GRADE 12 MATHEMATICS PAPER 1 MEMORANDUM - NSC PAST PAPERS AND MEMOS SEPTEMBER 2017

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GRADE 12 MATHEMATICS
PAPER 1 
NSC PAST PAPERS AND MEMOS
SEPTEMBER 2017

NOTE:

  • If a candidate answered a question TWICE, mark the FIRST attempt ONLY. 
  • Consistent accuracy applies in ALL aspects of the memorandum.
  • If a candidate crossed out an attempt of a question and did not redo the question, mark the  crossed-out attempt.
  • The mark for substitution is awarded for substitution into the correct formula. 

MEMORANDUM 

QUESTION 1 

?​

1.1  2?(? + 1) − 7(? + 1) = 0
(? + 1)(2? − 7) = 0
? = −1     or     ? = 72		 ? factors
? ?-value
? ?-value 		  (3) 
       
 1.2  ?2 − 5? − 1 = 0 
? = −? ± √?2 − 4??
             2?                                   
? = −(−5) ± √(−5)2 − 4(1)(−1)
                      2(1)
? = 5,19 or ? = −0,19		  ? substitution into correct formula
?? x- values		  (3)
       
1.3  4?2 + 1 ≥ 5?
4?2 − 5? + 1 ≥ 0
(4? − 1)(? − 1) ≥ 0
1.3 ans
x ≤ 1/4        or       ? ≤ 1    		 ? standard form
? factors
 ?? ≤ ¼
?? ≥ 1		  (4)
       
1.4 54?+3. 100−2?+1 = 50 000
54?+3. (52. 22)−2?+1  = 50 000
54?+3. 5−4?+2. 2−4?+2 = 50 000
55. 2−4?. 22 = 50 000
2−4? = 22
−4? = 2
? = − 1/2		 ?5−4?+2
?2−4?+2
?−4? = 2
? answer		  (4)
       
1.5 ? = 2? …………………..(1)
?2 + 2? − ? − ?2 = 36……….(2)
? = 2?
sub into (2)
(2?)2 + 2(2?) − ? − ?2 = 36
4?2 + 4? − ? − ?2 = 36
3?2 + 3? − 36 = 0
?2 + ? − 12 = 0
(? − 3)(? + 4) = 0
? = 3   or ? = −4
? = 6 or ? = −8		 ?substitution
?standard form
? factors
? y − values
? x − values(5)		 (5)
       
1.6 ?2 − ?? + ? − 1 = 0
∆ = ?2 − 4??
∆ = (−?)2 − 4(1)(? − 1)
∆ = ?2 − 4? + 4
∆ = (? − 2)2
∆ ≥ 0 roots are real and rational(perfect square)		 ?substitution
? simplification
? (? − 2)2
? conclusion(4) 		 (4)
      [23]
QUESTION 2
2.1.1 ?? = 4? − 1
483 = 4? − 1
484 = 4?
? = 121
121 terms in series		 ?  ?? = 4? − 1
? equating 483
?  answer(3)		  (3)
       
2.1.2 121
   ∑    (4? − 1)
?=1		 ? answer (2)
       
2.2.1  (? − 3) − (2? − 4) = (8 − 2?) − (? − 3)
−? + 1 = −3? + 11
2? = 10
? = 5		 ? setting up equation
? simplification
? answer		 (3)
       
2.2.2 … ; … ; … 6 ; 2 ; −2 ; … ; … ; …
?10 = 6        or              ?? = −4? + 46
? + 9? = 6                             ?1 = −4(1) + 46
? + 9(−4) = 6                        ?1 = 42?
= 42		 ? numerical values of?10; ?11; ?12
? difference  −4
?  a-value		 (3)
       
2.3

??2 + ??3 = −4
? + ?? = −1 
??2(1 + ?) = −4
?(1 + ?) = −1
∴ ?2 = 4
∴ ? = ±2

? + ?(2) = −1
∴ ? + 2? = −1
3? = −1
? = − 1/3 
First three terms: − 1/3  ; − 2/3  ; − 4/3           

 ? setting of equations
? common factor
? ? = 2
? value of a
? first three terms		 (6)
      [17
 QUESTION 3   
3.1

41 ; 43 ; 47 ; 53 ; 61 ; 71 ; 83 ; 97 ; 113 ; 131 
2        4       6        8        10     12      14      16       18
     2        2       2        2         2       2        2         2 
2? = 2            ? + ? = 2                ? + ? + ? = 41
? = 1                        ? = −1                            ? = 41
∴ ?? = ?2 − ? + 41

 ?  2nd difference
?  ? = 1
?  ? = −1
?  ? = 41
?  ?? = ?2 − ? + 41		 (5)
       
3.2

?41 = 412 − 41 + 41
?41 = 1681

Factors of 1681: 1 ; 41 and 1681
1681 is not a prime number

 ? ?41 = 1681
?  factors
? conclusion		 (3)
       
