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A-Level Tuition Singapore/A-Level Mathematics Tutition/H2 Mathematics Tuition/JC Mathematics Tuition/Mathematics Tutor
Please post your A-Level Mathematics/JC Mathematics questions here
Thank you
A-Level Tuition Singapore/A-Level Maths Tutition/H2 Maths Tuition/JC Maths Tuition/Maths Tutor
Hi JC Mathematics/A-level Mathematics students,
We intend to complete the A-level Mathematics syllabus by end May and in the month of June we will practise 2011 prelim questions from top 5 JCs.
Wishing you and your family a Blessed and Joyful Chinese New Year
JC/H2 Maths Tuition Singapore/A-Level Mathematics Tutition/H2 Math Tuition/JC Math Tuition/Math Tutor
Mathematics Tuition Singapore/A-Level Mathematics
Learning outcomes from Permutations and Combinations
1. Principles of Counting
Addition Principles – eg used to add up the various cases that lead to the same result.
Multiplication Principles – eg used to multiple up the various cases that lead to the same result.
2. Permutations is an ordered arrangements of objects.
Arrange n distinct objects in a row = n!
Arrange n distinct objects in a row which m of them are of the same kind – n!/m!
Arrange n distinct objects in a circle = n!/n = (n-1)!
Example : Arrange 8 people sitting in a round table = 8!/8 = 7!
If the chairs are number, nos of arrangement = 7!x8
3. Combination is a selection of objects in which the order or arrangements of the objects is NOT IMPORTANT.
Select r objects without replacement from n distinct objects = nCr
Technique 1
Considering cases and use of Permutations and Combinations Concurrently
Find the number of ways of forming 3-letter code words from the letters in the word SHOOT
Case 1: There is no O in the code word
Number of ways = 3! = 6
Case 2: There is one O in the code word
Number of ways = 1 (one O) x 3C2(choose 2 letters from S,H.T) x 3! (permutation of 3 letters) = 18
Case 3: There are two O in the code word
Number of way = 1 (two O) x 3C1 (choose 1 letter from S,H,T) x 3!/2!
(permutation of 3 letters with two O identical) = 9
Total = 6+18+9 = 33 (Additional Principle)
Try this !
Find the number of way of forming 4-letter code words in the word CUMUMBERS
More techniques on the WAY….
Techique 2 – Handle restrictions first ex fixed position or group together
Technique 3 – Slotting Method
Technique 4 – Use complement method if direct method is too tedious.
Technique 5 – For arrangement in circles with seat numbered, always handle the restrictionby first and lastly multiply the answer by the number of seats.
A-Level Mathematics Tuition Singapore/H2 Mathematics Tuition/JC Mathematics Tuition/Mathematics Tutor
Probability : Learning outcomes
1. P(A) = (number of successes)/(number of trial)
= p
= probability of success
2. P(A or B or both)
= P(A u B)
= P(A)+P(B)-P(A n B)
3. Conditional Probability of A occuring given B has occured
= P(A/B)
= P(A n B)/P(B)
4. Two events A and B are mutually exclusive if they cannot occur simultaneously. ie P(A n B) = 0
P(A u B) = P(A) + P(B)
5. An event A is said to be independent of an event B if the progability that A occur is not affected by the occurrance of B.
P(A n B) = P(A) x P(B)
P(A/B) = P(A)
P(B/A) = P(B)
Note that point (4) and (5) are tests to check if the 2 events A and B ae mutually exclusive or independent respectively.
6. Problem Solving Techiaues in Probability
A. Sample Space Diagrams or Tables
Commonly used when the sample space is finite and all outcomes can be listed out easily. This techniques is useful for questions involving outcomes of 2 tossed dice and so on.
B. Venn Diagrams
Commonly used when questions involve sets
C. Tree Diagram
Commonly used when the problem involves one experiment repeated many times or a sequence of processess.
D. Selection (Without Replacement) and Arrangements
Commonly used when the questions have some “element” of selection and /or arrangements
P(such arrangement) = (Number of such arrangements)/( Total number of arrangements)
P(such selection) = (Number of scush selections)/(Total number of of selections)
Note that this method is NOT APPLICABLE when questions involve selections WITH REPLACEMENT AND/OR ARRANGEMENTS
E. Reduced sample space for finding conditional probabilities if the outcome can be counted.
Practise questions will be given during the lesson.
