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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 

Related Items

QUESTION 1 

🗞​

1.1  2𝑥(𝑥 + 1) − 7(𝑥 + 1) = 0
(𝑥 + 1)(2𝑥 − 7) = 0
𝑥 = −1     or     𝑥 = 72		 🗞 factors
🗞 𝑥-value
🗞 𝑥-value 		  (3) 
       
 1.2  𝑥2 − 5𝑥 − 1 = 0 
𝑥 = −𝑏 ± √𝑏2 − 4𝑎𝑐
             2𝑎                                   
𝑥 = −(−5) ± √(−5)2 − 4(1)(−1)
                      2(1)
𝑥 = 5,19 or 𝑥 = −0,19		  🗞 substitution into correct formula
🗞🗞 x- values		  (3)
       
1.3  4𝑥2 + 1 ≥ 5𝑥
4𝑥2 − 5𝑥 + 1 ≥ 0
(4𝑥 − 1)(𝑥 − 1) ≥ 0
1.3 ans
x ≀ 1/4        or       𝑥 ≀ 1    		 🗞 standard form
🗞 factors
 🗞𝑥 ≀ ÂŒ
🗞𝑥 ≥ 1		  (4)
       
1.4 54𝑥+3. 100−2𝑥+1 = 50 000
54𝑥+3. (52. 22)−2𝑥+1  = 50 000
54𝑥+3. 5−4𝑥+2. 2−4𝑥+2 = 50 000
55. 2−4𝑥. 22 = 50 000
2−4𝑥 = 22
−4𝑥 = 2
𝑥 = − 1/2		 🗞5−4𝑥+2
🗞2−4𝑥+2
🗞−4𝑥 = 2
🗞 answer		  (4)
       
1.5 𝑥 = 2𝑊 






..(1)
𝑥2 + 2𝑥 − 𝑊 − 𝑊2 = 36


.(2)
𝑥 = 2𝑊
sub into (2)
(2𝑊)2 + 2(2𝑊) − 𝑊 − 𝑊2 = 36
4𝑊2 + 4𝑊 − 𝑊 − 𝑊2 = 36
3𝑊2 + 3𝑊 − 36 = 0
𝑊2 + 𝑊 − 12 = 0
(𝑊 − 3)(𝑊 + 4) = 0
𝑊 = 3   or 𝑊 = −4
𝑥 = 6 or 𝑥 = −8		 🗞substitution
🗞standard form
🗞 factors
🗞 y − values
🗞 x − values(5)		 (5)
       
1.6 𝑥2 − 𝑘𝑥 + 𝑘 − 1 = 0
∆ = 𝑏2 − 4𝑎𝑐
∆ = (−𝑘)2 − 4(1)(𝑘 − 1)
∆ = 𝑘2 − 4𝑘 + 4
∆ = (𝑘 − 2)2
∆ ≥ 0 roots are real and rational(perfect square)		 🗞substitution
🗞 simplification
🗞 (𝑘 − 2)2
🗞 conclusion(4) 		 (4)
      [23]
QUESTION 2
2.1.1 𝑇𝑛 = 4𝑛 − 1
483 = 4𝑛 − 1
484 = 4𝑛
𝑛 = 121
121 terms in series		 🗞  𝑇𝑛 = 4𝑛 − 1
🗞 equating 483
🗞  answer(3)		  (3)
       
2.1.2 121
   ∑    (4𝑛 − 1)
𝑛=1		 🗞 answer (2)
       
2.2.1  (𝑡 − 3) − (2𝑡 − 4) = (8 − 2𝑡) − (𝑡 − 3)
−𝑡 + 1 = −3𝑡 + 11
2𝑡 = 10
𝑡 = 5		 🗞 setting up equation
🗞 simplification
🗞 answer		 (3)
       
2.2.2 
 ; 
 ; 
 6 ; 2 ; −2 ; 
 ; 
 ; 

𝑇10 = 6        or              𝑇𝑛 = −4𝑛 + 46
𝑎 + 9𝑑 = 6                             𝑇1 = −4(1) + 46
𝑎 + 9(−4) = 6                        𝑇1 = 42𝑎
= 42		 🗞 numerical values of𝑇10; 𝑇11; 𝑇12
🗞 difference  −4
🗞  a-value		 (3)
       
