Maths

SENIOR CERTIFICATE EXAMINATION

FUNCTIONAL MATHEMATICS P1 STANDARD GRADE 2011 MEMORANDUM

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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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Functional Mathematics/SG/P1

3 SCE – Memorandum

DBE/2011

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]
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x = 4,01

(4)

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Functional Mathematics/SG/P1

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