Tampines Prelim 2009 Am 1 Solutions

1

Name:_________________________________(

) Class : Sec 4_____

TAMPINES SECONDARY SCHOOL PRELIMINARY EXAMINATION 2009 SECONDARY FOUR EXPRESS ADDITIONAL MATHEMATICS PAPER 1 3 Sept 2008

4038 / 1 2 hours

Additional Materials: Writing Paper

READ THESE INSTRUCTIONS FIRST Write your name, class and register number on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your calculator model on the top right hand corner of your answer script. Answer all questions. Write your answers on the separate writing papers 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 a scientific 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.

Setter: Ms CNF This document consists of 6 printed pages and 0 blank pages.

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

 b  b 2  4 ac 2a

Binomial Theorem n  n1  n n a b   a n 2 b 2  ...   a n  r b r  ...  b n ,  1  2 r

a  b n  a n  

n n! where n is a positive integer and    r ( n  r )!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 tan 2 A 

2 tan A 1  tan 2 A

1 1 ( 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 Formulae for ABC a b c   sin A sin B sin C sin A  sin B  2 sin

a 2  b 2  c 2  2bc cos A 

1 ab sin C 2

3

1

Show that the exact value of cos120° + sin135° =

1 2



1 2

[3]

cos(180  60)  sin(180  45)  cos 180 cos 60  sin 180 sin 60  sin 45

(M1 Award marks so long as students use “special angles” to get the surds.

1 1  ( 1)   0  2 (M1 2 1 1   (A1) 2 2

2

Solve, for x and y, the simultaneous equations

8x  4

[5] 3 y 2

1 81x    3

2 3 x  23 y  2 6 3x  3 y  6 x y2 Award

x  1 y  3

(B1

(M1)

 64

2 y 1

 27

34 x  31 2 y  33 4 x  2 y  2 (B1 2x  y  1

for solving sim eqns (whatever mtd)

(A2)

4

3

Given that M =

 1  4   , find M-1 and hence, solve the simultaneous equations 2  3   2p – 8r + 16 = 0 2p – 3r = - 1

1   3 4  M 1   5  2 1

(B1

 1  4  p    8         2  3  r    1 

(B1  p    r

1   3 4   8     5   2 1   1 

1  20     5  15   4     3 p4 (A2) r 3

(M1

If B1 above is not achieved, award (M1) if student used inverse matrix as mtd.

[5]

5

4

(i)

Find

d 2 ( x ln x  3 x) . dx

[2]

1  ln x 2 x   x 2    3 x

(B1  2 x ln x  x  3

(B1

4

(ii)

Hence, find

 x ln x

dx .

[4]

1

2 2 x ln x  x  3 dx  x ln x  3 x 

(B1 2  x ln x dx  x 2 ln x  3 x   x  3 dx

(M1) 4

4

1  2 1 2   x ln x d x  x ln x  3 x  x  3 x      1 2 2    1  

Award (M1) if student attempts to integrate (x – 3)

4

4

3 1 2 3  1 2 x ln x d x  x ln x  x  x  x 1  2 2 4 2 1 4

4

1 2 1 2 x ln x d x  x ln x  x 1  2 4 1  7.34

(A1)

Award (M1) for sub-ing values 4 & 1

6

5

(i)

Express

7x  6 in partial fractions. x (1  3 x)

7x  6 A B   x (1  3 x ) x 1  3 x

A6



(ii)

[3]

(B1 Award (M1) for mtd to arrive at A & B

B  25

7x  6 6 25   x (1  3 x ) x 1  3 x

(A1)

Hence, or otherwise, find the gradient of the curve y  where x = 2.

d 7x  6 dx x (1  3 x ) d 6 d 25   dx x dx 1  3 x 6 75  2  x 1  3x 2



Sub x = 2

6 75  2 2 1  3( 2) 2  1 .5 (A1)

gradient  

(M2)

7x  6 at the point x (1  3 x )

[3]

7

6

A particle starts from a point O and moves in a straight line with a velocity, v m/s, 2

given by v  2t  3t  2 , where t seconds is the time after leaving O. Calculate (i)

the acceleration of the particle at the end of 3.5 seconds,

[2]

(B1

a  4t  3 Sub x = 3.5

a  11 (ii)

(B1

the values of t when the particle is instantaneously at rest,

2t 2  3t  2  0 (t  2)( 2t  1)  0 t2 (iii)

s

(B1)

Award

[3]

for setting v = 0

(M1)

(A1)

the total distance travelled by the particle after 3.5 seconds.

2 3 3 2 t  t  2t  c 3 2

(M1)

When t = 0, s = 0 therefore, c = 0 **Deduct 1m if student did not explain why c = 0 Sub Sub

t = 2 t = 3.5

s  4

2 3

(M1)

2 2 Dist  4  4  3.2083  12.5 3 3

s  3.2083

(A1)

(M1)

[4]

8

7

The equation of a curve is y 

3 cos x . Find the x – coordinate, where 0 ≤ x ≤ π, 1  sin x

of the point at which the tangent to the curve is parallel to the line y = 9.