3.3 Consider the unit digits only
1 ; 3 ; 7 ; 3 ; 1 ; 1 ; 3 ; 7 ; 3 ; 1 ; 
groups of 5
49999998 = 9999999,6
      5
0,6  ×   5 = 3
?49999998 will end in 7		  ? unit digits
? groups of 5
? conclusion		 (3)
      [11]
QUESTION 4
4.1.1 ? = ?(1 + ?)?
? = 500 000 (1 +     7,2     )12?
                              1200
? = 500 000(1.006)12?		 ?sub into formula
? 12?		  (2)
       
4.1.2 ? = 500 000(1.006)12?
? = 500 000(1.006)12×5
? = ? 715894.21		 ? ? = 60
?  answer		 (2)
       
4.1.3

? = ?(1 + ?)?
1000000 = 500000(1.006)12?
12? =      log 2          
          log 1.006
12? = 115.8707581
? = 9,66 ?????

Will exceed R1 000 000 in 10 years.

 ? setting up equation
?  using logs
?  conclusion		 (3)
       
4.2.1 ?V =  10 000 [1 − (1 + 15/1200 ) −36]
                          15/1200                               
?? = ?288 472,67
???????/? = ?350 000 − ?288 472,67
???????/? = ?61 527,33		 ? ? and ?
?sub into ?? formula
? ?? formuleü ?? = ?288 472,67
? subtracting
? answer		 (5)
       
4.2.2 350 000    =  ? [1 − (1 + 18,5/1200 ) −60]
                              18,5/1200
? = ? 8 983,17		 ? ? = 18,51200
?? = −60
? substitution
? answer		 (4)
      [16]
QUESTION 5
5.1 ?(−3; 0) ? answer  (1)
       
5.2 ?(?) = ?2 + 3?
? = − ?
        2?
? = − 3
         2
? (− 3) = (− 3)2  + 3 (− 3)
       2          2               2
= − 9/4
? (− 3 ; − 9)
       2      4		 ?? = − 32   
?substitution 
?answer		 (3)
       
5.3 ?(−5) = 10     and  ?(−3) = 0
? =     10 − 0            
         −5 − (−3)?
= −5		 ? calculating ?(−5) and ?(−3)
?  substitution
? ?-value		 (3)
       
5.4 ? < −3 or / of ? > 0 ?answer  (2)
       
5.5

? (− 3 ; − 9)
        2     4
(− 3 − 2 ; − 9)
    2            4
(− 1 ; − 9)
    2      4             

 or


?(? − 2) = (? − 2)(? − 2 + 3)
?(? − 2) = ?2 − ? − 2
? = − (−1)   
          2(1)
? = − ½

 ? answer
? ?(? − 2) = ?2 − ? − 2
? ? = − 12		  (2)
       
5.6

?? = − 1/2 ? + 2 − (?2 + 3?)
?? = − 1/2 ? + 2 − ?2 − 3?
?? = −?2 − 7/2 ? + 2
?? = − (?2 + 7/2 ? − 2)
L? = − [(? +  7/4 )2   −  81/16  ]
L? = − (? + 7/4 )2  + 81/16

OR

?? = − 1/2 ? + 2 − (?2 + 3?)
?? = − 1/2 ? + 2 − ?2 − 3?
?? = −?2 − 7/2 ? + 2
??? = −2? − 7
??                 2
−2? − 7/2 = 0
? = − 7/4

? = − (− 7/4 ) 2  − 7/2  (−  7/4 ) + 2
? = 81/16
∴ ?? = − (? + 7/4)2   + 81/16

OR

? = − ?
2?
? = − [ − 7/2]
          2(−1)
? = − 7/4
? = − (− 7/4 ) 2  − 7/2  (−  7/4 ) + 2
? = 81/16
∴ ?? = − (? + 7/4)2   + 81/16

? ?(?) − ?(?) 
? standard form
?  completing the square
? answer

 

 

? ?(?) − ?(?)
? standard form
?  ? = − 7/4   
? ? = 81/16           

 

??(?) − ?(?)
? standard form/standaardvorm   
?  ? = − 7/4   
? ? = 81/16         

(4)
      [15]
QUESTION 6
6.1 ans 6.1 ? shape 
? y − intercept
? point on graph		 (3)
       
6.2 ?(?) = 2? ? answer (1)
       
6.3 ℎ−1? = 2−?
−? = log ?
        log 2
? = − log ? / ? = − log2 ? /? = log½ ?
         log 2         		 ? interchange ? and ?
? equation		 (2)
       
6.4 ? ≥ 0 ; ? ∈ ?   (1)
       
6.5 See 7.2.1 ??shape and x-intercept          (2)
       
 6.6 log1 ? = −321 −3(  )     = ? 2? = 8∴    0 < ? ≤ 8 ?? = 8 
?0 < ? ≤ 8		 (2)
      [11]
QUESTION 7
7.1 ? = 5? = 2 ?? = 5?? = 2 (2)
       
