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A level H1/H2 Maths/Physics Tuition

Making resolutions to better your grades and score As in year 2012 and beyond? Especially for those who are beginning JC studies in the year ahead, it is important to get a head start and pace yourself for the big major exams at the end of your marathon!

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Brief Profile of the Tutor
• Dynamic university grad with first class honours in Electrical Engineering from NUS
• 8 years of tutoring experience including students from RI and RJC
• Singapore’s Representative for 2001 Asian Physics Olympiad
===> Honourable Mention Award
• Excellence in A level Physics Award
===> Awarded only to top 10 scorers for A level Physics in Singapore
• Award winner of IME Poster Award
• Award winner for NUS Innovation Award
• 2 times NUS Dean’s List recipient
• 2001 A levels @ RJC with 4 As and 2 S papers (maths and physics) distinction
• 1999 O levels @ RI (E Maths -- A1, A Maths -- A1, Physics -- A1)
• Current Scholar working in an engineering company
• Trainer for Xinmin Sec Junior Physics Olympiad Team 2009, 2010 and 2011!
• Trainer for CHIJ Toa Payoh Junior Physics Olympiad Team 2011 ans 2012

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A lvl H2 Maths: Permutations Combinations

From http://www.sgforums.com/forums/2297/topics/409229

a) There is a committee of 5 boys and 3 girls. At a meeting, 2 boys are absent. Find the number of ways they can sit at a round table with 8 seats.

b) A deck of 10 cards is numbered from 0 to 9. Find the number of ways 5 cards can be chosen such that the sum of the chosen cards is larger than those not chosen.

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Answer:

a) Factorial 8 different seats, but 2 are "repeated" as empty seats
Hence, the answer is

(8 - 1)! / 2! = 2520

b) There are ¹⁰C₅ ways of choosing 5 cards without restriction

The 5 cards chosen is either bigger than or lesser than the cards not chosen.

The probability that the chosen cards are greater than the "unchosen" cards is equal to the probability that the chosen cards are lesser than the "unchosen" cards

Hence we can just take ¹⁰C₅ divide by 2 = 126

A lvl H2 Maths: Vectors

Question from http://www.sgforums.com/forums/2297/topics/407847

Relative to an origin O, the position vectors of points A and B are a and b respectively. Givent that angle AOB is 90deg, show that the position vector of the foot of the perpendicular from O to AB is

a + (|a|² / (|a|² + |b|²)) (b - a)

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Answer:

Using the concept of vector projection, (i.e. AF = |AF| * unit vector AB),

OF = OA + AF = a + |a| cos θ * (b - a) / |b - a|, where θ = angle between OA and BA
= a + |a| [ {a.(a - b)} / {|a| |a - b| }] * (b - a) / |b - a|
= a + {a.a - a.b} / (|a|² + |b|²) * (b - a)
= a + |a|² / (|a|² + |b|²) * (b - a) since a.a = |a|² and a.b = 0

(shown)