, x

 b  b 2  4ac 2a

Binomial expansion n  n  n (a  b) n  a n   a n1b   a n 2 b 2  ...   a n r b r  ...  b n , 1  2 r n n(n  1) ... (n  r  1) n! where n is a positive integer and     r!  r  r!(n  r )!

2. TRIGONOMETRY Identities

sin 2 A  cos 2 A  1 sec 2 A  1  tan 2 A cos ec 2 A  1  cot 2 A sin( A  B)  sin A cos B  cos A sin B cos( A  B)  cos A cos B  sin A sin B tan A  tan B 1  tan A tan B sin 2 A  2 sin A cos A

tan( A  B) 

cos 2 A  cos 2 A  sin 2 A  2 cos 2 A  1  1  2 sin 2 A 2 tan A 1  tan 2 A 1 1 sin A  sin B  2 sin ( A  B) cos ( A  B) 2 2 1 1 sin A  sin B  2 cos ( A  B) sin ( A  B) 2 2 1 1 cos A  cos B  2 cos ( A  B) cos ( A  B) 2 2 1 1 cos A  cos B  2 sin ( A  B) sin ( A  B) 2 2 tan 2 A 

Formulae for ABC a b c   sin A sin B sin C a 2  b 2  c 2  2bc cos A 

1 ab sin C 2

[Turn over

3 1

(a)

Solve the simultaneous equations. 8 x 2 y  4 ln( 2 x  3 y  1)  0

(b)

2

Solve 3 2 x1  6  3 x1 .

Find the value of k for which the following simultaneous equations has no solution.

[4]

[4]

[3]

2 x  ky  2 3 x  ky  y  4

3

4

Differentiate the following with respect to x (i)

ln 4 x  1 ,

[2]

(ii)

ex . x 1

[2]

3 , sin 2 x : cos y  39 : 25 and x and y are acute angles. 10  5 If x  , show that cos y  . 4 13

It is given that sin x cos x  (i)

[2]

Hence find the exact value of

5

(ii)

tan(2 x  y ),

[2]

(iii)

sin x.

[2]

(i)

Express

(ii)

6

7x  2

in partial fractions. (2 x  3)( x  1) 2  5  7x  2  dx. Hence evaluate   2  2( 2 x  3)( x  1) 2   

Given that y  3 x 3  6 x 2  5 x  7 dy find , (i) dx (ii) find the value of c for which x  y  c is a tangent to the curve, (iii)

show that y decreases as x increases.

[3] [3]

[1] [3] [2] [Turn over

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

4 7

The diagram (not drawn to scale) shows a trapezium OPQR in which PQ is parallel to OR and ORQ  90 . The coordinates of P and R are (4, 3) and (4, 2) respectively and O is the origin. y

Q

T P (4, 3) R (4, 2) O

8

(i)

Find the coordinates of Q.

[3]

(ii)

PQ meets the y-axis at T. Show that triangle ORT is isosceles.

[3]

(iii)

The point S is such that ORPS forms a parallelogram, find the coordinates of S.

[3]

(iv)

Find the area of the trapezium OPQR.

[2]

  Solve, for    x   , the equation cos 2 x    0.5 , leaving your answers 3  in terms of  .

[4]

Solve, for 0    180 , the equation 2 sec   4 cos   5 sin  .

[5]

(a)

(b)

9

x

Find the range of values of k for which the line y  k  3 x meets the curve y

3 at least once. 2 x

[4]

10 Find, in ascending powers of x, the first three terms in the expansion of (1  px) 6 . Given that the first two non-zero terms in the expansion of (1  px) 6 (1  qx) are 1 and 

7 2 x , find the possible values of p and q. 3

[5]

[Turn over

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

5 11 Variables x and y are related by the equation y  ab  x , where a and b are constants. The table below shows measured values of x and y. x y

1 470

2 190

3 80

4 30

5 12

(i)

Plot lg y against x and obtain a straight line graph.

[3]

(ii)

Use your graph to estimate the value of a and of b.

[4]

(iii)

On the same graph, draw the line representing the equation lg y – x = 2 [2]

and hence find the value of x for which ab  x  10 x  2 .

12 Q

ym

P

45

R

S

xm

T

A piece of wire of length 680 m is bent to form an enclosure consisting of a trapezium PQRS and a quadrant PST. Given in the figure PQ  y m, QRˆ S  45 and ST  x m. (i)

Show that the area A m2 of the enclosure is given by A  340 x 

2 1 2 x . 2

[4]

Given that x can vary, (ii)

find the stationary value of A,

[4]

(iii)

determine whether this stationary value is a maximum or a minimum.

