The following example illustrates how we can use GC in Differential equations, and avoid the need to do integration to solve differential equation.
Solution for part ii
Author: gjooheng
H2 maths tutor
Challenging graphing technique question 1
Methods to determine whether a line is on a plane
Set your GC in radian mode!
A student just asked me how come her GC gave a different answer from the solution provided.
The reason is because her GC is set in degree mode instead of radian. For h2 math students, it is advisable to set your GC in radian mode. Else, there might be error when trigo is involved in definite integrals, small angle approximation, parametric curve sketching and complex numbers.
I always set the GC in radian mode and use another calculator fx-96 SG for degree.
Challenging System of Linear Equations Problem 2
Challenging Alevel questions
These are the more challenging alevel questions mentioned in the examiner’s report:
2023 P1: 3,11
2023 P2: 3d, 5, 6b, 10b
2022 P1: 3b, 4c, 5, 9b i, 11d, 12c
2022 P2: 2b, 2c, 3c, 5c, 6, 9b, 11c, 11e
2021 P1: 4bii, 6c, 11
2021 P2: 2, 3b ii, 5c, 7b, 8c, 10e, 11
2020 P1: 4ii, 5, 9i, 10, 11
2020 P3: 3ii, 4iii b, 8ii, 9iv
2019 P1: 8b, 8c, 9, 10 iv,
2019 P2: 1 iii, 2i, 6i, 6ii, 7v, 8 iii, 9i, 10i
2018 P1: 6,9 iii, 10, 11
2018 P2: 6,7,9v
2017 P1: 5ii, 6, 7ii, 8, 9c, 11 iii
2017 P2: 1 ii, 2 ii, 3a, 4b, 7i, 9v, 9 vi
2016 P1: 3, 4, 6 iii, 7, 8 iii, 10 bii, 11 b
2016 P2: 2a ii, , 2b, 4bii, 7 iv, 9a
2015 P1: 3, 5 iii, 7iii, 11 ii explain why is maximum, 11 iii
2015 P2: 1 ii, 2ii, 3b, 4b iii (some students interpret wrongly), 9 ii, 9 iii, 10 iv, 11 iv
2014 P1: 2, 3, 4ii, 6 (a) (ii), 6 (b) (ii), 7 iii, 7 iv, 8 iii, 9 iii, 10 ii, 10 iv, 11 iii, 11 iv
2014 P2: 1 ii, 3 ii, 4 b ii, 6 ii, 6 iii, 10 ii
2013 P1: 2, 5, 6i, 6 iii, 8 i, 9 ii, 9 iii, 10 i, 10 iii, 11 ii, 11 iv
2013 P2: 1 ii, 2, 3 iii, 4iii, 6, 7 i, 8 iii, 9 i, 10 i, 11
2012 P1: 4, 5, 6 iii, 7 iii, 8 iii, 9 iii, 10, 11
2012 P2: 1 b, 3 ii, 3 v, 4, 6ii, 7 iii to v, 8 iii, 8 vii
2011 P1: 1, 3 iii, 4 i, 5 iii, 7 ii b, 8 c, 9 ii, 10 ii
2011 P2: 2 ii, 3 ii, 3 iii, 8 i, 9ii b, 10 iii, 11 ii
2010 P1: 1 ii, 6 iii, 6 iv, 7, 9, 10 i, 11 ii, 11 iii
2010 P2: 1 ii, 3 iii, 4v, 7iv, 7v, 8ii, 8iii, 10 iv
2009 P1: 1 ii, 2, 4, 10 iii, 11 iv
2009 P2: 1 i, 1 iii, 2 iii, 2 iv, 3, 4, 6 ii, 6 iii, 6 iv, 7, 8iii, 8iv, 9 iv, 11 i, 11 ii, 11 v
2008 P1: 1, 5 ii, 6a, 9 ii, 10 ii a, 10 ii b,
2008 P2: 1 iv, 2, 7, 8iii, 8 iv, 10 , 11 last part
2007 P1: 1, 2 i, 5 (hence part), 7i, 10 iii, 11
2007 P2: 2 iii, 2 iv, 3 iii, 4 ii b, 7 last part, 8, 9 (i) (b), 9 (ii) (c), 11
Common misconception on Volume of Revolution
These questions by students illustrate the common misconceptions on volume of revolution:
Question 1
Further queries by student
Misconception: Since the region is bounded on both sides of the Y-axis, the volume of revolution needs to be multiplied by 2 when integrate one side.
Clarification: The volume of revolution is the same whether one side of the region is rotated or both sides of the region are rotated. The swept volume of one side or both sides are the same.
Question 2
Solution
Notice that for part i, the volume needs to be multiplied by 2, whereas for part 2, there is no need.
Student’s queries on part ii:
Misconception: That the volume achieved by rotating only 180 degrees is half of the volume of rotating 360 degrees.
Clarification: Since the graph is bounded on both sides, just need to rotate by 180 degrees to obtain the full swept volume. If it is bounded by one side, then need to rotate by 360 degrees to obtain the full swept volume.
Challenging Probability problem 6
The owner of a restaurant counts the number of banquets received by the restaurant at the end of each week. The probability that the restaurant receives at least two banquets in a randomly chosen week is 0.3. A week is considered busy if the restaurant receives at least two wedding banquets in the week.
Calculate the probability that in a period of 6 consecutive weeks, the 6th week is the second busy week, given that there is at most one busy week in the first four weeks.
Ans: 0.166
Solution
Inequalities involving logarithm
Students often ask why sometimes need to flip the inequality sign when dealing with logarithm.
This is a question asked by a student:
For 0<x<1, both ln x and lg x are negative. When we divide by negative number, we need to change the sign.
For x>1, both ln x and lg x are positive. There is no need to change the sign when divide by ln x or lg x.
Challenging Vector problem 4
JC A Level H2 Math Tuition Singapore 
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