7.2 ? = 5 − ?
      ? − 2
? = −(? − 2) + 3
          (? − 2)
? =     3     − 1
       ? − 2		 ?? = 5−?
         ?−2
 ?? = −(?−2)+3
              (?−2)		 (2)
       
7.3 ?(5; 0)
? = ? − 3
? = ? + 3
?′(0 + 3; 5 − 3)
?′(3; 2)		 ?? = 3
?? = 2		 (2)
      [6]
 QUESTION 8  
 8.1

?(?) = −2?2 + ?
?′(?) = lim ?(? + ℎ) − ?(?)
          ℎ→0       ℎ
= lim     −2(? + ℎ)2 + ? − (−2?2 + ?)
  ℎ→0                   ℎ 
= lim   −2(?2 + 2?ℎ + ℎ2)  + ? + 2?2 − ?)
  ℎ→0                             ℎ
= lim −2?2 − 4?ℎ − 2ℎ2 + ? + 2?2 − ?
  ℎ→0                     ℎ
= lim −4?ℎ − 2ℎ2
 ℎ→0        ℎ
= lim ℎ(−4? − 2ℎ)
  ℎ→0         ℎ
= lim(−4? − 2ℎ)
   ℎ→0
= −4?  

  • Answer ONLY: 0 marks
  • Penalise 1 mark for incorrect use of formula. Must show ?′(?)
 ? formula
? substitution of (? + ℎ) 
? simplification to
(−4?ℎ − 2ℎ2) 
? common factor
? answer		  (5)
       
8.2 HDGF   (4)
      [9]
QUESTION 9
9.1 ?(?) = (? − 1)2(? + 3)
?(?) = ?3 + ?2 − 5? + 3 
?′(?) = 3?2 + 2? − 5
3?2 + 2? − 5 = 0
(3? + 5)(? − 1) = 0
? = − 5/3       or ? = 1
?(1) = 0
? (− 5/3 ) = 256
                   27		 ?  ?(?) = ?3 + ?2 − 5? + 3
?  ?′(?) = 0
? factors
?  ?-values
?  ?-values		 (5)
       
9.2 2567 ? shape 
? ? − intercepts
? ? − intercept
?  stationary points		 (4)
       
9.3 ?′′(?) = 6? + 2
6? + 2 = 0
? = − 1/3
? = 128 / 27 / 4,74 / 4 20/27 		 ? f′′(?) = 6? + 2 
?? = − 1/3
?  ? = 128 / 27 / 4,74 / 4 20/27 		 (3)
       
9.4 0 < ? < 25627 ??answer (2)
       
9.5 ?′(?) = 3?2 + 2? − 5
3?2 + 2? − 5 = −5
3?2 + 2? = 0
?(3? + 2) = 0
? = 0 or ? = − 2/3
? (− 2) = 175
       3      27         
y = −5? + ?
175  = −5 (− 2/3 ) + ?
 27                 3?
= 85
   27
? = −5? + 85
                27		 ? f′(?) = −5
? factors
? ? = − 2/3
? f (− 2/3) = 175
                     27
? substitution
? answer		 (6)
      [20]
  QUESTION 10    
 10.1 243 = 2(? × 2?) + 2(2? × ℎ) + 2(? × ℎ)
243 = 4?2 + 4?ℎ + 2?ℎ
243 = 4?2 + 6?ℎ 
ℎ =   243 − 4?2
            6?
ℎ = 81 − 2?
      2?      3		 ? TSA equation and sub
? simplification		 (2)
       
10.2 ? = 2? × ? × (81 − 2?)
                     (2?    3)
? = 81? − 4 ?3
             3		 ?sub into volume formula (1)
       
10.3 ?? = 81 − 4?2
??
81 − 4?2 = 0
?2 = 81
        4
? = 9 = 4.5
      2		 ? 81 − 4?2
? 81 − 4?2 = 0
? ?2 = 81
            4
? answer      		 (4)
      [7]
  QUESTION 11    
11.1 9 × 9 × 9 × 5 × 4 = 14580 ?9 × 9 × 9
?5 × 4
? 14580		  (3)
       
11.2.1     12!    = 119750400
   2! .2!		 ? 12!
?2!.2!
? 119750400		 (3)
       
11.2.2           10!           
           2!          = 1 = 0,015
119750400         66		 ?10!2!
? 119750400
? answer		 (3)
      [9]
11.3.1  tree ? first branch with values
? top part of second branch with values 
? bottom part of second branch with values 		  (3)
       
11.3.2 ?(?) = 2
           3		 ? ?(?) = 2/3 (1)
       
11.3.3 ?(?/?) = 1 × 1
               3     2
?(?/?) = 1
               6		 ? ?(?/?) = 1/6 (2)
      [15]
      TOTAL: 150
Last modified on Tuesday, 13 July 2021 06:32
Published in 2017 September Grade 12 Trial Examinations

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