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sritama

  • 4 years ago

prove, f(x)=xcos1/x (when x not = 2) =0 (when x=0) f(x) is continuous at the point x=0

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  1. saubhik
    • 4 years ago
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    is your function f(x)=xcos(1/x) (when x not 0) =0 when x=0 ?

  2. sritama
    • 4 years ago
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    yes yes. i am wrong xnot 0

  3. saubhik
    • 4 years ago
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    \[\lim_{x \rightarrow 0+}xcos(1/x)=\lim_{x \rightarrow 0-}xcos(1/x)=f(0)=0\] hence continuous at x=0.

  4. saubhik
    • 4 years ago
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    if you are wondering why \[\lim_{x \rightarrow 0+}xcos(1/x)=0\] then note that \[-1\leq \cos(1/x)\leq 1\ for\ all\ x.\] So \[\lim_{x \rightarrow 0}xcos(1/x)=\lim_{x \rightarrow 0}x\cdot a=a \lim_{x \rightarrow 0}x=0\] here a is any real value between -1 and 1.

  5. sritama
    • 4 years ago
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    thanks

  6. saubhik
    • 4 years ago
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    in which class u are?

  7. sritama
    • 4 years ago
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    class 11.u?

  8. saubhik
    • 4 years ago
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    me in college 1st year:) which city u?

  9. sritama
    • 4 years ago
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    kolkata

  10. saubhik
    • 4 years ago
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    whic school..i am in kolkata

  11. sritama
    • 4 years ago
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    khidirpur st.thomas

  12. saubhik
    • 4 years ago
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    cbse? if u need help regarding math problems u can ask me! i can suggest a great math book for classes 11 and 12!

  13. sritama
    • 4 years ago
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    oh , say the name.can u suggest some books for mathematical olympiad?

  14. sritama
    • 4 years ago
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    your college?

  15. saubhik
    • 4 years ago
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    yeah! I am in Jadavpur University (Mech Engg) for school to IIT level problems: Problems Plus In IIT Mathematics by Asit Dasgupta

  16. saubhik
    • 4 years ago
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    the book by asit dasgupta is great for the basics but for RMO,INMO u need to try some other books...

  17. sritama
    • 4 years ago
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    did u appear for RMO/INMO any time.

  18. saubhik
    • 4 years ago
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    yes one time in class 11:was not well prepared:( didnt get through...u need tough preparations for RMO...however if u r a genius u dont need much preparataions!

  19. sritama
    • 4 years ago
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    are u good in probability?it is tough to me.

  20. saubhik
    • 4 years ago
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    u can try me. :)

  21. sritama
    • 4 years ago
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    i have no hard problem at present.when i found a very hard problem,i would ask u .

  22. saubhik
    • 4 years ago
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    u can call me (if u want my cell no. u can ask me now) or u can mail me to saubhik.mukherjee@gmail.com...i dont come to openstudy often..so u can take my phone no. or gmail

  23. sritama
    • 4 years ago
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    ok.i will mail u.however do u have interest in astrophysics.it is my most fav.

  24. saubhik
    • 4 years ago
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    about me: i got 100 in maths in class 12 and 100 in class 10 CBSE. I love olympiad mathematics and want to help anyone interested in olympiad maths. about astrophysics:there is an olympiad for that and I dont know anything about it :)

  25. sritama
    • 4 years ago
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    wow.u r so brilliant!

  26. saubhik
    • 4 years ago
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    thanks.hope to help u soon :)

  27. sritama
    • 4 years ago
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    in triangle ABC,AB=AC,A is right angle.M & N are 2points on BC satisfying BM^2+CN^2=MN^2. prove that angle MAN=45.

  28. saubhik
    • 4 years ago
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    I proved it using coordinate geometry. Did not get a good pure geometry solution though. I can show you the proof. You can post this as a separate question.

  29. sritama
    • 4 years ago
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    show

  30. saubhik
    • 4 years ago
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    wait...i will attach a picture wait 2 min.

  31. saubhik
    • 4 years ago
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    see

    1 Attachment
  32. saubhik
    • 4 years ago
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    did u see the picture? The side AB is on y-axis and AC is on x-axis...

  33. sritama
    • 4 years ago
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    yes i did and understand.thanks a lot !!!but whenever u get pure geometry solution please post,30 MEDALS for u.

  34. saubhik
    • 4 years ago
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    i did not explain anything..u got it? u have to use the distance formula to obtain a relation and then use this relation in another equation which u will get by considering the angle between the two lines AM and AN, which u can get by considering the slopes of the two lines AM and AN. did u get this angle as 45 degrees?

  35. saubhik
    • 4 years ago
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    what's your e-mail id?

  36. sritama
    • 4 years ago
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    yes my elder sister helped me

  37. sritama
    • 4 years ago
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    it is chowdhurysaugata6@gmail.com ok

  38. saubhik
    • 4 years ago
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    ok...any more problems u can always ask me:)

  39. sritama
    • 4 years ago
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    i will.buy

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