Maths 14 Practice Quiz |Course:12th |Pages:05

Questions and Answers

1
ANSWERS
Mathematics P1/Wiskunde V1
2 14
Maths 14 Practice Quiz
DBE/November 2022
NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 6/VRAAG 6

6.1 A  P(1  i) n
8
 m  8
13 459  12 0001  
 400   subst into correct
 m 
8 formula
1    1,121...
 400 
m 8
1  1,121.. 1 
m 8
 1,121..
400 400
m
 0,0144...
400
m  5,78%  5,78 %
(4)
6.2
F

x (1  i)  1
n

i
0,075
 0,075 12  
1 0001    1 12

 12    12
F
0,075
12
 R12 421,22  answer

He won't be able to buy the computer because  conclusion

R13 000 – R12 421,22 = R578,78 (4)
OR/OF
He won't be able to buy the computer because
R12 421,22 < R13 000
6.3.1 Loan amount = 85% × R250 000
= R212 500  answer (1)

OR/OF OR/OF
Loan amount = R250 000 – (15% × R250 000)
= R212 500  answer (1)
6.3.2  0,13 
5
 0,13 
5

A  212 5001    A  212 5001  

 12   12 
A  224 262,53 answer

P
  
x 1  1  i)  n
i
  0,13  67  
x 1  1  ) 
  12   substitution into
224 262,53 
0,13 correct formula
12  – 67
 x  R4 724,96  answer (5)
[14]

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ANSWERS
Mathematics P1/Wiskunde V1 33 15 Maths 14
Maths 14 Practice
Practice Quiz
Quiz
DBE/November 2022
NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 7/VRAAG 7

7.1 f x   x 2  x
f x  h   f x 
f / x   lim
h 0 h
( x  h)  ( x  h )  ( x 2  x )
2
f x   lim
/ substitution into
h 0 h the formula
x  2 xh  h  x  h  x 2  x
2 2
 x 2  2 xh  h2  x  h
f / x   lim
h 0 h
2 xh  h  h
2
= lim
h 0
 2 xh  h 2  h
h
h(2 x  h  1)
 lim common factor
h 0 h
 f x   2 x  1
/
answer (5)
OR/OF OR/OF
f x   x 2  x
f x  h  ( x  h) 2  ( x  h)  x 2  2 xh  h 2  x  h  x 2  2 xh  h2  x  h
f x  h  f x   x 2  2 xh  h 2  x  h  x 2  x
 2 xh  h 2  h  2 xh  h 2  h

f x  h   f x 
f /
x   lim
h 0 h
2 xh  h 2  h substitution into
= lim the formula
h 0 h
h(2 x  h  1)
 lim common factor
h 0 h
 f x   2 x  1
/
answer (5)
7.2 f ( x)  2 x  3x  8 x
5 4
 10x 4

f / x   10 x 4  12 x 3  8   12x 3
8 (3)
7.3 g ( x)  ax 3  3x 2  bx  c  g ( x)  3ax  6 x  b
/ 2

g / ( x)  3ax 2  6 x  b 
g // ( x )  6ax  6 g // (1)  6a(1)  6  0
g // (1)  6a(1)  6  0 a 1
a  1
For concave up g // ( x)  0
6x  6  0 x>–1 (4)
x>–1

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Mathematics P1/Wiskunde V1
4 16 DBE/November 2022
NSC/NSS – Marking Guidelines/Nasienriglyne

OR/OF
Min gradient at (– 1 ; – 7) implies:
at x = – 1 - point of inflection
and g will be positive cubic
hence a > 0
Since g is concave up
x  1

OR/OF

. (– 1 ; – 7 )
.
Since g is concave up (– 1 ; y)
x  1 Since g is concave up
x  1

1
Answer only:
4

OR/OF
Min gradient at (– 1 ; – 7) implies:
at x = – 1 - point of inflection
and g will be positive cubic
hence a > 0
Since g is concave up
x  1

OR/OF

. (– 1 ; – 7 )
.
Since g is concave up (– 1 ; y)
x  1 Since g is concave up
x  1

1
Answer only:
4

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Mathematics P1/Wiskunde V1 5 17 DBE/November 2022
NSC/NSS – Marking Guidelines/Nasienriglyne

OR/OF OR/OF
Min gradient at (– 1 ; – 7) implies:
at x = – 1 - point of inflection  point of inflection
and g will be positive cubic
hence a > 0  a > 0
Since g is concave up
x  1
 x  1 (4)
OR/OF
OR/OF

. (– 1 ; – 7 )
pos graph
.
(– 1 ; y) point of inflection
Since g is concave up
x  1 Since g is concave up
x  1
 x  1 (4)
1
Answer only:
4

[12]

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