Logo
    • Education
      • Pre-School
      • Primary Schools Directory
      • Primary Schools Articles
      • P1 Registration
      • DSA
      • PSLE
      • Secondary
      • Tertiary
      • Special Needs
    • Lifestyle
      • Well-being
    • Activities
      • Events
    • Enrichment & Services
      • Find A Service Provider
      • Enrichment Articles
      • Enrichment Services
      • Tuition Centre/Private Tutor
      • Infant Care/ Childcare / Student Care Centre
      • Kindergarten/Preschool
      • Private Institutions and International Schools
      • Special Needs
      • Indoor & Outdoor Playgrounds
      • Paediatrics
      • Neonatal Care
    • Forum
    • ASKQ
    • Register
    • Login

    MathQA tutor - Ask your A-level Maths questions here!

    Scheduled Pinned Locked Moved Tertiary Education - A-Levels, Diplomas, Degrees
    32 Posts 8 Posters 15.5k Views 1 Watching
    Loading More Posts
    • Oldest to Newest
    • Newest to Oldest
    • Most Votes
    Reply
    • Reply as topic
    Log in to reply
    This topic has been deleted. Only users with topic management privileges can see it.
    • M Offline
      mathqa
      last edited by

      OK Lor:
      Hi,


      Please evaluate
      http://www.postimage.org/

      Thanks.
      Let u=tan(x). Rewrite it in term of u, it would become an integral of an elementary function in form of

      http://lh6.ggpht.com/_nr85VD4DdiA/TPjKaT7U5HI/AAAAAAAAAE0/jSSE2KNWb9E/s800/elementary-function-integral.png\">

      1 Reply Last reply Reply Quote 0
      • O Offline
        OK Lor
        last edited by

        mathqa:
        OK Lor:

        Hi,


        Please evaluate
        http://www.postimage.org/

        Thanks.

        Let u=tan(x). Rewrite it in term of u, it would become an integral of an elementary function in form of

        http://lh6.ggpht.com/_nr85VD4DdiA/TPjKaT7U5HI/AAAAAAAAAE0/jSSE2KNWb9E/s800/elementary-function-integral.png\">

        Hi mathqa, thanks 😄 .

        1 Reply Last reply Reply Quote 0
        • O Offline
          OK Lor
          last edited by

          Hi,

          Pls help on (ii):
          A curve has parametric equation x = 2t – 1, y = 1/(t² + 1).
          (i) Prove that the equation of the tangent at the point with parameter t is (t² + 1)² y + tx = 3t² - t + 1.
          (ii) The tangent at point where t = 3 meets the curve again at the point where t = q. Find the value of q.
          Ans: -4/3

          Thanks.

          1 Reply Last reply Reply Quote 0
          • F Offline
            FrekiWang
            last edited by

            OK Lor:
            Hi,

            Pls help on (ii):
            A curve has parametric equation x = 2t – 1, y = 1/(t² + 1).
            (i) Prove that the equation of the tangent at the point with parameter t is (t² + 1)² y + tx = 3t² - t + 1.
            (ii) The tangent at point where t = 3 meets the curve again at the point where t = q. Find the value of q.
            Ans: -4/3

            Thanks.
            Assuming (i) has been proven.

            tangent at t=3 has equation
            [(3^2+1)^2]y+3x=3*3^2-3+1
            simplified, we have 100y+3x=25

            to find the point of intersection between the curve and the straight line, we need to solve
            100y+3x=25 and x=2t-1, y=1/(t^2+1)

            By subsititution,
            100/(t^2+1)+3(2t-1)=25
            100+(6t-3)(t^2+1)=25(t^2+1)
            100+6t^3-3t^2+6t-3=25t^2+25
            6t^3-28t^2+6t+72=0
            3t^3-14t^2+3t+36=0

            since we know t=3 is a repeated root as a tangent, (t-3)^2 must be a factor, (or t^2-6t+9 must be a factor)

            after doing a long division, we have
            (3t+4)(t-3)^2=0

            the other root is t=-4/3 (when 6t+8=0)

            1 Reply Last reply Reply Quote 0
            • E Offline
              equink
              last edited by

              I’ve been trying to figure this out but I still don’t understand.

              Why is dx/dy proven to be 1/(dy/dx)? I know that dx/dy cannot be treated as a fraction because it simply isn’t a fraction. Please help,thanks so much!

              1 Reply Last reply Reply Quote 0
              • A Offline
                albertong
                last edited by

                [quote]
                equink posted: I've been trying to figure this out but I still don't understand.
                Why is dx/dy proven to be 1/(dy/dx)? I know that dx/dy cannot be treated as a fraction because it simply isn't a fraction. Please help,thanks so much![/quote]@mathstuition88 answered:

                Good question! A rigorous proof of that would require analysis (a math course in university), but one can prove it briefly using chain rule,

                (dy/dx)(dx/dy)=dy/dy=1, hence dx/dy=1/(dy/dx).

                Hope this helps!

                Post and get your questions answered here: http://fbl.me/Tt

                1 Reply Last reply Reply Quote 0
                • S Offline
                  Studentx
                  last edited by

                  deleted.

                  1 Reply Last reply Reply Quote 0
                  • S Offline
                    student101
                    last edited by

                    Hi , Any expert in perm&comb in H2 Maths can explain to me clearly how the below questions red circled highlighted parts can get. I did not figure out how that step get . Please help

                    http://i59.tinypic.com/vhe2h2.jpg\">

                    1 Reply Last reply Reply Quote 0
                    • G Offline
                      gjooheng
                      last edited by

                      exactly one yellow tile and 2 blue tiles in the same row can happen in the 4th or 3rd row, or both rows.


                      So using the venn diagram property n(AUB) = n(A)+n(B)-n(AnB)
                      The circled part is n(AnB)

                      http://i59.tinypic.com/2uiacsk.jpg\">

                      1 Reply Last reply Reply Quote 0
                      • S Offline
                        student101
                        last edited by

                        gjooheng:
                        exactly one yellow tile and 2 blue tiles in the same row can happen in the 4th or 3rd row, or both rows.


                        So using the venn diagram property n(AUB) = n(A)+n(B)-n(AnB)
                        The circled part is n(AnB)

                        http://i59.tinypic.com/2uiacsk.jpg\">
                        Thanks sir for your answer. Easy to understand with your diagram. Great!!

                        1 Reply Last reply Reply Quote 0

                        Hello! It looks like you're interested in this conversation, but you don't have an account yet.

                        Getting fed up of having to scroll through the same posts each visit? When you register for an account, you'll always come back to exactly where you were before, and choose to be notified of new replies (either via email, or push notification). You'll also be able to save bookmarks and upvote posts to show your appreciation to other community members.

                        With your input, this post could be even better 💗

                        Register Login
                        • 1
                        • 2
                        • 3
                        • 4
                        • 4 / 4
                        • First post
                          Last post



                        Online Users

                        Statistics

                        2

                        Online

                        210.7k

                        Users

                        34.2k

                        Topics

                        1.8m

                        Posts
                        Popular Topics
                        New to the KiasuParents forum? Tips and Tricks!
                        Choosing and Evaluating Primary Schools
                        DSA 2026
                        PSLE Discussions and Strategies
                        How much do you spend on the kids' tuition/enrichments?
                        SkillsFuture + anything related to upskilling/learning something new!

                          About Us Contact Us forum Terms of Service Privacy Policy