[1]

END OF PAPER

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

6 2009 Prelim Exam Additional Mathematics Paper 1 (4 Express / 5 Normal) 1 (a) ln( 2 x  3 y  1)  0 8 x 2 y  4 [8] 2x  3y  1  e0  1 2 3( x 2 y )  2 2 2x  3y  2 M1, M1 3 x  6 y  2 ........(1)  2, 4 x  6 y  4 .......(2) A1 (2)-(1), x = 2 Fr (1), 3(2)  6 y  2 A1 2 y  3 2 x  1 x  1 (b) 3 63

3 2 x  3 1  6  3 x  3 Let y  3 x ,

1 2 y  3y  6  0 3 y 2  9 y  18  0 ( y  3)( y  6)  0

M1

y  3 or y  6

M1

3 x  3 or 3 x  6 x 1 2 [3]

or

x

lg 6  1.63 lg 3

 2  k  x   2         3 k  1 y   4 

A2 M1

1

3 [4]

(a)

2  k  No solutions implies   does not exist  3 k  1 2  k    0  det  3 k  1 Same gradient for 2 lines 2(k  1)  (3k )  0 OR 2 3  2k  2  3k  0 k k 1 2 k    0.4 5 1 y  ln 4 x  1  ln( 4 x  1) 2 dy 1 4  dx 2 4 x  1 2  4x  1

(b)

y

A1

M1 A1

ex x 1

dy ( x  1)(e  x )  e  x  dx ( x  1) 2 

M1

 xe  x ( x  1)

2

or

x x

e ( x  1) 2

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

M1

A1

7 4 [6]

(i)

3 10 3 2 sin x cos x  5 3 sin 2 x  5 sin 2 x 39  Given cos y 25 25 cos y   sin 2 x 39 25 3   39 5 5  ( shown ) 13   x   2 x  (2nd quad ) 4 2 4 cos 2 x   5 3 tan 2 x   4

M1

Majority could get to the ratio for sin 2x but could not show their working clearly while trying to find the ration for cos y

M1

No method mark for 2nd part if pupil jumped from ratio 39:25 to ratio 3 5 : without showing 5 13 any working

sin x cos x 

y acute  1st quad 12 sin y  13 12 cos y  5

3

Pupils who could apply method correctly get the quadrant wrong for angle 2x, hence resulted in wrong 'sign' for the ratios

5 4

13

12

5

tan 2 x  tan y 1  tan 2 x tan y

(ii)

tan(2 x  y ) 

(iii)

3 12  4 5   3 12  1     4 5 33  56 4 cos 2 x   5 4 1  2 sin 2 x   5 9 sin 2 x  10 3 sin x  ( x in 1st quad ) 10 

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

M1

A1

M1

A1

8 5 [6]

(i)

7x  2 (2 x  3)( x  1)

2



A B C   2 x  3 x  1 ( x  1) 2

Quite well done M1

7 x  2  A( x  1) 2  B ( 2 x  3)( x  1)  C ( 2 x  3)

let x  1, 7(1)  2  C (2  3) C 1

let x 

3 3 3  , 7( )  2  A  1 2 2 2  A2

2

M1 (at least 2 correct values found)

let x  0, 2  A  3B  3C B  1 7x  2



(2 x  3)( x  1)

(ii)

2



2 1 1   2 x  3 x  1 ( x  1) 2

  7x  2 2  2(2 x  3)( x  1) 2  dx   1 5 2 1 1     dx 2 2 2 x  3 x  1 ( x  1) 2

A1

5

M1

Many could not handle 1  ( x  1) 2 after integrating the 1st 2 terms in ln, e.g.

5



1 1  ln( 2 x  3)  ln( x  1)   2 x  1 2



1 1  1  ln 7  ln 6    ln 1  ln 3    2 6  3 

 0.710 (3sf )

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

M1 A1

1

 ( x  1)2

 ln( x  1) 2

9 6 [6]

(i)

(ii)

y  3 x 3  6 x 2  5 x  7 dy  9 x 2  12 x  5 dx x  y  c  y  x  c gradient of tangent = –1 dy  9 x 2  12 x  5  1 dx

B1

M1

9 x 2  12 x  4  0 (3 x  2)(3 x  2)  0 2 x 3 3

M1 2

4 2 2 2  y  3   6   5   7  5 9 3 3 3 1 c  x y  6 9

(iii)

dy  9 x 2  12 x  5 dx 4 5   9 x 2  x   3 9  2 2  4 2    2  5  2   9  x  x         3 9   3    3     9 x  

A1

Very badly done. Majority have no idea what to do

M1

2

2  1 3

dy 0 dx hence y decreases as x increases for all real values of x,

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

dy  0, dx they end up with the eqn  9 x 2  12 x  5  0 which gives imaginary root. Coincidently it is the 2 same value x  in 3 their calcuator screen but with the small symbol "xy" which represent imaginary value!!

Many pupils let

A1

10 7 (i) [11]

(ii)

Gradient of PQ = gradient of OR= 0.5 1 Eqn of PQ: y  3  ( x  4) 2 1 y  x  5 --------(1) 2 Gradient of QR =  2 Eqn of QR: y  2  2( x  4) y  2 x  10 ------(2) (1)=(2) 1  2 x  10  x  5 2 5 x5 2 x2 y  2(2)  10  6  Q (2, 6) In eqn (1), let x  0 , y = 5,  OT  5units RT  (4  0) 2  ( 2  5) 2

Well done!

