Hi students,
In the footsteps of eagle, I will look into mathematical content :)
Questions and responses will be consolidated in this thread.
Kindly highlight the difficulty you faced and answer accompanying the question. Do not simply expect us to spoonfeed you as learning is a very personal experience.
Questions regarding IB maths and AP Calculus could also be posted here.
Cheers,
Wen Shih
Posted 30 Jan in reply to a query about Getting Ready for JC:
Hi Xavolon,
For H2 Maths, relevant knowledge from O-level E & A Maths are:
1. Proportion and variation
2. Perimeter, area and volume
3. Linear & quadratic inequalities
4. Sum and product of roots of a quadratic equation
5. Binomial expansion
6. Long division, factor and remainder theorems
7. Surds and indices
8. Logarithmic and exponential functions
9. Solving of simultaneous equations (both linear, one linear and one non-linear)
10. Matrices
11. Graphs of circles, parabolas, logarithmic, exponential, trigonometric, y = 1/x, y = 1/x^2, y^2 = kx and y = |ax+b|
12. Differentiation techniques and its applications
13. Integration techniques and area of a region
14. Coordinate geometry, simple geometrical results, similarity, congruency
15. Trigonometric identities
16. Sequences and patterns
17. Vector geometry
18. Set theory
19. Probability
20. Mean and standard deviation
You may also obtain further details on pages 12 and 13 of the syllabus document at:
http://www.seab.gov.sg/aLevel/2013Syllabus/9740_2013.pdf
Please note that some content I have listed above are not stated in the syllabus. From my experience, such content are necessary to extend to A-level H2 Maths.
Thanks.
Cheers,
Wen Shih
By LastRide7 on 25 Jan 2012:
Solution I posted on the same day:
Hi,
We try to make the denominator of q real, which is done by multiplying q top and bottom by e^(-iθ/2) giving
e^(iθ/2) / [ e^(-iθ/2) - e^(iθ/2) ].
Since e^(iθ/2) + e^(-iθ/2) = 2 cos (θ/2), the above expression simplifies to
[ cos (θ/2) + isin (θ/2) ] / [ -2 cos (θ/2) ], so that
real(q) = -1/2.
I have written about this type of problem before, so you may find it useful to read it:
http://sgforums.com/forums/2297/topics/356377?page=3#post_10226565
Thanks.
Cheers,
Wen Shih
By knt on 23 Jan 2012:
Solve the DE (sec x)(dy/dx) = e^(sin x) + y given that y = 1 when x = 0.
Solution I posted on 23 Jan 2012:
Hi,
I believe it is assessed under H3 Maths where the integrating factor method is to be used to solve the question.
We rewrite the DE as
(dy/dx) - (cos x) y = (cos x) e^(sin x).
Integrating factor is e to the power of integral of -cos x dx, leading to
e^(-sin x).
Multiplying both sides of the DE, we obtain
e^(-sin x) (dy/dx) - (cos x) e^(-sin x) y = cos x
=> d/dx [ y e^(-sin x) ] = cos x.
Integrating both sides wrt x, we obtain
y e^(-sin x) = integral of cos x dx
= sin x + c.
When x = 0 and y = 1, c = 1.
Now y e^(-sin x) = sin x + 1 is the particular solution, from which we arrive at
y = (sin x + 1) e^(sin x).
Cheers,
Wen Shih