Page 3 - Spotlight A+ SPM Additional Mathematics Form 4 & 5

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Formative Zone ...”
Chapter 2 Quadratic Functions Additional Mathematics Form 4
Example 6 Solving quadratic inequalities
Given the roots of quadratic equation 1. A quadratic inequality is a function where
The tagging is located at the 2x 2 + (p − 2)x + (q + 1) = 0 are −1 and − 3 , find its degree is 2 and uses the inequality symbols CHAP
which are less than, greater than, less than or
2
the values of p and q.
Solution: equals and greater than or equal. Symbol 2
Inequality
end of the example guides Sum of roots = −1 +  − 3 2  Less than 
2 
−  p − 2 = − 5 2 Greater than  <
Less than or equal to
the students to answer the p − 2 = 5 Greater than or equal to  Chapter 10 Index Numbers Additional Mathematics Form 4
2 2
p − 2 = 5 2. Solving a quadratic inequality means to find the
corresponding questions in Product of roots = −1  − 3 2  3. There are three methods to determine the 1. Find the index number or price index of the 7. The table below shows the price (RM) and the
10.1
range of the values of x that satisfies the inequality.
p = 7
range of values of x that satisfy the quadratic
Formative Zone. q + 1 = 3 2 inequality which are: following quantity or price as 2014 is the base price indices of four items, P, Q, R and S used
(a) Graph sketching
year. C1
in making a dress.
2
(b) Number line
q + 1 = 3 (c) Tabulation (a) Year Price of a tile (RM) Price (RM) Price index for
q = 2 4. The range of values of x can be obtained by 2017 44 Item Year Year the year 2019
Thus, the values of p = 7 and q = 2. considering two types of quadratic inequalities. 2014 55 2016 2019 year 2016
based on the
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Try question 9 to 11 in Formative Zone 2.1 (a) (x − a)(x − b)  0
(b) (x − a)(x − b)  0 (b) Number of visitor P 3.40 x 125
Year
Example 7 A Graph sketching method 2019 (thousand) Q 2.50 3.20 y
24.7
1. In graph sketching method, we have to consider R z 2.30 115
Given that one of the roots of quadratic 2014 19 FORMATIVE ZONE
equation 2x 2 − kx + 54 = 0 is three times the two forms of quadratic equations which are S 2.85 2.80 98.25
other root, find the values of k. y = (x − a)(x − b) and y = −(x − a)(x − b). Find the values of x, y and z. C3
Solution: 2. For quadratic equation y = (x − a)(x − b), the 2. The price of a school bag in the years 2019 and
2017 were RM63.90 and RM45 respectively.
Let the roots be α and 3α. graph is as follows: Find the price index of the school bag for the 8. The price indices of a tube of facial cleanser
Sum of roots = α + 3α y > 0 year 2019 based on the year 2017. C1 in the years 2017 and 2014 based on the year
C3 Questions to test
−k x 2012 are 135 and 120 respectively. Calculate
−  2  = 4α a b 3. The table below shows the price of two type the price index in the year 2017 based on the
y < 0 of food, vegetables and chicken in the year year 2014.
k = 4α (a) The values of a and b are the roots of the 2015 and 2018.
2
quadratic equation.
shown in the table below. students' understanding
α = −k 8 …❶ (b) Thus, Type of food Price (RM) per kg 9. In a boutique, the total number of customers
who came to the boutique are recorded as
Product of roots = α(3α) (i) x  a or x  b when (x − a)(x − b)  0 2015 2018 C3
(ii) a  x  b when (x − a)(x − b)  0
54 = 3α2 3. For quadratic equation y = −(x − a)(x − b), the Vegetable 3.20 3.52 Year 2012 2014 2016
2
3α 2 = 27 …❷ graph is as follows: Form 5 Chicken 6.00 7.20 Number of 3 000 at the end of each
Substitute ❶ into ❷. Chapter 5 Probability Distribution Additional Mathematics customers 3 930
y > 0
k 2 x Calculate the price index of vegetable and (hundred)
3  8  = 27
subtopic.

SPM Simulation HOTS Questions y < 0 a EXAMINER’S b chicken in the year 2018 based on the year (a) Find the index number of the number of
2015. C1
k 2
COMMENT
customers that came to the boutique in CHAP
the year 2014 based on the year 2012.
Paper 1 64 = 9 (a) The values of a and b are the roots of the 4. The number of accident in the year 2017 is (b) Given the index number of the number of 10
k 2 = 576
k = 24 or k = −24 2. The diagram below shows the graph of a
1. A bank auditor claims that credit card’s quadratic equation. 1 840 cases while 1 564 cases in the year
(b) Thus,

