![N0. SOLUTION MARKS
3 1
y = x3 − x 2 + 2
3
(a) dy
= x2 − 2 x = 3 K1 Equate and solve
dx
x2 − 2 x − 3 = 0 quadratic
equation
( x + 1) ( x − 3) = 0
x = −1 , 3
2
x = −1 y=
3
x=3 y=2
2
−1, and ( 3, 2 ) N1 N1
3
(b) Equation of normals :
1
mnormal = −
3 K1 Use mnormal to form
2 1 1 equations
y− = − ( x + 1) y−2=− ( x − 3)
3 3 3
1 1 1 N1 N1
y=− x+ or equivalent y = − x+3 or equivalent
3 3 3
6
4
y
(a)
P1 Modulus sine
shape correct.
2 y = 3sin 2 x − 1
P1 Amplitude = 3
[ Maximum = 2
1
and Minimum =
-1]
O π π 3π 2π x P1 Two full cycle in
-1 2 2 0 ≤ x ≤ 2π π
3x
y = 1−
-2
2π P1 Shift down the
graph
3](https://image.slidesharecdn.com/addmaths2-100920222130-phpapp02/85/Add-Maths-2-2-320.jpg)
![N0. SOLUTION MARKS
6
(a) K1 for using vector
(i) uuu uuu uuu
r r r triangle for a(i) or
OD = OC + CD
a(ii)
= 6a + 12b
% % N1
(ii) uuu uuu uuu
r r r
AB = OB − OA
1 uuu uuu
r r
= OD − OA
2
= 3a + 6b − 3a
% % % N1
= 6b
%
OR
uuu 1 uuu
r r
AB = CD = 6b [ K1 N1 ]
2 %
(b)
uuur
AE
uuu uuu
r r
= AB + BE K1 for using vector
uuu
r
1 uuu
r triangle and BE
= 6b + h OD
% 2
= 6b + h ( 3a + 6b )
% % %
a + kb = 3ha + ( 6 + 6h ) b
K1
% % % %
k = 6 + 6h K1 for equating
3h = 1 coefficients
1
1 = 6 + 6 correctly
h= 3
3
=8 N1 N1
8
5](https://image.slidesharecdn.com/addmaths2-100920222130-phpapp02/85/Add-Maths-2-4-320.jpg)
![N0. SOLUTION MARKS
10.
(a) Equation of str. line PQR :
1 K1
m= −
2
1 N1
y= − x+1
2
(b) 1 K1 solving
2x + 6 = − x+1
2 simultaneous
equation
P( –2, 2) N1
(c) 1( x) + 2(−2) 1( y ) + 2(2) K1 Use the ratio
=0 or =1
1+ 2 1+ 2 rule
R( 4, –1) N1
(d)
(i) 1 K1 Use distance
( x − 4) 2 + ( y + 1) 2 = ( x + 2) 2 + ( y − 2) 2
2 formula
4 [ x2 – 8x + 16 + y2 + 2y +1 ] = x2 + 4x + 4 + y2 – 4y +4
x2 + y2 – 12x + 4y + 15 = 0 N1
(ii) Substitute x = 0, y2 + 4y + 15 = 0 K1 Substitute x = 0
b2 – 4ac = (4)2 – 4(1)(15) 2
and use b – 4ac
= – 44 < 0 to make a
conclusion
⇒ No real root for y,
⇒ The locus does not intercept the y-axis. N1 if b2 – 4ac = -44
9](https://image.slidesharecdn.com/addmaths2-100920222130-phpapp02/85/Add-Maths-2-8-320.jpg)
Add Maths 2
- 1. MARKING SCHEME ADDITIONAL MATHEMATICS PAPER 2 SPM TRIAL EXAMINATION 2010 N0. SOLUTION MARKS 1 x = 10 − 2 y P1 K1 Eliminate x y 2 + (10 − 2 y ) y = 24 y 2 − 10 y + 24 = 0 K1 Solve quadratic ( y − 4) ( y − 6) = 0 equation y=4 or y=6 N1 x=2 or x = −2 N1 5 2 (a) k=6 P1 (b) Mid point 23 , 28 , 33 , 38 , 43 P1 (i) Mean = ∑ fx = 1 × 23 + 4 × 28 + 7 × 33 + 5 × 38 + 3 × 43 K1 Use formula and calculate ∑f 1+ 4 + 7 + 5 + 3 685 = = 34.25 N1 20 (ii) Varian = ∑ fx2 − x 2 ∑f K1 Use formula and 1 × 232 + 4 × 282 + 7 × 332 + 5 × 382 + 3 × 432 calculate = − 34.252 20 24055 = − 34.252 20 N1 = 29.69 (iii) Median , m 1 1 K1 Use formula and 2N −F 2 (20) − 5 calculate = L+ C = 30.5 + 5 fm 7 N1 = 34.07 8 2