2.3

𝑎𝑟2 + 𝑎𝑟3 = −4
𝑎 + 𝑎𝑟 = −1 
𝑎𝑟2(1 + 𝑟) = −4
𝑎(1 + 𝑟) = −1
∎ 𝑟2 = 4
∎ 𝑟 = ±2

𝑎 + 𝑎(2) = −1
∎ 𝑎 + 2𝑎 = −1
3𝑎 = −1
𝑎 = − 1/3 
First three terms: − 1/3  ; − 2/3  ; − 4/3           

 🗞 setting of equations
🗞 common factor
🗞 𝑟 = 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 
2𝑎 = 2            𝑎 + 𝑏 = 2                𝑎 + 𝑏 + 𝑐 = 41
𝑎 = 1                        𝑏 = −1                            𝑐 = 41
∎ 𝑇𝑛 = 𝑛2 − 𝑛 + 41

 🗞  2nd difference
🗞  𝑎 = 1
🗞  𝑏 = −1
🗞  𝑐 = 41
🗞  𝑇𝑛 = 𝑛2 − 𝑛 + 41		 (5)
       
3.2

𝑇41 = 412 − 41 + 41
𝑇41 = 1681

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

 🗞 𝑇41 = 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
𝑇49999998 will end in 7		  🗞 unit digits
🗞 groups of 5
🗞 conclusion		 (3)
      [11]
QUESTION 4
4.1.1 𝐎 = 𝑃(1 + 𝑖)𝑛
𝐎 = 500 000 (1 +     7,2     )12𝑛
                              1200
𝐎 = 500 000(1.006)12𝑛		 🗞sub into formula
🗞 12𝑛		  (2)
       
4.1.2 𝐎 = 500 000(1.006)12𝑛
𝐎 = 500 000(1.006)12×5
𝐎 = 𝑅 715894.21		 🗞 𝑛 = 60
🗞  answer		 (2)
       
4.1.3

𝐎 = 𝑃(1 + 𝑖)𝑛
1000000 = 500000(1.006)12𝑛
12𝑛 =      log 2          
          log 1.006
12𝑛 = 115.8707581
𝑛 = 9,66 𝑊𝑒𝑎𝑟𝑠

Will exceed R1 000 000 in 10 years.

 🗞 setting up equation
🗞  using logs
🗞  conclusion		 (3)
       
4.2.1 𝑃V =  10 000 [1 − (1 + 15/1200 ) −36]
                          15/1200                               
𝑃𝑣 = 𝑅288 472,67
𝑑𝑒𝑝𝑜𝑠𝑖𝑡/𝑜 = 𝑅350 000 − 𝑅288 472,67
𝑑𝑒𝑝𝑜𝑠𝑖𝑡/𝑜 = 𝑅61 527,33		 🗞 𝑖 and 𝑛
🗞sub into 𝑃𝑣 formula
🗞 𝑃𝑣 formuleÃŒ 𝑃𝑣 = 𝑅288 472,67
🗞 subtracting
🗞 answer		 (5)
       
4.2.2 350 000    =  𝑥 [1 − (1 + 18,5/1200 ) −60]
                              18,5/1200
𝑥 = 𝑅 8 983,17		 🗞 𝑖 = 18,51200
🗞𝑛 = −60
🗞 substitution
🗞 answer		 (4)
      [16]
QUESTION 5
5.1 𝐎(−3; 0) 🗞 answer  (1)
       
5.2 𝑓(𝑥) = 𝑥2 + 3𝑥
𝑥 = − 𝑏
        2𝑎
𝑥 = − 3
         2
𝑓 (− 3) = (− 3)2  + 3 (− 3)
       2          2               2
= − 9/4
𝑃 (− 3 ; − 9)
       2      4		 🗞𝑥 = − 32   
🗞substitution 
🗞answer		 (3)
       
5.3 𝑓(−5) = 10     and  𝑓(−3) = 0
𝑚 =     10 − 0            
         −5 − (−3)𝑚
= −5		 🗞 calculating 𝑓(−5) and 𝑓(−3)
🗞  substitution
🗞 𝑚-value		 (3)
       
5.4 𝑥 < −3 or / of 𝑥 > 0 🗞answer  (2)
       
5.5

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

 or


𝑓(𝑥 − 2) = (𝑥 − 2)(𝑥 − 2 + 3)
𝑓(𝑥 − 2) = 𝑥2 − 𝑥 − 2
𝑥 = − (−1)   
          2(1)
𝑥 = − Âœ

 🗞 answer
🗞 𝑓(𝑥 − 2) = 𝑥2 − 𝑥 − 2
🗞 𝑥 = − 12		  (2)
       