2018. Calculate the index number of accident
balances are normally distributed with a mean binomial distribution of X. occur in the year 2018 based on the year 2017. customers in the year 2016 based on the
Try question 9 to 11 in Formative Zone 2.1
of RM2 870 and a variance of RM810 000. (i) a  x  b when (x − a)(x − b)  0 Describe the answer. C2 year 2014 is the same as the index number
What is the probability a randomly selected P (X = x) (ii) x  a or x  b when (x − a)(x − b)  0 in the year 2014 based on the year 2012.
credit card holder has a card balance less than 5. The number of workers in the year 2018 is Estimate how many customers came at the
C4
RM2 500? 2.1.2 2.1.3 0.432 25 2 106 compared to 1 950 in the year 2015. After boutique on 2016?
Examiner’s comment: three years, the index number of workers in the 10. During a dry season, water level in a lake is
year 2020 based on 2018 is 132.6. In what year,
Let X is the credit card’s balance 0.288 the increment of workers is higher? C2 25 m on February 2019 and 27.3 m on March
m = 2 870, s 2 = 810 000 0.216 2019. C3
s = 900 6. The index number of annual income of James (a) Find the water level of the lake on March
So, X ~ N(2 870, 900). 0.064 CHAP. decreases by 6.2% from the year 2017 to the as February is a base month.
Given X is less than RM2 500. x 5 year 2018. If his annual income in the year 2017 (b) If the water level of the lake on April (March
Standardised variable X to Z, 0 0 1 2 3 was RM48 750, what is James’s annual income is a base month) were twice than (a), find
the water level of the lake in April 2019.
in the year 2018? C2
P(X , 2 500) = P ( Z , X – m ) s Find C4
= P ( Z , 2 500 – 2 870 ) (a) the value of probability of ‘success’, 201
(b) P(1 < X , 3).
900
= P(Z , –0.411) Examiner’s comment:
= P(Z . 0.411) (a) Let p is probability of ‘success’
= 0.3405 q is probability of ‘failure’
From the graph, P(X = 3) = 0.216
3 C 3 p 3 q 0 = 0.216
p 3 = 0.216
From standard normal distribution table, p = 0.6
f(z) Thus, the value of probability of ‘success’
is 0.6.
SPM SIMULATION HOTS QUES 0TIONS = 0.288 + 0.432 SPM MODEL PAPER
(b) P(1 < X , 3) = P(X = 1) + P(X = 2)
= 0.72
z
–0.411
Paper 1
3. In normally distributed, the mean and standard
z 1 1 Paper 2 deviation of length of fish is 11 inches and Time: 2 hours
Substract
4 inches respectively. C5 Section A
Provide a complete solutions 0.3409 4 (a) What is the percentage of the length of fish Instruction: Answer all questions
(64 marks)
0.4
are longer than 14 inches?
(b) If 200 fish are randomly selected, how many
function of y = f(x).
Calculator fishes has the length less than 9 inches? 1. (a) Diagram 1 shows a part of graph for a (ii)
with the examiner's comments Examiner’s comment: y y = f(x)
Check the answer by using scientific
calculator.
Let X is the length of fish
1. Press MODE MODE and choose 1 Given m = 11, s = 4, so X ~ N(11, 4)
(a) Given the length of fish is longer than
which is SD.
for the SPM Simulation HOTS 14 inches, X . 14. 0 2 x
2. Press SHIFT 3 and choose 3
Change variable X to Z
represents P(z . a). P(X . 14) = P( Z . X – m s ) –3
questions. 3. Insert 0.411 and the screen will display = P( Z . 14 – 11 ) Diagram 1
4
0.34054 .
= P(Z . 0.75)
SPM MODEL PAPER (ii) has an inverse function or not. [2 marks] domain 0 < x < 4 and its inverse function
= 0.2266 State whether the function of f 2. (a) Diagram 2 shows a graph of function f for
(i) is a discrete or continuous,
335 (b) A function f is derived by f : x ˜ a , x ≠ 0 f –1 . y A (4, 12)
x
such as a is a constant. Given f –1 (2) = 2 , find
(i) the value of a,
(ii) f 17 (8). [3 marks] f
Answer: f –1 B
(a) (i)
Form 0 x
5
Additional Mathematics Chapter 7 Linear Programming –4
(ii) Based on the graph, determine SPM MODEL PAPER
Diagram 2

