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SENIOR CERTIFICATE EXAMINATION

FUNCTIONAL MATHEMATICS P1 STANDARD GRADE 2011 MEMORANDUM

MARKS: 150

This memorandum consists of 8 pages.

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Functional Mathematics/SG/P1

2 SCE – Memorandum

DBE/2011

VRAAG 1/QUESTION 1

1.1.1 5

3x + 2

25

2x – 1

.5

x-3

.5

1.1.4 4  - 81 + 1250 1 1 2(-) 4() =2 –3 +1 = 2– 1 – 3 3 + 1 1 =  – 27 + 1 1 = – 25 1 1.2.1 4.3x = 36 (5) ∴4.3x = 36 4 4 ∴3x = 9 1 ∴3x = 32 1 ∴x = 2 1 1.2.2 16-x = 2 1 4(- x) =2 ∴2 (3) ∴ 2-4x = 2 ∴-4x = 1 1 ∴-4x = 1 -4 -4 ∴ x = - 1 (5) (3) [23] (3) (4)

-

= 53x + 2. 5x - 3. 5 52(2x – 1) 1 = 53x + 2. 5x - 3. 5 54x – 2 1 =5 3x + 2 + x – 3 + 1 – 4x + 2

= 52 1 = 25 1 1.1.2 3x + 3x + 2 3x + 1 = 3x + 3x .32 3x .3 1 = 3x(1 + 32) 3x.3 1 = 10 of/or 3 1 3

__ ___ ___ 1.1.3 4√ 8 + 3√ 18 – √ 32 ____ ____ _____ = 4√4 x 2 + 3√9 x 2 – √16 x 2 1 1_ 1_ 1_ = 8√2 + 9√2 – 4√2 _ = 13√2

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3 SCE – Memorandum

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VRAAG 2/QUESTION 2 2.1.1 2log60 – 2log2 – 2log3 = log 602 – log22 – log32 1 = log 3 600 1 4x9 1

2.2.2

logx 64 = 6 ∴ ∴ ∴ x6 = 64 1

x6 = 2 6 1 x = 2 1 (3)

2.3 = log 3 600 36 = log 100 1 = 2 1 2.1.2 log327 – log216 – log41 1 3 = log33 – log224 – 0 1 = 3log33 – 4log22 1 = 3–4 = -1 1 2.1.1 log 4x = 3 ∴10 = 4x 1 ∴4x = 1 000 1 ∴ x = 250 1 of/or log 4x = 3log1010 1 ∴log 4x = log10103 ∴log104x = log101000 ∴ ∴ 4x = 1000 x = 250 1 1 (3) 3

3x – 1 = 28 ∴log 3x – 1 = log 28 1 ∴(x – 1)log 3 = log 28

(5)

∴ x – 1 = log 28 1

log 3

∴ x – 1 = 3,033103…..1 ∴ x = 4,01 of/or (4) 3x – 1 = 28 ∴ x – 1 = log 28 11

log 3

∴ x – 1 = 3,033103…..1 ∴ 2.4 log 45 = log (3 x 3 x 5) 1 = log 3 + log 3 + log 5 1 = a+a+b = 2a + b 1 (3) [22] Please turn over

x = 4,01

(4)

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4 SCE – Memorandum

DBE/2011

VRAAG 3/QUESTION 3 x -2 x 2  x 3  2 x /9 2.3 3.1

-1   

0 1 1 2

1 2 3 6

2 4 9 18
27 26 25

3 8 27 54

y

f(x) = 2

x
24 23 22 21 20 19

- y-afsnit/yintercept - vorm/shape g(x) = 3x - y-afsnit/yintercept - vorm/shape h(x) = 2.3x - y-afsnit/yintercept - vorm/shape (12) 3.2 k(x) = log3x - y-afsnit/yintercept - vorm/shape (2) 3.3.1 x = 2,8 (2) 3.3.2 x = 0,4 11 (2) 3.3.3 x = 1 11 (2)

h(x)

g(x)

18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 -5 -4 -3 -2 -1 -1 -2 -3

f(x)

A

x
4 5

B

1

C

2

3

[20]
k(x) Copyright reserved Please turn over

Functional Mathematics/SG/P1

5 SCE – Memorandum

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VRAAG 4/QUESTION 4 4.1.1 T2 – T1 = T3 – T2 ∴x – 3 – (8 – 2x) = 2x – 4 – (x – 3) ∴ ∴ ∴ 3x – 11 = x – 1 2x = 10 x = 5 (4) ∴ ∴ – : 4.2.1 a + 4d = 90 a + 10d = 36 – 6d = 54 ∴ d=–9 4.2.2 a + 4(-9) = 90 a = 90 + 36 a = 126 T2 = 117 (3) T3 =108 (3) [25] (4) ………. ………

4.1.2 (a) T1 = 8 – 2(5) = – 2 T2 = 5 – 3 = 2 T3 = 2(5) – 4 = 6 (b) d = 6 – 2 =4 (c) T20 = a + 19d = – 2 + 19(4) = – 2 + 76 = 74 (d) Sn = ∴S12 = ∴ ∴ ∴ n

(2)

(3)

/2[2a + (n – 1)d] /2[2(-2) + (12 – 1)4]

12

= 6[-4 + 11(4)] = 6[-4 + 44] = 6[40] (6)

∴S20 = 240

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VRAAG 5/QUESTION 5 5.1 Sn = a(r n – 1) r–1 ∴ S8 = 7[(-3)8 – 1] –3–1 = 7[6561 – 1] –4 = – 11 480 5.2.1 ar5 = 256 ar = 16 …….. …….. (5) 5.4 Tn = 4(3)n – 1 T1 = 4(3)1 – 1 = 4(3)0 = 4 T2 = 4(3)2 – 1 = 4(3)1 = 12 T3 = 4(3)3 – 1 = 4(3)2 = 36 (3) [20]

__ ÷ : r4 = 16 4 ∴ r = ± √16 ∴r=2 5.2.2 a(2) = 16 ∴ 2a = 16 ∴ a =8 5.3 Tn = ar n – 1 ∴192 = 3(2)n – 1 ∴192 = 3(2)n – 1 3 3 ∴ 64 = 2n – 1 ∴ 2 6 = 2n – 1 ∴ ∴ 6=n–1 n=7 (6) (2) r>0 (4)

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VRAAG 6/QUESTION 6 6.1 lim ( –x2 + 2)
x→2

6.4.2 f '(x) = 4x + 3 6.4.3 f '(-2) = 4(–2) + 3 = –5 (2) 6.4.4 (–2 ; 8) m = –5 y – y1 = m(x – x1) ∴y – 8 = –5(x + 2) ∴ ∴ y = –5x – 10 + 8 y = – 5x – 2

(2)

= – (2)2 + 2 = –4 + 2 =...
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