5.6

𝐿𝑀 = − 1/2 𝑥 + 2 − (𝑥2 + 3𝑥)
𝐿𝑀 = − 1/2 𝑥 + 2 − 𝑥2 − 3𝑥
𝐿𝑀 = −𝑥2 − 7/2 𝑥 + 2
𝐿𝑀 = − (𝑥2 + 7/2 𝑥 − 2)
L𝑀 = − [(𝑥 +  7/4 )2   −  81/16  ]
L𝑀 = − (𝑥 + 7/4 )2  + 81/16

OR

𝐿𝑀 = − 1/2 𝑥 + 2 − (𝑥2 + 3𝑥)
𝐿𝑀 = − 1/2 𝑥 + 2 − 𝑥2 − 3𝑥
𝐿𝑀 = −𝑥2 − 7/2 𝑥 + 2
𝑑𝐿𝑀 = −2𝑥 − 7
𝑑𝑥                 2
−2𝑥 − 7/2 = 0
𝑥 = − 7/4

𝑊 = − (− 7/4 ) 2  − 7/2  (−  7/4 ) + 2
𝑊 = 81/16
∎ 𝐿𝑀 = − (𝑥 + 7/4)2   + 81/16

OR

𝑥 = − 𝑏
2𝑎
𝑥 = − [ − 7/2]
          2(−1)
𝑥 = − 7/4
𝑊 = − (− 7/4 ) 2  − 7/2  (−  7/4 ) + 2
𝑊 = 81/16
∎ 𝐿𝑀 = − (𝑥 + 7/4)2   + 81/16

🗞 𝑔(𝑥) − 𝑓(𝑥) 
🗞 standard form
🗞  completing the square
🗞 answer

 

 

🗞 𝑔(𝑥) − 𝑓(𝑥)
🗞 standard form
🗞  𝑥 = − 7/4   
🗞 𝑊 = 81/16           

 

🗞𝑔(𝑥) − 𝑓(𝑥)
🗞 standard form/standaardvorm   
🗞  𝑥 = − 7/4   
🗞 𝑊 = 81/16         

(4)
      [15]
QUESTION 6
6.1 ans 6.1 🗞 shape 
🗞 y − intercept
🗞 point on graph		 (3)
       
6.2 𝑞(𝑥) = 2𝑥 🗞 answer (1)
       
6.3 ℎ−1𝑥 = 2−𝑊
−𝑊 = log 𝑥
        log 2
𝑊 = − log 𝑥 / 𝑊 = − log2 𝑥 /𝑊 = logÂœ 𝑥
         log 2         		 🗞 interchange 𝑥 and 𝑊
🗞 equation		 (2)
       
6.4 𝑊 ≥ 0 ; 𝑊 ∈ 𝑅   (1)
       
6.5 See 7.2.1 🗞🗞shape and x-intercept          (2)
       
 6.6 log1 𝑥 = −321 −3(  )     = 𝑥 2𝑥 = 8∎    0 < 𝑥 ≀ 8 🗞𝑥 = 8 
🗞0 < 𝑥 ≀ 8		 (2)
      [11]
QUESTION 7
7.1 𝑑 = 5𝑝 = 2 🗞𝑑 = 5🗞𝑝 = 2 (2)
       
7.2 𝑊 = 5 − 𝑥
      𝑥 − 2
𝑊 = −(𝑥 − 2) + 3
          (𝑥 − 2)
𝑊 =     3     − 1
       𝑥 − 2		 🗞𝑊 = 5−𝑥
         𝑥−2
 🗞𝑊 = −(𝑥−2)+3
              (𝑥−2)		 (2)
       
7.3 𝐎(5; 0)
𝑊 = 𝑥 − 3
𝑥 = 𝑊 + 3
𝐎′(0 + 3; 5 − 3)
𝐎′(3; 2)		 🗞𝑥 = 3
🗞𝑊 = 2		 (2)
      [6]
 QUESTION 8  
 8.1

𝑓(𝑥) = −2𝑥2 + 𝑝
𝑓′(𝑥) = lim 𝑓(𝑥 + ℎ) − 𝑓(𝑥)
          ℎ→0       ℎ
= lim     −2(𝑥 + ℎ)2 + 𝑝 − (−2𝑥2 + 𝑝)
  ℎ→0                   ℎ 
= lim   −2(𝑥2 + 2𝑥ℎ + ℎ2)  + 𝑝 + 2𝑥2 − 𝑝)
  ℎ→0                             ℎ
= lim −2𝑥2 − 4𝑥ℎ − 2ℎ2 + 𝑝 + 2𝑥2 − 𝑝
  ℎ→0                     ℎ
= lim −4𝑥ℎ − 2ℎ2
 ℎ→0        ℎ
= lim ℎ(−4𝑥 − 2ℎ)
  ℎ→0         ℎ
= lim(−4𝑥 − 2ℎ)
   ℎ→0
= −4𝑥  