(b) (i) (i) the domain of f –1 ,
(ii) the coordinate of point B on the graph
of f –1 that corresponding with the point
A on the graph of f.
[2 marks]
Paper 2 SPM format questions according
(i)
1. On the given grid below, show the region that 4. By using the given grid below, Answer:
C2 satisfies all the following inequalities. C3 (a) show that the region is bounded by all the
x > 3, y > 1 and x + y < 5 following inequalities. to the latest SPM 2021
x > 2, y > x and x + y < 6
y y (ii)
Summative Zone 7 6 5 7 6 assessment format cover all the
4 5 418
3 4
3 chapters in Forms 4 and 5.
2
CHAP. 1 2
7
Questions of various levels of 3 4 5 6 7 x 1
2
1
–1 0
–1 –1 0 1 2 3 4 5 6 7 x
–1
thinking skill are provided to (b) The point P with coordinates (x, y) lies inside
2. A green grocer sells bananas and apples. In one
day, he sells
C3
I up to 80 bananas, the region R. x and y are integers. Write down
II up to 90 apples, the coordinates of all the points of R whose
evaluate the understanding 5. The cost of a book is 50 cents and a pen is
III not more than 110 fruits.
coordinates are both integers.
If x be the number of bananas sold and y be
the number of apples sold, show the region that

RM1.30. A student wants to buy x books and
satisfies these inequalities and label the region C4 y pens based on the following conditions: ANSWERS Complete answers
http://bit.ly/2ORHlke
as R.
of each chapter. 3. The diagram below shows the R region that I II The total number of books and pens bought
At least three pens must be bought.
C2
satisfied all three linear inequalities.
must not more than 12.
FORM 4
y III The amount of money spent is at most 9. f(x) 5. – 4 5
RM10. 13 3
10 Write down the three inequalities other than 6. 10
Chapter 1 Functions
8 x > 0 and y > 0 that satisfy all the above 9 7. g(x) = 3x – 2
conditions. 1.1
6 R 3 8. f(x) = x – 2
6. By using the same graph, draw all the inequalities: (a) The relation is a function x 9. (a) 15
1.
4 C3 because each object has –6 – 3 0 5 (b) 13 5
y > x – 1, x > 2 and 2x + y > 8 only one image. 2
2 10. h = –3k
Shade the region that bounded by the (b) The relation is not a Range of f is 0  f(x)  13.
function because there is
x inequalities. Hence, state the minimum value of one object does not have 10. (a) q = 1, p =1 11. p = 12, q = 5 2
0 1 2 3 4 5 y in the region. any image. (b) 3 x x
2. (a) h(x) = |10 – x| or (c) –10 12. (a) 2x + 1 (b) 3x + 1
Define all the three linear inequalities. h(x) = |x – 10| 11. (a) 3 (c) 2x + 1 x (d) 3x + 1 x
(b) h(x) = x 2 – 1 (b) 5 25
3. (a) A function. (c) 2 13. k = 1 , h = 11
(b) A function. 2 2
(c) Not a function.
3 or 1.7321
391 4. (a) {a, b, c, d} 12. (a) (b) 2 or 1.4142 14. (a) RM13 400
(b) RM449 675
FORM 4 ANSWER 5. (a) 7 13. (a) 1 (b) 2 (ii) 3 1. (a) –4 1.3 (b) –2 ANSWER
(b) {–2, 0, 2, 4}
2
(c) a, b, c, d
(d) –2, 0, 2
14. 3
(b) {–3, –2, 2, 7}
15. (a) 3
(d) 2
(c) 4
(b) (i) –12
6. (a) 0, 2, 6, 8
(b) 7, 1, 10
(b) – 5 and 5
3. (a) –7
(b) –2
(c) {7, 1, 10}
(d) 7 16. (a) (i) 5 4 12 (ii) 11 2. (a) –2 (b) 4
4. (a) Has an inverse function
(e) 8 17. (a) (i) 15 because each element in set
P matched with only one
7. (a) Domain = {–2, 0, 2, 4} 4 (b) Does not have an inverse Complete answers are
Codomain = {4, 6} (ii) 3 element in set Q.
Range = {4, 6} 4
(b) Domain of f is –1  x  3. function because from
Codomain of f is 1  f(x)  3. (b) – 31 and 33 horizontal line test, the
Range of f is 1  f(x)  3. 4 4 line cuts the graph on two
points.
8. (a) f(x) 18. –3 and 9 5. (a) Has an inverse function g. provided. Scan the
10 19. (a) 4 4 (b) Does not has an inverse
5 (b) 0  g(x)  10 6. function g. QR Code provided to get
4 y
1.2
x 11
–2 0 4 3 1. (a) fg(x) = 3x – 1 y = x
3 3 g(x)
(b) gf(x) = x – 5
(b) f(x) 2. (a) f 2 (x) = 36x – 7 6 the steps to the solution.
5 (b) g 2 (x) = 16 + 9x 3
(c) gf(x) = 18x + 1 2
3 (d) fg(x) = 18x + 23 –2 0 2 3 6 11 x
3. (a) 111 (b) 731 –2 g –1 (x)
x 4. (a) –4 (b) – 17 Domain of g –1 (x) is 2  x  11.
0 5
1 5 2 Range of g –1 (x) is –2  g –1 (x)  3.
2
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Prelims Spotlight Add Math Tg4&5.indd 5 23/04/2021 2:10 PM