Pierce Prelim 2009 Am1 & Solution

Additional Mathematics 4038 /Prelim 2009/ Sec 4E AM P1/ Mark Scheme Qn #

Solution

1

 9     5 7 

ab 7 9



5 7

9 5 7



Qn # 4

2

(i)

5 7 5 7

1

1 11  7    10  3  1  Rearranging equations,  y  7 x  13  3 y  11x  9   1 7  y  13          3 11 x   9 

9(5  7 ) 25  7 5 7  2

=



2

5 7   9        5 7   2 

2

25  10 7  7 4 5  8 7 2 5  a  8, b   2 

2

(i)

(ii)

3

1 sin(90   ) 1  cos 3

cosec (90   ) 

1 1  2 sin 2   cos 2 1 1 2 1 3 1 sin   2 2 3 1 1 sin   2 3

f(x) = 9 x 4  5ax 3  7bx 2  5 x  6 x – 1 is a factor  f(1) = 0 9 + 5a – 7b + 5 + 6 = 0 5a – 7b = 20 ……….….. (1) f(1) = 40 9 – 5a – 7b – 5 + 6 = 40 5a + 7b = 50 ……...…….. (2) Solving, a = 3, b = 5

Solution  1 7  A =     3 11 11  7  1 A-1 =    11  21  3  1 

 y  1 11  7 13         x  10  3  1  9   8     3 Point of intersection is (3, 8)

1

2 (ii) 1

1 5(a)

(i)

7 x  y  13 14 x  2 y  5 They are parallel lines and have no point of intersection. OR A singular matrix will be obtained, resulting in no inverse. 2x 2  9x  5  0 (2 x  1)( x  5)  0 1 x5 2 2  x  3 x 2  0 or 3 x 2  x  2  0 

1 (ii)

(b)

1 1

1

1 1

1

(k  3 )(k  3 )  0  3k 3

2

1

x 2  2kx  3 always positive  b 2  4ac  0 4k 2  12  0  k 2  3  0

1

1

(3 x  2)( x  1)  0 1 2 x  1, x  1 3

1

1 1

1

1

Additional Mathematics 4038 /Prelim 2009/ Sec 4E AM P1/ Mark Scheme

6

(i)

1 20 cos 20  20  40 sin   2  400 cos   400 cos sin 

A

 400 cos   200 sin 2

(ii)

(iii)

1 1

(i)

8

 

2

 2 6  6C1 ( 2) 5 x 2  6C2 ( 2) 4 x 2  ...

(ii)

2

 64  192 x 2  240 x 4  ... n  1  1  x   2  1

2

 1   1   1 nC1   x   nC2   x   ...  2   2  1 n(n  1)  1 2   1  nx   x   ... 2 2 4  1 n(n  1) 2  1  nx  x  ... 2 8 1 3 Comparing with 1  nx  x 2 +… 2 2 n(n  1) 3  8 2 2 n  n  12 (n  4)(n  3)  0 n4

1

1

2 6 1

1

1

2  x 

 



 288 x 2  ... Coefficient of x 2 = 288

2 sin 2   sin   1  0 (2 sin   1)(sin   1)  0 1 , sin   1 2  5     or    (  acute) 6 6 6



 ... 

 96 x 2  192 x 2  ...

1 1

sin  

4

3 2  1  1  nx  x  ... 2 2   3   64 x 2   192 x 2 1  ... 2 



 sin   1  2 sin 2   0

7

2 6

2

1

1    400  cos  sin 2  2   dA For maximum area, 0 d 400( sin   cos 2 )  0

n

2  x  1  12 x     64  192 x  240 x

1

1

1

(i)

LHS 

sin A  cosec A cos A  sec A

1 sin A  1 cos A  cos A 2  sin A  1    sin A     cos 2 A  1     cos A  sin 2 A  1 cos A   sin A cos 2 A  1  cos 2 A cos A   sin A  sin 2 A  cot 3 A sin A 

 RHS

1

1

1

Additional Mathematics 4038 /Prelim 2009/ Sec 4E AM P1/ Mark Scheme

(ii)

10(a)

sin A  cosec A  5 cos 3 A cos A  sec A

p  log 2 q 32 log 2  log 2 32  log 2 q q

cot 3 A  5 cos 3 A cos 3 A  5 cos 3 A 3 sin A cos 3 A  5 cos 3 A  0 3 sin A  1  cos 3 A  3  5   0  sin A  1 cos 3 A  0 or 5 0 sin 3 A cos A  0 ( NA) 1 or sin 3 A  5 sin A  0.5848 When sin A = 0.5848,  = 35.8 A = 35.8 or 144.2

9

(i)

 5 p (b)

2 log 2 x 2 log 2 4

log 2 x( x  3)  2 ( x  1)( x  4)  0 x  1, x  4 x  1 (  x  0)

1

(ii)

e x (3e x  2)  5

ARQ = x (s in opp seg)

1

(3e x  5)(e x  1)  0 5 or e x  1 3 5 e x  ( e x  0) 3 5 x  ln 3  0.511 ex 

1

ABS = QRS = x 1

 ABS and QRS are similar 1

 SA  SR  SB  SQ By tan-sec thm,

ST 2  SB  SQ SA  SR  SB  SQ

1

1

3e 2 x  2e x  5  0

ST 2  SA  SR

1

x  3x  4  0

1

AS BS  QS RS

1

2

ABS = x (s in alt seg)

ASB is common

(iii)

log 2 ( x  3)  2 log 4 x  log 6 36

log 2 ( x  3)  log 2 x  2

1

 AQ = RQ (ARQ is isosceles)



(i)

log 2 ( x  3) 

But RAQ = x (vert opp s)

(ii)

1

1

Let SAP = x

ARQ = RAQ

1

1 1

1

1

1

Additional Mathematics 4038 /Prelim 2009/ Sec 4E AM P1/ Mark Scheme

11

(a)

11

3x 2  5x  1  0 5 3 1 Product of roots,   3 New sum, (   )  (    )  0 New product, (   )(   )

Sum of roots,    

1 1 1 (c)

    2   2    2  ( 2   2 )



 2  (   ) 2  2  4  (   ) 4 25  3 9 13  9



1

2



13 0 9 9 x 2  13  0

New equation: x 2  or

(b)

1 1

x 2  8x  m  7  0 Let , 3 be roots. Sum of roots, 4 = 8  = 2 Product of roots, 32 = m + 7 m = 3(4) – 7 = 5 x 2  x  (1  2 )  0 Sum of roots, tan A  tan B  1 Prod of roots, tan A tan B  1  2 tan A  tan B tan( A  B)  1  tan A tan B 1  1  (1  2 ) 1  2

1 1

1

1

Additional Mathematics 4038 /Prelim 2009/ Sec 4E AM P1/ Mark Scheme

12

(i) ln x ln y

0.69 3.53

1.38 5.40

Table Graph (correct axes, correct points) (ii)

(iii)

y  ax

1.79 6.27 (1) (2)

n

ln y  ln a  n ln x

(1)

Y-intercept = 1.8 ln a  1.8

(1)

a  e1.8  6.05 Gradient = 2.5 n = 2.5

(1)

n

(1)

4

ax  x  y  x4

ln y  ln x 4  4 ln x Plot graph of Y  4 ln x Graph x = 1.2

(1) (1)

2.08 7.00

2.30 7.52