Home > MathOnFridays > MathOnFridays–Week 52 : Doubts from TS-6

MathOnFridays–Week 52 : Doubts from TS-6

We will be discussing doubts from TS-6 in MathOnFridays sessions planned for next week. Please let me know the question numbers as comments you want me to cover in these sessions.

Categories: MathOnFridays
  1. PRAKHAR GUPTA
    January 3, 2013 at 7:48 pm

    paper 2 aecg 47, 42,41

  2. Mudit
    January 3, 2013 at 9:15 pm

    paper-1 (aecg)
    Q-68,88

    paper-2(aceg)
    Q-45,56,59

  3. apaar
    January 3, 2013 at 10:13 pm

    paper-1 aecg 90,71
    paper 2- aecg 56,47,45,42

  4. Ojasvi Monga
    January 3, 2013 at 10:25 pm

    Paper-1: AECG: Q.83,84
    Paper-2: AECG: Q.55,57

  5. Bhavya Gupta
    January 4, 2013 at 2:29 am

    paper 2 ACEG Q. 59

  6. Akash Gupta
    January 4, 2013 at 7:13 am

    paper 2 : ACEG Q. 59,
    Paper 1 : ACEG Q.74,64

  7. Akash Gupta
    January 4, 2013 at 7:39 am

    Akash Gupta :
    paper 2 : ACEG Q. 59,
    Paper 1 : ACEG Q.74,64

  8. Kshitij Aggarwal
    January 4, 2013 at 9:51 am

    paper 1- AECG; Q77,71
    paper 2- ACEG; Q52,44,57

  9. Prashant
    January 4, 2013 at 10:02 am

    paper-1 aecg 64,,72,76,90,
    paper-2 aecg 42,47,49,56

  10. January 5, 2013 at 3:43 am

    paper-1 aceg 68,82,
    paper-2 aceg 44,52,59

  11. charubansal12
    January 5, 2013 at 3:45 am

    paper-1 aceg 68,82
    paper-2 aceg 44,59,52

  12. Aashi Jain
    January 7, 2013 at 4:52 am

    Paper-1 AECG 67, 83

  13. January 7, 2013 at 9:58 am

    paper 1 aceg 62. 67,74,75,

    paper 2 aecg 42,43,47,49,57,59

  14. Rishabh Chhabra
    January 7, 2013 at 11:15 am

    Paper 1 : ACEG : 74, 82
    Paper 2 : ACEG : 45, 52, 59

  15. Sagar Batra
    January 7, 2013 at 7:24 pm

    Bhaiyya please solve the following question which came in TS 4
    No one in av centre is able to solve
    Q45 paper1 aceg
    From a point R(5,8) two tangents RP and RQ are drawn to a given circle S=0 whose radius is 5. If circumcentre of the triangle PQR IS (2.3),then the equation of circle s=0 is?

    Bhaiyya i know you have limited time but one question wont matter you but it does matter me so please cover up this question. THANKS

    • Abhishek gupta
      January 8, 2013 at 7:25 am

      its easy , you have to use a graph paper!

  16. anshul jain
    January 8, 2013 at 1:22 am

    Sagar
    i hope you won’t mind if i answer ur question
    the circumcentre of triangle PQR will be the mid point of R and centre of the circle s
    because the quadrilateral formed by PQR and centre of s will be a cyclic quadrilateral
    and thus you can find the equation

  17. ayush
    January 8, 2013 at 10:13 am

    bhaiya plz take a class on questions of determinants or upload a recording as you told that you will do it in december.In the two classes of matrices there is not any question of determinant discussed by you.

  18. Anupam
    January 8, 2013 at 2:44 pm

    paper 1 aecg Q 61,63,74

  19. Anupam
    January 8, 2013 at 4:03 pm

    paper 2 aecg Q 46 ,58,59

  20. Arpit
    January 9, 2013 at 9:56 am

    Bhaiya You asked to raise question from TS Paper 1

    SET ACEG
    Q: 83
    In this ques, they have used differentiation to solve the equation
    but on integrating there is a difference of a constant, so the LHS = RHS + constant which got removed during differentiation.
    Plz check it out.

  21. Ojasvi Monga
    January 9, 2013 at 10:02 am

    Bhaiya, it was a really nice class……..
    I think this is what most of us expected as it helps in understanding the questions which we couldnt understand from the given solutions….

    And those questions which we have already understood ,well, you really make them a lot more simpler 🙂
    Wish we had such classes for other subjects as well 🙂

  22. Abhinav
    January 9, 2013 at 10:04 am

    bhaiya, i dont know about others.. but i feel that the class was good..but it was conducted very late..so i had already gone through their solutions…so there was no much to learn…
    the thrill of the class would have been doubled, or even tripled..if it was conducted earlier..
    sorry it was too much cynicism..

    • January 10, 2013 at 9:49 am

      Point very well taken. In future, I will try to keep it early.

  23. Abhishek gupta
    January 9, 2013 at 10:13 am

    bhaiya in ts-6 paper 1 class
    in ques 72 AECG (where we have to intigrate and find alpha)
    it came like ‘cube root of (1-alpha) = 64’
    then how can the value be (+- 4) i think it should only be +4
    and ans should be only 5! But that is not in options!!

    • January 10, 2013 at 9:49 am

      You are right. I did a mistake. But answer is correct. Actually there should be a modulus over the area in left hand side. Due to this modulus, we will get +/- on right hand side. Hence two answers.

  24. Lallan tiwari
    January 9, 2013 at 10:23 am

    bhaiya Plz discuss ques –
    TS-6 Paper 2 ACEG
    Q- 41,44,45,46,49,,51,54,59

  25. aastha garg
    January 11, 2013 at 4:00 am

    bhaiya, i attended the session but missed a few questions coz of internet problem. it wud be great if u upload video of this class. thanks

  26. Rohan Taneja
    January 11, 2013 at 7:03 am

    Please upload TS 6 paper 1 video. Missed it due to preboards. Thank you.

  27. January 11, 2013 at 8:07 am

    Bhaiya.. I think u must not make these videos of TS 6 Qs discussion as public….
    Instead, u can provide us with video links on Edmodo!!

  28. vibhor97
    January 11, 2013 at 8:07 am

    Bhaiya.. I think u must not make these videos of TS 6 Qs discussion as public….
    Instead, u can provide us with video links on Edmodo!!

  29. Amanraj gupta
    January 11, 2013 at 10:00 am

    bhaiya in the ques no. 43 ts 6 paper 2
    the value of k>0 so how it can satisfy K= R-{0,1}

    • January 16, 2013 at 11:52 am

      You have a point. It should be only positive values excluding 1.

  30. Arpit
    January 11, 2013 at 10:03 am

    Arpit :
    Bhaiya You asked to raise question from TS Paper 1
    SET ACEG
    Q: 83
    In this ques, they have used differentiation to solve the equation
    but on integrating there is a difference of a constant, so the LHS = RHS + constant which got removed during differentiation.
    Plz check it out.

    Bhaiya what I meant by this was that on differentiating we got b=4a.
    Now we replace b=4a in the integral, we get LHS = RHS + 1/4.
    So, solving by taking derivatives and equating gives a wrong ans.

    Plz reply…..

    • Arpit
      January 11, 2013 at 10:05 am

      And bhaiya plz don’t stop taking these classes as they’re very useful…..

  31. January 12, 2013 at 12:02 am

    bhaiya please upload the recording of week 52

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