1. Differential equation
(i) h = 32m
(ii) ≈ 46.9 years
2. Vectors
(i) ≈ 73.4°
(ii) λ = 3/7 or 1
Point = ( 17/7 , 1/7 , -58/7 )
(iii) -36x + 2y – 11z = 4
3. Functions
a)(i) Graph of y=k , k∈R , cuts at most on 1 point.
f (x) is a 1 – 1 function, f inverse exists
(ii) f -¹ : x → √(1 – 1/x) , x > 0
b) y≤ 1 – (1/2)√3 or y≥ 1 + (1/2)√3
Rg = ( -∞, 1 – (1/2)√3] ∪ [ 1 + (1/2)√3 , ∞ )
4. Mathematical induction / Method of difference
b)(i) 2 = A(2r+3) + B(2r+1)
(ii) (1/3) – (1/(2n+3))
(iii) n≥ 1498.5
smallest n = 1499
5. Sampling methods
(i) Sampling frame is not available, therefore the manager is unable to define the appropriate subgroups required for stratified sampling.
(ii) Survey 25 people each for different age groups (5-20, 21-40, 41-60, 61-80). The manager can survey the customers going in and out of the supermarket to obtain the required data.
6. Binomial distribution
(i) ≈ 0224
(ii) ≈ 0.149
(iii) ≈ 0.825
7. Poisson distribution
(i) Average number of errors per page is constant
Errors occurring are independent of each other
(ii) ≈ 0.165
(iii) least n = 4
8. Hypothesis Testing
p-value = 0.264618 > 0.1
Do not reject Ho. There is insufficient evidence at the 10% level of significance to doubt his claim.
9. Probability
(i) = 0.4
(ii) 0.185
(iii) Lowest = 0.165
Highest = 0.33
10. Correlation and Regression
i)
(ii) a. ≈ -0.9807
b. ≈ -0.9748
c. ≈ -0.9986
(iii) √h and P
P = 34.8 – 0.266√h
(iv) 34.8 – 0.266√h
11. Permutations and Combinations
(i) 10080
(ii) 10079
(iii) 720
(iv) 5760
12. Normal Distribution
(i) 0.01267
(ii) 0.05238
(iii) 0.742
A-LEVEL H2 MATH JUNE HOLIDAYS INTENSIVE REVISION
A-LEVEL H2 PHYSICS JUNE HOLIDAYS INTENSIVE REVISION
JUNIOR COLLEGE / A-LEVEL TUITION:
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Suggested Answers to 2015 A Level H2 Math 9740 PAPER 1
1. System of linear equations / Graphing techniques
(i) a = -3.953 , b = -5.187 , c = 7.303
(ii) x = -0.589
(iii) y = 5.19x + 7.3
2. Graphing techniques / Inequalities
i)
(ii) -1.73 < x < 0.414 or x > 1.73
3. Integration
(i) When n→∞ , sum of area of rectangles ≈ area of function
(ii) 3/4
4. Application of differentiation
Maximum area = (1/32) d²
5. Graphing techniques / Transformation
(i) Translation of +3 units parallel to the x-axis,
Scaling by scale factor of 1/4 parallel to the y-axis
ii)
iii)
6. Maclaurin’s series / Binomial expansion
(i) 2x – 2x² + (8x³/3)
(ii) -(104/27)
7. Vectors
(i) OC = (3/5)a
OD = (5/11)b
(ii) line BC : r = (3/5)λa + (1 – λ)b
line AD : r = (1 – μ)a + (5/11)μb
(iii) AE : ED = 11:9
8. AP/GP
(i) T = { T∈R : 59 ≤ T ≤ 77 }
(ii) t = { t∈R : 63.8 ≤ t ≤ 74.5 }
9. Complex numbers
a) w = a ± (a/√3)i
b)(i) z = 2e^(-π/10) , 2e^(3π/10) , 2e^(-π/2) , 2e^(7π/10) , 2e^(-9π/10)
(ii) 4 sin ( π/5)
10. Application of integration
i) A1/A2 = (2-√2)/(√2 -1)
= √2
iii) Volume = (π³/16√2) + (π²/2√2) – √2 π
11. Application of differentiation and integration
i) dy/dx = (-3sin³Θ + 6sinΘ cos²Θ)/(3sin²Θ cosΘ)
= 2 cot Θ – tan Θ (shown)
ii) ( (2√2)/(3√3) , ( 2 /√3 ) ) maximum point
iii) Area ≈ 0.884
iv) a = (3√2)/(2)
A-LEVEL H2 MATH JUNE HOLIDAYS INTENSIVE REVISION
A-LEVEL H2 PHYSICS JUNE HOLIDAYS INTENSIVE REVISION
JUNIOR COLLEGE / A-LEVEL TUITION:
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MUST-KNOW PSLE MATH PROBLEM 1
PSLE MATH JUNE HOLIDAY INTENSIVE REVISION
PSLE SCIENCE JUNE HOLIDAY INTENSIVE REVISION
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Question:
The ratio of Eric’s money to Kim’s money was 3 : 1. After Eric gave Kim $60, the ratio of Eric’s money to Kim’s money became 3 : 5. How much did Eric have at first?
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Solution using the Storyline-Balance Method:
E : K
At start: 3u : 1u
Story: – 60 + 60
At end: 3u – 60 1u + 60
Final ratio: 3 : 5
So 5 sets of (3u – 60) = 3 sets of (1u + 60)
15u – 300 = 3u + 180
15u – 3u = 180 + 300
12u = 480
1u = $40
So at first Eric had 3u = 3 x 40 = $120
Ans: $120
Note:
The above is just one method. Another method requires you to notice that in the above problem, the total number of units remain the same as it is an INTERNAL TRANSFER or TOTAL UNCHANGED situation or concept. However, using the Storyline-Balance method, you don’t need to know what has been unchanged.
