Tips & Challenges

Trend and recommended learning for H2 maths 9758 syllabus

Trends from 2017 and 2018 alevel:

  1. Solving in terms of a and b.
  2. Solve inequality involving modulus in exact form.
  3. Secondary school syllabus like remainder theorem appear in N2017. R-formula appear in specimen paper.
  4. Using integration techniques to evaluate arc length given the formula.
  5. Applications of 1st order Differential equation: Motion with resistance proportional to velocity came out in N2017.
  6. Applications of 2nd order Differential equation: Electric circuits came out in N2018.

Recommended Learning:

  1. Applications of integration such as arc length, surface area of revolution and centroid. Applications of integation
  2. Applications of 1st order DE such as motion, population growth, orthogonal trajectories, mixture problems, Torricelli’s Law, Newton’s Law of cooling. 1st order DE applications
  3. Applications of 2nd order DE such as vibrating springs and electric circuits Applications of 2nd order DE
  4. Odd/even functions, floor/ceiling functions

How to sketch both cartesian and parametric graphs on the same diagram using GC

Example 1

DHS J1 2018 Aug Test

DHS 2018 J1 Aug test

Suppose we want to sketch the curve C and the line y=-2x+2 on the same graph

DHS 1

Note how the line y-2x+2 is entered.DHS 2

Adjust the Tmax to 1%  more so the graph displays correctly.DHS 3

Example 2

2011 alevel P1 Q11

2011 alevel P1 Q112011 GC12011 GC2

Finding sequence of transformation

Usually the sequence of transformation can be found by the replacement method. However, it can be tough to find the replacement for complicated functions. For such cases, we can use the general linear transformation c f(bx+a)+d to find the sequence.

cf(bx+a)+d = Translate by a units in the negative X direction, then scale by a factor of 1/b parallel to the X-axis, then scale by a factor of c parallel to the Y-axis, then translate by d units in the positive Y direction.

Example

NYJC 2016 prelim

Solution

NYJC 2016. alterativeJPG

Rounding in Inequality

Example 1: 2015 P1 8ii

2015 P1 Q8

Solution

RI

Examiner’s report 

exam report

Example 2: 2015 P1 2ii

2015 P1 Q2

Examiner’s report

exact round

Comments

Graphical approach will yield  x> 1.732. Notice Cambridge answer  is not rounded to x>1.74 to satisfy the non-rounded inequality but rounded to 3 s.f.

Conclusion

Considering example 1 and example 2, we conclude that in real life applications questions, when rounding off your answer to 3 s.f, round off such that the rounded off answer still satisfies the non-rounded off inequality.

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