(B1 dt (1  sin x )( 3 sin x )  3 cos x (  cos x )  dx (1  sin x ) 2

(B1

 3 sin x  3 sin 2 x  3 cos 2 x  (1  sin x ) 2 3  3 sin x  (B1 (1  sin x ) 2

dt 0 Award dx sin x  1   2  x (A1) 2

(M1)

dy for setting dx = 0

[5]

9

8

(a)

Given that sin A  

12 4 , cos B   and that A and B are in the same 5 13

[5]

quadrants, find each of the following without the use of calculators: (i)

tan A

(ii)

4 3

sin( A  B)

(iii)

1 sec B 2

(B1

(i)

tan A 

(ii)

sin( A  B )  sin A cos B  cos A sin B

 4  12   3  5               5  13   5  13  33  (A1 65

(iii)

1 1  sec  B    2  cos  1 B    2  1  sec B   2 

1 

1 26

1  sec  B    5.10 2 

(A1

(M1)

for either one of these values

cos 2 B  2 cos 2 B  1 1  cos B  2 cos 2  B   1 2 

cos B  1 1  cos  B    2 2   12    1 1  13  cos  B     2 2 

1 1  cos  B    26 2  Since B is in the 3rd

(M1)

10

1 1  cos  B    26 2 

quad, (b)

Solve the equation

sin 4 x  sin 2 x  cos 3x  0

for 0 < x < 2π.

2 cos 3 x sin x  cos 3 x  0

(B1 cos 3 x ( 2 sin x  1)  0 cos 3 x  0

(M1) sin x 

 2   5 7 3 11 x , , , , , 6 2 6 6 2 6

 6  5 x , 6 6



x 

  5 7 3 11 , , , , , 6 2 6 6 2 6

1 2



(A1

**Accept answers in rad to 3 s.f

(M1)

[4]

11

9

The line 3x – 2y – 1 = 0 intersects the curve 9x2 + y = 7 at points A and B. Find (i)

the coordinates of A and B,

[3]

(M1)

Perform Sim Eqns :

5 3  ,  and 6 4

 1,2  (A1)

(A1) (ii)

the distance AB. 2

5  3    1    2  6  4   3.31 (A1)

10

(a)

[2] 2

Find the range of values of c for which

3x 2  9 x  c  2  0 (9) 2  4(3)(c  2)  0 < 0

c  8.75 (b)

(M1)

3 x 2  9 x  c > 2.

[3]

(B1 Award

(M1)

for discriminant

(A1) 2

Find the range of values of k for which the equation x  kx  3 x  k will have real roots.

( k  3) 2  4(1)(k )  0 ≥ 0

Award

(M1)

for discriminant

[3]

12

k 2  10k  9  0 ( k  9)(k  1)  0 k 1

(M1)

k 9 6

11

(a)

2  Find the first three terms in the expansion of  x   . Hence, find the x 

(A1)

6

2  coefficient of x 4 in the expansion of ( 2  3 x ) x   . x  2

2

 2  2 ( x )  C1 ( x)     6C2 ( x) 4     ...  x  x 6

6

5

 x 6  12 x 4  60 x 2  ... **

(A1)

Penalise 1m for omitting the dots……

6

2  2  3x  x   x   2  3 x 2 x 6  12 x 4  60 x 2  ... (M1)



Award (M1) for expanding using Bin thrm

2



 24 x 4  180 x 4  ...  156 x 4  ...

 coefficien t  156

(A1)

Award (M1) if student used the above expansion to find coefficient of x4

[4]

13

8

(b)

1   3  . Find the term independent of x, in the expansion of  2 x  4x 5  

r

1   1   8 Cr ( 2 x 3 )8  r   5  8Cr 28  r x 24  3 r   r 5 r   4x   4 x  24  3r  5r  0 r  3 (M1)

(M1) 3

8

3 8 3

C3 ( 2 x )

1   1  15    56 ( 32 x )    5  15   4x   64 x   28 (A1)

(M1)

[4]

14 12

[4] (0, 4)

Variables x and y are related by the equation 2 px  qy

 3xy .

When a graph of

1 1 against is drawn, the resulting line has a gradient of – 2.5, and passes y x through (0, 4). Calculate the value of p and of q.

2 px  qy  3 xy

÷ xy

1 1 2 p   q   3  y  x 1 1 2 p    q   3 x  y

1 q 1 3    y 2p  x  2p



q   2. 5 2p

 q  5p q

15 8

(A1)

3 4 2p 3 p  (A1) 8

(M2)

15 13

O y

4x

A diagram above shows a piece of grassland, made up of a rectangle and a quadrant. The length and breadth of the rectangle is 4x m and y m respectively. The centre of the quadrant, O is the midpoint of the length of the rectangle. Given that the perimeter of the grassland is 8 m, (i)

show that the area of the grassland is given by the formula A  16 x  16 x 2  x 2 .

y  4 x  y  2 x  2 x  x  8

y  4  4x 

1 x 2

1 A  4 xy   ( 2 x) 2 4 1   A  4 x 4  4 x  x   x 2 2   A  16 x  16 x 2  2x 2  x 2 A  16 x  16 x 2  x 2

(A1)

(B1 (M1)

(M1)

[4]

16 Given that x can vary, (ii)

find the stationary value of A,

[4]

dA  16  32 x  2x (B1 dx 0  16  32 x  2x Award (M1) for setting dA/dx to be zero 16 x (M1) or 0 . 41793 32  2 2

 16   16   16   A  16   16     32  2   32  2   32  2   3.34 (A1) (iii)

2

determine whether this stationary value is a maximum or minimum.

d2A  16  32 x  2 2 dx

 max

Award (B1) for both d2A/dx2 & conclusion that its max ** Do NOT award any marks once dA/dx is not correct

End of paper

[1]