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Suggested Answers to 2014 A Level H2 Math 9740 PAPER 2
NEW!
H2 PHYSICS PRACTICAL
H2 CHEMISTRY PRACTICAL
H2 BIOLOGY PRACTICAL
The following is a suggested solution by our H2 and H1 Math Tutor, Mr. Teng. Please note that it is a suggested solutions and was rushed out. Mr. Teng omitted several presentation steps out of convenience. There might be alternative solutions at times too. Feel free to comment and discuss. Please let us know if there are mistakes too. Thanks in advance.
PLEASE NOTE THAT STUDENTS ARE EXPECTED TO WRITE STATE YOUR DISTRIBUTIONS CLEARLY. We omitted for convenience.
Disclaimer: This suggested solutions is provided free by Mr. Teng & Singapore Learner. Both parties do not profit from this. We simply believe that this will help current students and future students to learn from the mistakes of others. Thanks.
ERRATA & Remarks:
6(ii) I miscounted and was careless.
Case 1: A plays midfield, B sits out of attacker =>3c1 x 8c4 x 4c1 x 5c4
Case 2: A sits out midfield, B plays attacker => 3c1 x 8c4 x 4c2 x 5c3
Case 1 + Case 2 =1 6800
(iii) I doubled counted indeed. Thanks for pointing out. Should be
Case 1: Midfielder A plays midfield => 8c4 x 3c1
Case 2: Midfielder A plays defender => 8c3 x 3c2
Case 3: Midfielder A sits out => 8c4 x 3c2
3c1 x 5c4 x (8c4 x 3c1 + 8c3 x 3c2 + 8c4 x 3c2) = 8820
8ai. I drew the left side of the quadratic. It should be the right side. my bad.
9. I initially used Z-Test for this question. But T-test is definitely more appropriate as we are given as unknown population variance. I changed it to T-test in the following solutions, but understand that some teachers consider Z-Test to be acceptable too. I should highlight that nobody has access to the actual answer scheme; we only see the marker’s report.
(ii), (iii) Set notations should be used. {t>3.48} and {0<k^2<0.423}
11. Please take note that I used CLT here, so cc is not required.
– I understand some students misunderstand that CLT is used to find the mean of any distribution. This is a terrible misconception. CLT is a limiting theorem, and we apply it to approximate a non-normal distribution to a normal distribution, given n is sufficient. And they can be sum of independent random variables, sample sum, or sample mean. You can read more from the link below.
– Why I used CLT approximation instead of poisson approximation (Poisson approximation is not incorrect either): I used continuity correction in qn 7 alr, so given examiner psychology, I would think they want to be impressed with something different. That’s why I picked CLT approximation. Do note that this does not mean that doing poisson approximation is wrong; so long as you perform cc precisely.
I do hope this clarifies, and I’m sorry it came late as I was ill and had jc1 classes yesterday too.
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Suggested Answers to 2014 A Level H2 Math 9740 Paper 1
MOCK EXAMS FOR SCIENCE PRACTICAL
The following is a suggested solution by our H2 and H1 Math Tutor, Mr. Teng. Please note that it is a suggested solutions and was rushed out. Mr. Teng omitted several presentation steps out of convenience. There might be alternative solutions at times too. Feel free to comment and discuss. We will try our best to answer.
Errata:
Qn3 I misread it. Answer should be 2/3 and need not be evaluated.
Qn5i I copied wrongly off GC. answer should -3+4i. It shouldn’t affect the back parts.
Qn7 I copied wrongly off GC. So rounding off, the answers should α=1.885(3dp)
Advice and tips from Mr Teng for H2 Math exam 2014 paper 2
Please focus on tomorrow’s paper 2 and not let paper 1 affect you too much.
Errata:
Qn3 I misread it. Answer should be 2/3 and need not be evaluated.
Qn5i I copied wrongly off GC. answer should -3+4i. It shouldn’t affect the back parts.
Qn7 I copied wrongly off GC. So rounding off, the answers should α=1.885(3dp)
Much thanks for those that point it out.
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Suggested Answers to 2013 A Level H2 Math 9740 Paper 2
NEW! H2 MATH TOPICAL CRASH COURSE (Sign Up Now!)
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The above solutions had been worked out by Mr Teng.
Errata:
For qn 3(I) line 7, the denominator of the third derivative of f(x) should be of power 4 instead of 2.
For qn 4(I), was some scanning error. The answer is 40.4.
For qn11(III), I made a terrible mistake and forgot the third case for which exactly three letters same AND two digits same. The permutation for that will be (26×9).
11(iv) I made the mistake of double counting and arranged after using the permute function. The following is the full solutions
Case 1: Exactly one vowel and 2 different consonants and exactly one even digit
(5C1)(21C2)(3!)(4C1)(5C1)(2!) =252000
Case 2: Exactly one vowel and 2 same consonants and exactly one even digit
(5C1)(21C1)(3!/2!)(4C1)(5C1)(
P(exactly one vowel & exactly one even digit) = (252000+12600)/(26^3*9^2) = 0.186 (3SF)
Disclaimer: We welcome fair comments and discussion. The answers were rushed out and done while Mr Teng was in between classes too. So there are errors definitely, so many thanks for all who point them out. 🙂
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