- 2. N0. SOLUTION MARKS 3 1 y = x3 − x 2 + 2 3 (a) dy = x2 − 2 x = 3 K1 Equate and solve dx x2 − 2 x − 3 = 0 quadratic equation ( x + 1) ( x − 3) = 0 x = −1 , 3 2 x = −1 y= 3 x=3 y=2 2 −1, and ( 3, 2 ) N1 N1 3 (b) Equation of normals : 1 mnormal = − 3 K1 Use mnormal to form 2 1 1 equations y− = − ( x + 1) y−2=− ( x − 3) 3 3 3 1 1 1 N1 N1 y=− x+ or equivalent y = − x+3 or equivalent 3 3 3 6 4 y (a) P1 Modulus sine shape correct. 2 y = 3sin 2 x − 1 P1 Amplitude = 3 [ Maximum = 2 1 and Minimum = -1] O π π 3π 2π x P1 Two full cycle in -1 2 2 0 ≤ x ≤ 2π π 3x y = 1− -2 2π P1 Shift down the graph 3
- 3. N0. SOLUTION MARKS 4 3x (b) 3sin 2 x − 1 = 1 − 2π or N1 For equation 3x y = 1− 2π 3x Draw the straight line y = 1− K1 Sketch the 2π straight line Number of solutions = 5. N1 7 5 (a) Common ratio, r=4 N1 (b) 1 K1 A6 = π ( 32 ) 1 T6 = ar = π ( 4 ) 2 5 5 4 OR 4 N1 = 256π = 256π (c) S6 − S 2 K1 Use S6 or S2 1 ( π 46 − 1 1 ) ( π 42 − 1 ) K1 Use S6 - S2 = 4 −4 4 −1 4 −1 N1 = 341.25π −1.25π = 340π 6 4
- 4. N0. SOLUTION MARKS 6 (a) K1 for using vector (i) uuu uuu uuu r r r triangle for a(i) or OD = OC + CD a(ii) = 6a + 12b % % N1 (ii) uuu uuu uuu r r r AB = OB − OA 1 uuu uuu r r = OD − OA 2 = 3a + 6b − 3a % % % N1 = 6b % OR uuu 1 uuu r r AB = CD = 6b [ K1 N1 ] 2 % (b) uuur AE uuu uuu r r = AB + BE K1 for using vector uuu r 1 uuu r triangle and BE = 6b + h OD % 2 = 6b + h ( 3a + 6b ) % % % a + kb = 3ha + ( 6 + 6h ) b K1 % % % % k = 6 + 6h K1 for equating 3h = 1 coefficients 1 1 = 6 + 6 correctly h= 3 3 =8 N1 N1 8 5
- 5. N0. SOLUTION MARKS 7 (a) x 1 2 3 4 5 6 N1 6 correct log10 y 0.65 0.87 1.08 1.30 1.52 1.74 values of log y log10 y (b) K1 Plot log10 y vs x Correct axes & uniform scale N1 6 points plotted correctly N1 Line of best-fit 0.43 0 x (c) log10 y = ( k log10 A ) x + log10 A P1 (i) x = 2.6 N1 (ii) y-intercept = log10 y K1 A = 2.69 N1 gradient = k log10 A K1 gradient k= log10 A * = 0.51 N1 10 6
- 6. N0. SOLUTION MARKS 8 (a) P(5, 1 ) P1 Q(1, 0 ) P1 (b) 1 ∫ (4y + 1) dy ∫ x dy 2 A= K1 use 0 K1 correct limit 4y3 1 = + y 3 0 K1 integrate correctly 7 = OR equivalent 3 N1 5 x −1 (c) V= π∫ dx K1 integrate 4 1 π ∫ y 2 dx K1 correct limit π x2 5 = − x 4 2 1 K1 integrate correctly = 2π N1 10 7
- 7. N0. SOLUTION MARKS 9 (a) 4 K1 Use ratio of cos ∠ POQ = 10 trigonometry or equivalent ∠ POQ = 1.16 rad. N1 (b) ( 2π – 1.16 ) rad P1 PQ = 10 ( 2π – 1.16 ) K1 Use s = rθ = 51.24 cm N1 (c) P1 10 2 − 4 2 = 9.17 cm 1 K1 Area of trapezium POQR = ( 6 + 10 ) × 9.17 * 2 = 73.36 cm2 1 K1 Use formula Area of sector POQ = (10) 2 (1.16) 2 1 A = r 2θ = 58 cm2 2 Area of shaded region = 73.36 – 58 K1 = 15.36 cm2 N1 10 8
