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  • #797

    admin
    Member

    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

    • This topic was modified 6 years, 1 month ago by  admin.
    • This topic was modified 6 years, 1 month ago by  admin.
    • This topic was modified 6 years, 1 month ago by  admin.
    #962

    admin
    Member

    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

    • This reply was modified 6 years, 1 month ago by  admin.
    • This reply was modified 6 years, 1 month ago by  admin.
    • This reply was modified 3 years ago by  admin.
    #1015

    admin
    Member

    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.

    #1069

    admin
    Member

    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!

    #1079

    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!

    #1084

    Micole Soh
    Participant

    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

    #1107

    How do I change an equation in scalar product form into parametric form?

    #1115

    admin
    Member

    A-Level/JC Mathematics Tuition Singapore by JC Maths Tutor

    Hi Jeremy,

    If possible please post the question.

    Thanks

    #1116

    What are the possible ways to solve a simultaneous equation? (for complex numbers)

    For example:
    iz+2w=1
    4z+(3-i)w*= -6

    #1118

    admin
    Member

    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

    #1185

    Adeline Ong Min Yi
    Participant

    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

    #1231

    admin
    Member

    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

    #1232

    admin
    Member

    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

    #1265

    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

    #1267

    admin
    Member

    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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