Multiplying Matrices

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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~

2012 ADD MATH PROJECT WORK PART 1

PART 1

a)      

b)      Polygon is defined as a closed, 2-dimensional shape made up of three or more straight line segments connected end to end to end. Polygons have been known since ancient times. The regular polygons were known to the ancient Greeks, and the pentagram, a non-convex regular polygon (star polygon), appears on the vase of Aristophonus, Caere, dated to the 7th century B.C. Non-convex polygons in general were not systematically studied until the 14th century by Thomas Bredwardine. In 1952, Shephard generalized the idea of polygons to the complex plane, where each real dimension is accompanied by an imaginary one, to create complex polygons.

c)      The four different methods of finding the area of a triangle is stated below:

Method 1:

                     

                    

If you know base (b) and height (h) of the triangle, the following formula can be applied.

Area = ½ x b x h

Method 2:

          a

c

           b

If you know three sides (a, b and c) of the triangle, Heron’s Method can be applied.

s = (a+b+c) / 2

Area =Ö s (s-a) (s-b) (s-c)

Method 3:

 

             

If you know two sides (a and b) and the included angle (q), the following formula can be applied.

Area = ½ x a x b x sinq

Method 4:

 

If you know coordinate of the three vertices, the following formula can be applied.

       

Method 5:

If you know equation (y=mx+c) of the three lines, integration can be applied.

Area =      (mx₁ + c) dx +    (mx₂ + c) dx -      (mx₃ + c) dx

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PART 2

a)      Cost needed to fence the herb garden      = RM20.00 per metre x 300m

= RM6000.00

b)      (Refer to Microsoft Excel attached.)

c)      In order for the enclosed area to be maximum, the shape of the triangle must be equilateral. Therefore, the length of both p and q must be 100m.

d)     (i) 50

(ii) p α 1/q

                                                                                          

        

(iii) A² = B² + C² – 2 x B x C x cos (a). This theorem is known as ‘Cosine Rule’.

PART 3

a)      Suggestion 1: Square

Suggestion 2: Pentagon

Suggestion 3: Hexagon

Suggestion 4: Heptagon

Suggestion 5: Octagon

 

b)      i) Aloe vera - The local name is known as 'Lidah Buaya'. Due to its exceptional healing properties, Aloe Vera is also known as "the little of the desert", "the plant of immortality", "the first aid plant" and "the miracle plant". It is also one of the most important crude drug history and is still extensively used in modern medicine. It contains the same painkilling and anti-inflammatory compound as in Aspirin. Aloe vera is also a nutrient-rich-plant, containing more than 200 active-components, vitamins, minerals, essential amino acids, enzymes and other plant chemicals which is said to gently strengthen, sustain and encourage cellular activity in body.

 

Tongkat Ali - It has become popular for its testosterone-enhancing properties. Because of that, it is included in certain herbal supplements for bodybuilders. Historically, it has been used by the folk medicine in its countries of origin as a libido enhancer and to treat various sexual dysfunctions. Numerous scientific studies performed by Malaysian Universities, including University Science Malaysia (USM) have confirmed its effects on enhancing sexual characteristics in animal models. Currently, it is being researched for its possibilities as an anti-cancer supplement.

 

Agrimony - It is not commonly used today, but has its place in traditional herbal medicine. This herb is safe for use for minor ailments in most healthy people. Like most herb simples, the uses to which it is put are remarkably varied. The English use it to make a delicious "spring" or "diet" drink for purifying the blood. It is considered especially useful as a tonic for aiding recovery from winter colds, fevers, and diarrhea. Agrimony contains tannin and a volatile essential oil.

ii) (Logo to be created by your own creativity.)

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GOOD LUCK TO ALL SPM CANDIDATES YEAR 2012