MATHEMATICS P1 SEPTEMBER 2019 PREPARATORY EXAMINATION MARKING GUIDELINE ...

MATHEMATICS P1

SEPTEMBER 2019 PREPARATORY EXAMINATION

MARKING GUIDELINE

NATIONAL SENIOR CERTIFICATE

MARKS: 150 TIME: 3 hours

GRADE 12

This marking guideline consists of 13 pages.

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Mathematics P1

QUESTION 1 1.1.1 = 0 = 4

2 NSC Marking Guideline

September 2019 Preparatory Examination

A 0 A 4

(2)

1.1.2

2x2 + 5x -1 = 0 - b b2 - 4ac

x= 2a

- (5) (5)2 - 4(2)(-1) =

2(2) = 0,19 or - 2,69

Astandard form

CAsubstitution in correct formula

CACAanswers

(4)

(penalize 1 mark if rounding off is incorrect-once here for entire paper)

1.2.1 - 2 = 2 -

( - 2)2 = (2 - )2 - 2 = 4 - 4 + 2 2 - 5 + 6 = 0 ( - 2)( - 3) = 0 = 2 = 3

n/a

OR

A squaring both sides

CAstandard form

CAfactors

CAanswers and rejecting

(4)

OR

1.2.2

- 2 = 2 - - 2 0 2 - 0 2 2 = 2 2 - = 2 - = 2 2 - - 2 = 0 ( + 1)( - 2) = 0

= -1 = 2

1.3 -22 + 5 0

5 2 ( - 2) 0

5 0 or 2 OR

+

A Aeach inequality

Aboth inequalities

(4)

CAanswer

Answer only FULL MARKS

letting 2 - =

CA2 - = 2

CAstandard form

CAfactors

CAanswers

MAX 3marks if four solutions

(4)

arrived at

AA factors

CA 0

CA

5 2

(4)

OR

+

+

0

2,5

OR

0

-

If graphical Solution is used:

2,5

AA 2 marks for graph

-

CACA 2 marks for answer

-

(4)

Mathematics P1

1.4

2+1 + 2 = 3+2 - 3

2(2 + 1) = 3(9 - 1)

2(3) = 3(8)

2-3 = 3-1

- 3 = 0 - 1 = 0

= 3 = 1

+ = 4

QUESTION 2

2.1 48 ; 63

2.2

8

15

24

3 NSC Marking Guideline

September 2019 Preparatory Examination

Afactorising

CAsimplifying

CAexponential form

CAeach exponent and equal to 0

CAx and y value

CAanswer

(6)

If - 3 = 0 - 1 = 0 is

missing then maximum 5/6 marks

[24]

Aanswers

(1)

35

1D

7

9

11

2D

2

2

Aa value

2 = 2 3 + = 7

= 1 = 4

CAb value

+ + = 8 = 3

CAc value

= 2 + 4 + 3

CAanswer

(4)

OR

OR

2 = 2

= 1

Aa value

1 + 2 - 1 = 8 + 2 - 7 =

3=c = 2 + + 3 8 = 1 + + 3

CAc value

= 4 = 2 + 4 + 3

CAb value

(4)

CAanswer

Mathematics P1 OR

4 NSC Marking Guideline

OR

September 2019 Preparatory Examination

Tn

=

T1

+

(n

- 1)d1

+

(n -1(n 2

- 2) d2

= 8 + (n -1)(7) + (n -1(n - 2) (2) 2

= 8 + 7n - 7 + n2 - 3n + 2

= n2 + 4n + 3

Aformula Asubstitution into correct formula

CAsimplifying

CAanswer

(4)

OR

Tn

=

n -12a

2

+

(n

-

2)d + T1

= n -12(7) + (n - 2)(2)+ 8

2

= n -114 + 2n - 4+ 8

2

= n -12n +10+ 8

2

= (n -1)(n + 5) + 8

= n2 + 4n - 5 + 8

= n2 + 4n + 3

OR Aformula Asubstitution into correct formula CAsimplifying

(4)

CAanswer [5]

QUESTION 3

3.1 T2 - T1 = T3 - T2 + 5 - 2 + 3 = 2 + 7 - - 5

- + 8 = + 2 = 3

3.2 Pattern is 3 ; 8 ; 13 ; . ..

3.3.1

Sn

=

n 2a

2

+

(n

-1)d

S120

=

1202(3) +119(5)

2

= 36060

x = k + 1 and y = k + 2

A equating differences

CAsimplifying CAanswer

(3)

CAa = 3 and d = 5

CAsubstitution into formula

CAanswer

(3)

Ax ? value Ay ? value

(2)

Mathematics P1

3.3.2

= + ( - 1) = 3 + 5 = + ( + 1) = 3 + ( + 1)(5) = 8 + 5 + = 11 + 10

5

September 2019 Preparatory Examination

NSC Marking Guideline

CAsubstitution into nth term

CA3 + 5

CA8+5

CAanswer

(4)

[12]

QUESTION 4

4.1.1 4.1.2

4.2

15 ; 5

S

=

a 1- r

=

15 1- 1

3

= 45 = 22,5 2

3 0 ; 3 0 ;

3 2

1 3 3 2; 2 ;2

1 = 2 ; = 3 -1 = 40,53

1 2

(3)-1

=

81 2

3

-1

32

=

34.

1

32

- 1 1 9 2 = 42 = 2

- 1 = 9

= 10

OR

AAboth terms

(2)

CAsubstitution of common ratio

CAanswer (2)

Alisting terms

CA12

(3)-1

=

81 2

3

-1

CA3 2

=

34

.

1

32

CA

-1 2

=

4

1 2

=

9 2

CAanswer

(5)

OR

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