  • Answer ONLY: 0 marks
  • Penalise 1 mark for incorrect use of formula. Must show 𝑓′(𝑥)
 🗞 formula
🗞 substitution of (𝑥 + ℎ) 
🗞 simplification to
(−4𝑥ℎ − 2ℎ2) 
🗞 common factor
🗞 answer		  (5)
       
8.2 HDGF   (4)
      [9]
QUESTION 9
9.1 𝑓(𝑥) = (𝑥 − 1)2(𝑥 + 3)
𝑓(𝑥) = 𝑥3 + 𝑥2 − 5𝑥 + 3 
𝑓′(𝑥) = 3𝑥2 + 2𝑥 − 5
3𝑥2 + 2𝑥 − 5 = 0
(3𝑥 + 5)(𝑥 − 1) = 0
𝑥 = − 5/3       or 𝑥 = 1
𝑓(1) = 0
𝑓 (− 5/3 ) = 256
                   27		 🗞  𝑓(𝑥) = 𝑥3 + 𝑥2 − 5𝑥 + 3
🗞  𝑓′(𝑥) = 0
🗞 factors
🗞  𝑥-values
🗞  𝑊-values		 (5)
       
9.2 2567 🗞 shape 
🗞 𝑥 − intercepts
🗞 𝑊 − intercept
🗞  stationary points		 (4)
       
9.3 𝑓′′(𝑥) = 6𝑥 + 2
6𝑥 + 2 = 0
𝑥 = − 1/3
𝑊 = 128 / 27 / 4,74 / 4 20/27 		 🗞 f′′(𝑥) = 6𝑥 + 2 
🗞𝑥 = − 1/3
🗞  𝑊 = 128 / 27 / 4,74 / 4 20/27 		 (3)
       
9.4 0 < 𝑘 < 25627 🗞🗞answer (2)
       
9.5 𝑓′(𝑥) = 3𝑥2 + 2𝑥 − 5
3𝑥2 + 2𝑥 − 5 = −5
3𝑥2 + 2𝑥 = 0
𝑥(3𝑥 + 2) = 0
𝑥 = 0 or 𝑥 = − 2/3
𝑓 (− 2) = 175
       3      27         
y = −5𝑥 + 𝑐
175  = −5 (− 2/3 ) + 𝑐
 27                 3𝑐
= 85
   27
𝑊 = −5𝑥 + 85
                27		 🗞 f′(𝑥) = −5
🗞 factors
🗞 𝑥 = − 2/3
🗞 f (− 2/3) = 175
                     27
🗞 substitution
🗞 answer		 (6)
      [20]
  QUESTION 10    
 10.1 243 = 2(𝑥 × 2𝑥) + 2(2𝑥 × ℎ) + 2(𝑥 × ℎ)
243 = 4𝑥2 + 4𝑥ℎ + 2𝑥ℎ
243 = 4𝑥2 + 6𝑥ℎ 
ℎ =   243 − 4𝑥2
            6𝑥
ℎ = 81 − 2𝑥
      2𝑥      3		 🗞 TSA equation and sub
🗞 simplification		 (2)
       
10.2 𝑉 = 2𝑥 × 𝑥 × (81 − 2𝑥)
                     (2𝑥    3)
𝑉 = 81𝑥 − 4 𝑥3
             3		 🗞sub into volume formula (1)
       
10.3 𝑑𝑉 = 81 − 4𝑥2
𝑑𝑥
81 − 4𝑥2 = 0
𝑥2 = 81
        4
𝑥 = 9 = 4.5
      2		 🗞 81 − 4𝑥2
🗞 81 − 4𝑥2 = 0
🗞 𝑥2 = 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 𝑃(𝐹) = 2
           3		 🗞 𝑃(𝐹) = 2/3 (1)
       
11.3.3 𝑃(𝑀/𝐻) = 1 × 1
               3     2
𝑃(𝑀/𝐻) = 1
               6		 🗞 𝑃(𝑀/𝐻) = 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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