ADDITIONAL MATHEMATICS PAPER 2 SECTION A Q1 Q2 Q3 Q4 Q5 Q6 chapter Chap 4 /f4 Chap 5 /f5 Chap 1/f5 Chap 3 /f5 Chap 6 /f4 Chap 7/f4 topic Simultaneous equations Trigonometric functions Progressions Integration Coordinate geometry Statistics marks 5 7 6 7 7 8

SECTION B Q7 Q8 Q9 Q10 Q11

Chapter Chap 2 /f5 Chap 9 /f4 Chap 4 /f5 Chap 8 /f5 Chap 8 /f4

Topic Linear law Differentiation vectors Probability distributions Circular measures

Marks 10 10 10 10 10

SECTION C Q12 Q13 Q14 Q15

Chapter Chap 9 /f5 Chap 10 /f4 Chap 10 /f5 Chap 11 /f4

Topic Motion along a straight line Solution of triangles Linear programming Index numbers

Marks 10 10 10 10

1

X 2y = 7 X = 7 + 2y [P1] (7 + 2y)y (7 +2y)= 9y [K1] 2y2 4y 7 = 0 Y= [K1] = 3.1213 or -1.1213 [N1] X = 13.2426 or 4.7574 [N1]

2

(a)

Shape of cosine graph [P1] Amplitude = 3 [P1] 1 cycle for 0x2 [P1] Graph shifted up by 1 unit [P1] (b) 6 cos x =4 - 3x 3 cos x = 1+3cos x = 2 +1 Y = 3 - x [N1] Drawing a straight line [k1] Number of solution = 2 [N1]

3

(a) V1 = (3)2(4) =36 [P1] d = 18 T17 = 36 +(17-1)(18 ) [K1] = 324 (b) Sn = { 2(36 n2+ 3n -180 = 0 (n-12)(n + 15) = 0 [K1] n = 12 [N1]

[K1]

4

a (i) (ii) = =[ = = 24 [N1] [k1] [K1] [P1]

(b) g(x) = [K1] = x2 8x + c [must have c] Substitute x = 1 and g(x) = 0 0 = 12 8(1) + c [k1] C=7 g(x) = x2 8x + 7 [N1]

5

(a) Equation of AC: 3y = 2x 15 y= mAC = m2 = y (- 7 ) = y= x(x ( - 3 )) [N1] [K1] [P1] x m2 = - 1

(b) (i) Equation of AC : y = When x = 0, y = - 5 B = (0, - 5) [N1] (ii) 2 A(- 3, - 7) [K1] X=

7 C(x,y)

B(0, - 5)

Y=2 C = ( ,2)

[N1]

6

(a) Height of the bars proportional to the frequencies. [N1] Label the class boundaries/ The midpoints of the classes/ The class interval correctly [N1] Method of finding mode [K1] Mode = 25.5 26.5 [N1] (b) Mean = = = = = 42172.5 Standard deviation = = = = 11.43 [N1] [k1] [K1] [K1]

7

(a) x 3 4 Log10y 0.41 0.52 [N1] Plot log10y against x [K1] 6 points plotted correctly [N1] Line of best fit [N1]

5 0.61

6 0.69

7 0.80

8 0.90

(b) Log10y = (log10h)x log10k [P1] (i) m = log10 h [K1] h = 1.25 [N1] (ii) c = - log10 k [K1] k = 0.74 0.78 [N1] (iii) y = 2.40 [N1]

8

(a) Y = x3 6x2+9x + 1 = 3x2 12x +9 [N1] At A(2,3), x = 2 = 3(2)2 12(2) + 9 [K1] = - 3 [N1] (b) m1x(-3) = - 1 [K1] m1 = y (3) = (x 2) [k1] y= x+ (c) At turning point, =0 3x2 12x +9 = 0 X2 4x + 3 = 0 (x 3)(x -1) = 0 X = 1 or 3 At P (3,1), x=3. At Q, x=1, Y = 13 6(1)2 + 9(1)+1 = 5 [K1] Q=(1,5) [N1] = 6x 12 At Q(1,5), = 6(1) 12 [K1] = - 6