Example of Solutions for Paper 1

1	Given f(x) = x + 3 and g(x) = 6x2 + 3. Diberi f(x) = x + 3 dan g(x) = 6x2 + 3. gf(x) = g(f(x))      = g(x + 3)      = 6(x + 3)2 + 3      = 6(x2 + 6x + 9) + 3      = 6x2 + 36x + 57 gf(−3) = 6(−3)2 + 36(−3) + 57      = 3

2	α + β = −4 αβ = −4 New equation: Sum of roots, (1 − 9α) + (1 − 9β) = −9(α + β) + 1      = −9(−4) + 1      = 37 Product of roots, (1 − 9α)(1 − 9β) = 81αβ − 9α − 9β + 1      = 81(−4) − 9(−4) + 1      = −287 Equation: x2 − 37x − 287 = 0

3	−6x2 − nx − 6 = x −6x2 + (−n − 1)x − 6 = 0 The equation has two equal roots b2 − 4ac = 0 (−n − 1)2 − 4(−6)(−6) = 0 n2 + 2n + 1 − 144 = 0 n2 + 2n − 143 = 0 (n − 11)(n + 13) = 0 n = 11 or n = −13

4	f(x) = y = 4x2 + 8x + k + 2 The equation does not have real roots b2 − 4ac < 0 (8)2 − 4(4)(k + 2) < 0 64 − 32 − 16k < 0 −16k < −32 k > 2

5	y = x2 − 2hx − 2h    = x2 − 2hx + ()2 − ()2 − 2h    = (x − h)2 − h2 − 2h Minimum value = − h2 − 2h −h2 − 2h = −3 h2 + 2h − 3 = 0 (h − 1)(h + 3) = 0 h = 1 or h = −3

6	−3 logx a − 5 logx b + logx ab = −logx a3 − logx b5 + logx ab = logx = logx

7	logx 128 = − 128 = x x = 128    = 2    = 4

9	PQ = 5 = 5 = 5 Squaring both sides, (6a − 4)2 + 42 = 52 (6x − 4)2 = 25 − 16 (6x − 4)2 = 9 6x − 4 = ±3 a = or a =

10	Gradient of EF = = 1 (1)m2 = −1 m2 = −1 Midpoint of EF = (, ) = (−6, 6) Equation: y − 6 = −(x + 6) −y + 6 = x + 6 x + y = 0

11	Rearrange the data, 41   59   62   78   85   87   89 First quartile = 59 Third quartile = 87 Interquartile range = 87 − 59 = 28

12	(3x2 + 2x + 4) = f(x) = [3x2 + 2x + 4]    = 3(2)2 + 2(2) + 4 − [3(2)2 + 2(2) + 4]    = 20 − (20)    = 0

13	30° 54' = 30.9 × = 0.5394 rad.

14	0.6π rad. = 0.6π × = 108°

15

Step 1: Langkah 1: Let δx be a small increment in x and δy be the corresponding small increment in y. Biar δx menjadi tokokan kecil pada x dan δy menjadi tokokan yang sepadan pada y. Step 2: Langkah 2: y = x2 + 7 ----- (1) y + δy = (x + δx)2 + 7 y + δy = x2 + 2x(δx) + (δx)2 + 7 ----- (2) Step 3: Langkah 3: (2) − (1), δy = 2x(δx) + (δx)2 Step 4: Langkah 4: = 2x + (δx) = (2x + (δx))       = 2x Therefore, Maka, = 2x

16	y = −x4 + 2x3 = −4x3 + 6x2 At point (2, 0), = −4(2)3 + 6(2)2    = −8 Therefore, the gradient of the tangent at point (2, 0) = −8

17	(a) sin ∠SOU = = 0.5 ∠SOU = 30°    = 30 ×    = 0.5237 rad. ∠SOT = 2 × 0.5237    = 1.0474 rad. (b) Area of sector SOT = r2θ    = (16)2(1.0474)    = 134.07 cm2 OU =    = 13.86 cm Area of triangle OST = × 13.86 × 16 = 110.88 cm2 Area of the shaded region = 134.07 − 110.88 = 23.19 cm2

18	(a) BC→ = BA→ + AC→    = −2j~ + 7i~    = 7i~ − 2j~ (b) AD→ = AB→ + BD→    = AB→ + BC→    = 2j~ + (7i~ − 2j~)    = i~ + j~

19	Price index, I = × 100 178.3 = × 100 P1985 =    = RM2300

20	Food Makanan Price index, I Indeks harga, I Weightage, W Pemberat, W WI A × 100 = 91 8 728 B × 100 = 82 2 164 C × 100 = 195 9 1755 D × 100 = 183 8 1464 ∑W = 27 ∑WI = 4111 Composite index, =    =    = 152.3

21	(a) Sn = ( n + 1) S9 = [(9) + 1]    = (10)    = 45 (b) T9 = 45 − [(8) + 1]    = 45 − 36    = 9

22	(a) T2 = 30 ar2 = 30 −−−− (1) T2 + T3 = 35 30 + ar2 = 35 ar2 = 5 −−−− (2) (2) ÷ (1), = r = Substitute r = into (1), a()1 = 30 a = 180 (b) Sum to infinity = = = 180 × = 216

23	= + 8 = 8() − 8 X = Y =

24	(a) ∫ f(x) dx = −∫ f(x) dx    = −3 (b) ∫ [3 + 5f(x)] dx = ∫ 3 dx + 5∫ f(x) dx = [3x] + 5(3) = (27 − 6) + 15 = 36

25	OH = 3HF OH→ = OF→    = (OE→ + OG→)    = (8x~  + 10y~)    = 6x~  + y~