St Gabriels Prelim 2009 Am P2 Solutions

SAINT GABRIEL’S SECONDARY SCHOOL 2009 ‘O’ LEVEL ADDITIONAL MATHEMATICS PRELIMINARY EXAMINATION Secondary Four Express Paper 2

Marking Scheme 1(i)

2(i) amplitude = 1 (B1) 2 Period = (B1) 3 (ii) points : (B1) curve : (B1)

1200 1199e kt  1 1200 66  1199e k ( 5)  1  1200  e  5k    1  1199  66   1200    5k  ln   1  1199   66   k = 0.849 (3dp) N

y 4

3

(M1A1) 2

(ii)

1200 N = 962.76 1199e 0.849(10)  1 1

 total no of students = 963.

(M1A1)

1200 (iii)  0.35  1200 1199e 0.849t  1  1200  e  0.849t    1  1199  420   1200    0.849t  ln   1  1199   420 

t  7.62  number of days = 8 days.

(M1A1)

1 x 2

y = e  ln x , x > 0 1 dy 1 2 x 1  e  (B1,B1) dx 2 x dy For stationary points, 0 dx 1 1 2x 1  (B1) e  2 x 1 1 x Consider y  e 2 , it is an increasing function. 2 1 Consider y  , it is an decreasing function. x Since one is an increasing function & the other is an decreasing, there will only be 1 point of intersection. (B1) 3

1

0.5

1

 6

1.5

 3

2

 2

   ,1 2 

(iii) min turning point =

2x  1 = sin 3x π 2x  1 = sin 3x + 2 π  number of solution = 1.

2 3

(B1)

(iv)

(B1) (B1)

Con’t 3 1

d 2 y 1 2x 1  e  2 dx 2 4 x 1

(B1,B1)

x

Since x > 0, e 2 > 0, and 

1 >0 x2

d2y 0 dx 2

 it is a minimum point.

(B1)

x

 only 1 stationary point for y = e 2  ln x .

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x

2

5x  6 5x  6 4(i)  2 ( x  2)( x  4) ( x  2) 2 ( x  2) A B C =   (B1) 2 ( x  2) ( x  2) ( x  2) 5x + 6 = A(x +2)(x 2) + B(x  2) + C(x + 2)2 let x = 2,  C = 1 (B1) let x = 2,  B = 1 (B1) let x = 0. A = 1 (B1) 

5x  6 1 1 1    2 2 ( x  2)( x  4) ( x  2) ( x  2) ( x  2)

(ii)



5x  6 dx 3 ( x  2)( x 2  4) 5

5

1   =  ln( x  2)   ln( x  2) (B3) x2  3 5

1   x2 = ln   x  2 x  2  3 3 1 1 1 = ln   ln   0.819 (A1) 7 7 5 5 6a 2x + 2y = 9 --------- (1) 3x1 = 32  3y x  1 = 2 + y --------- (2) x=3+y (B1) sub (1) into (2) 23+y + 2 y = 9 (23)(2y) + 2y = 9 Let z = 2 y 8z + z = 9 (M1, A1) 9z = 9 z = 1,  y = 0, x = 3 (B1)

6b

radius 2 = (1  3 ) 2 = 2 + 2 3 1 2 2 r h   (8  3 3 ) (B1) 3 3 2(8  3 3 ) 4  2 3 h=  42 3 42 3 28  8 3 4 = 7  2 3 cm

=

(M1,A1)

(B1)

5(i) 3x2 + px + 120 = 0    = 3 ------------ (1)  = 120  3 = 40 -----------(2) Sub  = 3 +  into (2) (B1) (3 + )  = 40 2 + 3  40 = 0 ( + 8)(   5) = 0   =  8 or  = 5  = 5 or  = 8 (rejected) (M1,A1) p +=  3 p 5  8 =   p = 39 (B1) 3 (ii)  +  = 13  = 40 (2 + 2) = ( + )2  2 = (13)2  80 = 89 (M1,A1) 2 2 () = 40 = 1600 (B1) 2 eqn is x  89x + 1600 = 0 (B1) 7(i) ADO = ACO (B1) (s in the same segment, chord = OA) DAC = DOC (B1) (s in the same segment, chord = DC) Or DEA = CEO (vertically opp ) [any of the 2 reasons] triangles AED & CEO are similar. (B1) (ii)

OC AD  , using similar triangles. EC ED

 OC  ED = AD  EC

(M1,A1)

(iii) ADO = ACO ACO = OAC (isos. ) (B1) ODC = OAC (B1) (s in the same segment, chord = OC)  ADC = ADO + ODC = OAC + OAC = 2  OAC (shown) (B1)

2009 AM 4E5N Prelim P2

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2009 AM 4E5N Prelim P2

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11(i)

2009 AM 4E5N Prelim P2

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