- 8. N0. SOLUTION MARKS 10. (a) Equation of str. line PQR : 1 K1 m= − 2 1 N1 y= − x+1 2 (b) 1 K1 solving 2x + 6 = − x+1 2 simultaneous equation P( –2, 2) N1 (c) 1( x) + 2(−2) 1( y ) + 2(2) K1 Use the ratio =0 or =1 1+ 2 1+ 2 rule R( 4, –1) N1 (d) (i) 1 K1 Use distance ( x − 4) 2 + ( y + 1) 2 = ( x + 2) 2 + ( y − 2) 2 2 formula 4 [ x2 – 8x + 16 + y2 + 2y +1 ] = x2 + 4x + 4 + y2 – 4y +4 x2 + y2 – 12x + 4y + 15 = 0 N1 (ii) Substitute x = 0, y2 + 4y + 15 = 0 K1 Substitute x = 0 b2 – 4ac = (4)2 – 4(1)(15) 2 and use b – 4ac = – 44 < 0 to make a conclusion ⇒ No real root for y, ⇒ The locus does not intercept the y-axis. N1 if b2 – 4ac = -44 9
- 9. 10 N0. SOLUTION MARKS 11 (a) µ = 80, σ = 12 65 − 80 X −µ P ( X ≥ 65 ) = P ( Z ≥ ) K1 Use Z = 12 σ = P ( Z ≥ − 1.25 ) = 1 – 0.1056 K1 Use 1 – Q(Z) = 0.8944 N1 (b) 0.1056 × 4000 K1 = 422 or 423 N1 (c) 200 P1 = 0.05 4000 Q( Z ) = 0.05 Z = 1.645 K1 Find value of Z m − 80 m−µ = − 1.645 K1 Use 12 σ K1 Use negative value m = 60.26 g N1 10 10
- 10. N0. SOLUTION MARKS 12 (a) - 5 ms-1 N1 (b) v<0 K1 2 t - 4t - 5 < 0 K1 (t – 5) (t +1) < 0 0<t<5 N1 v (c) 8 7 7 6 P1 (for shape ) 4 2 -5 0 0 2 5 6 10 15 t -2 2 6 -4 P1 min(2,-9) , (6,7) -5 -5 -6 &(0,-5) must be -8 seen -9 -9 -10 -12 (d) Total dis tan ce 5 6 K1 for = ∫ vdt ∫ + vdt 5 6 0 5 ∫ 0 and ∫ 5 5 6 t3 t3 = − 2t 2 − 5 t + − 2 t 2 − 5 t 3 0 3 5 K1 (for Integration; either one) 5 3 216 5 3 = − 2 (5 ) 2 − 5 (5 ) − ( 0 ) + 2( 36 ) − 30 − − 2 (5 ) 2 − 5 (5 ) 3 3 3 K1 (for use and summation) 1 1 = −33 + −30 − (−33 ) 3 3 2 = 36 m 3 N1 11
- 11. 10 N0. SOLUTION MARKS 13 P09 (a) × 100 = 125 K1 60 N1 P09 = RM 75 (b) (125 × 4 ) + (120 m) + (80 × 5 ) + 150 m + 450 K1 (i) 120 = 12 + 2 m K1 (use formula) 1440 + 240 m = 1350 + 270 m N1 m = 3 (ii) 100 P07 = RM 30 × K1 120 N1 = RM 25 (c) 120 + (120 × 0.15) = 138 K1 K1 (125 × 4 ) + (138 × 3) + (80 × 5 ) + (150 × 6 ) I 10 / 07 = 18 N1 = 123 12
- 12. 10 N0. SOLUTION MARKS 14 (a) y ≥ 200 N1 x+y ≤ 800 N1 4x + y ≤ 1400 N1 (b) y 1000 4x + y = 1400 900 800 700 600 (200,600) 500 400 R 300 y = 200 200 100 x + y = 800 x 100 200 300 400 500 600 700 800 900 • At least one straight line is drawn correctly from inequalities K1 involving x and y. N1 • All the three straight lines are drawn correctly • Region is correctly shaded N1 (c) (i) 650 N1 (ii) Maximum point (200, 600) N1 Maximum profit = 20(200) + 6(600) K1 = RM 7600 N1 13
- 13. 10 N0. SOLUTION MARKS 15 (a) TQ2 = 92 + 62 – 2(9)(6)cos56o K1 TQ = 7.524 cm N1 (b) sin ∠QTR sin 56 0 K1 = 6 7.524 ∠QTR = 41o 23’ N1 (c) 1 K1 42.28 = ( RS )(6 )sin 56 o 2 RS = 17 ST = 17 − 9 (or ST + 9 in formula of area) K1 = 8 cm N1 (d) 1 Area ∆ QTR = (9)(6) sin 56 0 2 K1 2 = 22.38 cm Area of quadrilateral PQTS = 2(42.28) – 22.38 K1 = 62.18 cm2 N1 10 END OF MARKING SCHEME 14