M1

M1

A1 M1

OK

M1

RT  25  5

Since OT = RT = 5 units  ORT is isosceles. (iii) Let S (a, b) Midpoint of RS = Midpoint of OP a  4 b 2  4 3 ,     ,  2   2 2  2 a  4  4 & b  2  3 a  8 b 1 Hence coordinates of S (8,1)

A1

M1

M1

A1

(iv) Area of trapezium OPQR 1 0 4 2 4 0  20 3 6 2 0 1  24  4  24  6 2 1   50 2 

 25units 2

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

Many equate gradient of OP = gradient of RS instead of their midpoint

OK

M1

A1

11 8 [9]

(a)

 cos( 2 x  )  0.5 3 Basic angle =

  x   7  5   2x   3 3 3

   5 7 5  , , , , 3 3 3 3 3 3  2 x  0, ,  ,   , 3 3 2 sec   4 cos   5 sin  2  4 cos   5 sin  cos  2  4 cos 2   5 sin  cos

M1

 cos 2  1  5 2  4   sin 2 2   2 5 2  2 cos 2  2  sin 2 2 5 sin 2  2 cos 2 2 sin 2 2  2 cos 2 5 4  tan 2  5

M1

A2 M1

3x  k 

Common mistakes from step 3 onwards: cos (4 cos  5 sin  )  2

cos  2 or 4 cos  5 sin   2

M1

Basic angle = 38.66 o  2  38.66 or 218.66   19.3 or 109.3 (1 dec pl ) 9 [4]

Very badly done, only a couple could list down all the ans in radian in terms of  accurately

 3

 2x 

(b)

M1

M1 A1 Many pupils let D > 0. If end up with the correct solving method up to k < –12 or k > 0, all 3 method marks awarded

3 2x

6 x  3 x 2  2 k  kx  3 3 x 2  ( k  6) x  3  2k  0 For line to meet curve at least once implies one or more real roots, therefore discriminant  0

M1 M1

( k  6) 2  4(3)(3  2k )  0 k 2  12k  36  36  24k  0 k 2  12k  0 k ( k  12)  0 k  12 or k  0

-12

0

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

M1 A1

able to identify a, b & c for discriminant & reduce it to get k (k + 2)

12 10 [5]

Many pupils let coefficient of x = 1 instead of 0, i.e. q  6p 1

 6  6 (1  px) 6  1    px   ( px) 2  ... 1  2  1  6 px  15 p 2 x 2  ...

A1 Many end up with only 1 set of 1 ans, i.e. p  & q  2 3

6

(1  px) (1  qx)  (1  6 px  15 p 2 x 2  ..)(1  qx)  1  qx  6 px  6 pqx 2  15 p 2 x 2  ...

 q  6 p  0  q  6 p........(1) 7 6 pq  15 p 2   .................(2) 3 Subt (2) into (1) 7 6 p (6 p )  15 p 2   3 7  36 p 2  15 p 2   3 1 p2  9 1 p 3 from (1), q  2

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

M1 M1

A1 A1

13 11 [9]

(i)

Quite a number could not define the scale appropriately, hence missing out the lg y–intercept

y  ab  x lg y  (  lg b) x  lg a

B3

Draw straight line with appropriate scale lg y lg y = x+2

(0,3.08)

Commonly seen: y  ab  x lg y   x lg a  lg b

(iii) – B1

gradient = –lg a intercept = lg b (3.2, 1.8)

(0,2)

(i) Graph – B2

lg y = (–lg b) x + lg a

0.8

(ii)

x

3.08  1.8 0  3 .2  lg b  0.4 (accept  0.45 to  0.36) b  2.51 gradient 

lg a  3.08 ( accept 3 to 3.2)

M1 A1 M1

3.08

(iii)

a  10  1202.26  1200 (3sf )

A1

Draw lg y = x + 2

B1

y  ab  x  10 x 2 lg y  ( x  2) lg 10 lg y  x  2 From the intersection point, x = 0.8

B1

AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

Many left blank

14 12 [9]

Q

(i)

P

Badly done or left blank

y x

x

45

sin 45  1 2

y

M

R



x

S

x QR

tan 45 

x RM

x RM RM  x

x QR

1

QR  2 x Given perimeter = 680 m 2x 2 x  2x  y   y  680 4  2 y  680  2 x  2 x  x 2 2  y  340  x x x 2 4  Area, A  

T

M1

M1

1 2 x 2 x  xy  2 4

 1 2 2 x  x 2 x  x 340  x  x    2 2 4 4 

1 2 2 2 x 2 x 2 x  340 x  x  x2   2 2 4 4 2 2 1 2  340 x  x  x 2 2 2 1 2  340 x  x 2 dA  340  2  1 x  0 dx 340 x  140.8326 2 1 x  141 m 3sf 

M1





(ii)







2  1  340   340   A  340    2  2  1  2 1 d2A

M1

M1

Many could get the method mark but not the accuracy mark, check their calculator skill!

2

A  23941.54  23 900 m2 (iii)

A1

M1 A1

 ( 2  1)  0

dx 2 Hence A is maximum. AMKSS 4E/5N Additional Math Prelim Exam 2009, 4038/01

B1

left bracket out! d2A   2 1 dx 2