Cheers!
A Level Maths Tuition/H2 Maths Tuition/JC Maths Tuition/ Maths Tutor
How to integrate 1/(1+cosx) ?
= 1/(1+2Cos sq (x/2) -1)
= 1/(2Cos sq (x/2))
= 1/2 Sec sq (x/2)
After integrate
= tan(x/2) + C
Cheers!
A Level Mathematics Tuition Singapore/H2 Maths Tuition/JC Maths Tutor
How do you do this P&C question: N85/I/3 (about the placing of 4 objects)?
i) You need to consider cases in this part of question
Case 1 : a is in the right box A and the rest wrong
Nos of ways = 1 (a in box A) x 2 (c or d in box B) x 1 (d in box C) x 1 (b in box D) = 2
Case 2 : b is in the right box B and the rest wrong
Nos of ways = 2
Case 3 : c is in the right box C and the rest wrong
Nos of ways = 2
Case 4 : d is in the right box D and the rest wrong
Nos of ways = 2
Total nos way = 8
How do I change an equation in scalar product form into parametric form?
A-Level/JC Mathematics Tuition Singapore by JC Maths Tutor
Hi Jeremy,
If possible please post the question.
Thanks
What are the possible ways to solve a simultaneous equation? (for complex numbers)
For example:
iz+2w=1
4z+(3-i)w*= -6
A-Level Mathematics Tuition Singapore/JC Maths/H2 Math Tuition and Tutor
Treat z and w as a variable,
From eqn (1) make z = (1-2w)/i
Sub this into the second eqn.
let w = x+iy and conj w = x-iy
Solve the eqn as usual
From JC Maths Tutor
How do you do these questions?
Integration
2010 IJC Prelim/P1/11 (a)(ii),(c)
2010 JJC Prelim/P1/11 (ii),(b)
2010 MJC Prelim/P2/5 (ii)
Solutions posted in the Math Resource
A-Level Mathematics Tuition Singapore/JC Maths/H2 Math Tuition and Tutor
Hi Wei Kiat
Q8 solution
To show coefficient
C3 = (−1)(−1) + (−2)(1) + 3(4) =11
Cr = (-1)(-1)^r + (-2)(1)+3(r+1)
= (-1)^r+1 +3r + 1
C3 =11, Cr = (-1)^r+1 +3r + 1
From A-Level Tutor Singapore
A-Level Mathematics Tuition Singapore/JC Maths/H2 Math Tuition and Tutor
Hi Valerie
Q1 soln
iw+2z = 3+4i ——-(1)
i^3w+(1-i)z = 2-i
-iw+(1-i)z = 2-i ——(2)
(1)+(2)
2z+(1-i)z = 5+3i
(3-i)z = 5+3i
z = (5+3i)/(3-i)
using GC
z = 6/5 +7/5i
sub z into equation (1) u can get w
z1 – z2 = 3i
(1+ai) – (a+bi) = 3i
1+a +(a-b)i = 3i
Equating the real and img nos
1+a = 0 therefore a = -1
a-b = 3
-1-b = 3
b = -4
To prove 1/z1 + 1/z2 = (7-i)/10
LHS 1/(1-i)+ 1/(-1-4i)
= RHS
(z1^2 – z2^2)/(Z1Z2)
= (z1-z2)(z1+z2)/(z1z2)
= (3i)(7-i)/10
= (-3+21i)10
From A level tutor
How to solve for e^6=-64? (Complex Numbers)
Should it be Z^6 = 64
= 64e^2kpie
Therefore z = 2e^(kpie/3) k=0, +-1, +-2, 3
From A-level Mathematics Tuition Tutor Singapore
A-Level Mathematics Tuition Singapore/JC Maths/H2 Math Tuition and Tutor
Hi Wei Kiat
Q11 Solution
Method 1: Graphical
Sketch any appropriate graph(s), with equation(s) stated
State intersection points and/or equation of vertical asymptote.
x root 5 (2.236)
From A level tutor
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