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mukushla

  • 3 years ago

ok this is mine \[f(x)=\sqrt[3]{x-1}+\sqrt[3]{3-x}\]find all \(x \in \mathbb{R}\) such that \(f(x) \in \mathbb{N}\)

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  1. sauravshakya
    • 3 years ago
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    f(x)=2 and 1

  2. mukushla
    • 3 years ago
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    how?

  3. mukushla
    • 3 years ago
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    @nipunmalhotra93 @sauravshakya post complete solution ... i'll check it later

  4. sauravshakya
    • 3 years ago
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    Take it to the form a^3+b^3

  5. sauravshakya
    • 3 years ago
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    |dw:1347716779899:dw|

  6. sauravshakya
    • 3 years ago
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    Wll that be enough

  7. sauravshakya
    • 3 years ago
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    So, f(x) can only be either 1 or 2

  8. sauravshakya
    • 3 years ago
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    So the hard part is to find x

  9. sauravshakya
    • 3 years ago
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    Was that correct @mukushla

  10. sauravshakya
    • 3 years ago
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    x=2

  11. mukushla
    • 3 years ago
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    thats correct...complete it :) can u find the exact range of f(x) ?

  12. sauravshakya
    • 3 years ago
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    |dw:1347720322437:dw|

  13. nipunmalhotra93
    • 3 years ago
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    well, here's my solution. |dw:1347734655828:dw||dw:1347734803553:dw|

  14. nipunmalhotra93
    • 3 years ago
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    |dw:1347736319687:dw|

  15. nipunmalhotra93
    • 3 years ago
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    |dw:1347736404280:dw|

  16. nipunmalhotra93
    • 3 years ago
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    so three answers in total.... :)

  17. mukushla
    • 3 years ago
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    Fantastic work @nipunmalhotra93 :)

  18. mukushla
    • 3 years ago
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    i think the answer is only x=2

  19. sauravshakya
    • 3 years ago
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    |dw:1347769714174:dw|

  20. sauravshakya
    • 3 years ago
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    I think that also works

  21. mukushla
    • 3 years ago
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    that gives f(x)=1 ?

  22. sauravshakya
    • 3 years ago
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    YEP

  23. sauravshakya
    • 3 years ago
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    And is solved the equation

  24. mukushla
    • 3 years ago
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    emm... http://www.wolframalpha.com/input/?i=%5Csqrt%5B3%5D%7Bx-1%7D%2B%5Csqrt%5B3%5D%7B3-x%7D%3D1

  25. sauravshakya
    • 3 years ago
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    WHAT! so my solution was wrong

  26. mukushla
    • 3 years ago
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    because the range of that function is\[[\sqrt[3]{2},2]\]

  27. sauravshakya
    • 3 years ago
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    Cant it be zero as x---->infinity

  28. sauravshakya
    • 3 years ago
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    Oh I got it there is a problem with my solution

  29. sauravshakya
    • 3 years ago
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    So if (x)^1/2 + y^1/2 =2........i and x^1/2 - y^1/2 =4.......ii

  30. sauravshakya
    • 3 years ago
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    It has no solution........ though we get the value of x and y..... if we solve those equation

  31. sauravshakya
    • 3 years ago
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    Same thing happened to my equations

  32. sauravshakya
    • 3 years ago
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    right? @mukushla

  33. mukushla
    • 3 years ago
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    where that equations come from?

  34. sauravshakya
    • 3 years ago
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    Actually the equations I got was: (x-1)^1/3 - (x-3)^1/3 =1...........i (x-1)^2/3 +{(x-1)(x-3)}^1/3 + (x-3)^2/3=2........ii

  35. sauravshakya
    • 3 years ago
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    Solved those two equations

  36. sauravshakya
    • 3 years ago
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    Let x-1=A and x-3=B

  37. sauravshakya
    • 3 years ago
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    Square first equation

  38. sauravshakya
    • 3 years ago
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    |dw:1347770764883:dw|

  39. nipunmalhotra93
    • 3 years ago
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    @mukushla actually I think that wolfram uses the domain of \[\sqrt[3]{x}\]to be only non negative real nos. I have noticed that it gives complex answers when we try to plug in negative reals in the cube root function http://www.wolframalpha.com/input/?i=%5Csqrt%5B3%5D%7B-1%7D

  40. mukushla
    • 3 years ago
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    yeah thats right...have u checked ur solutions?

  41. nipunmalhotra93
    • 3 years ago
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    yup! when I plug in x=.98165, calculate cube root of 1-x and then put a minus sign with it. That works fine.

  42. nipunmalhotra93
    • 3 years ago
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    similarly, when I put x=3.0183 , I calculate x-3 and put minus sig with the cube root of x-3 to get \[\sqrt[3]{3-x}\] That's the only possible way to check it I think.

  43. sauravshakya
    • 3 years ago
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    @nipunmalhotra93 Actually u and I got the same answers....|dw:1347775568533:dw| But And the substituted the value |dw:1347775753579:dw| in f(x) I got 1

  44. sauravshakya
    • 3 years ago
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    @mukushla can u PLZ CHECK it

  45. sauravshakya
    • 3 years ago
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    I think the range of f(x) is|dw:1347775870157:dw|

  46. mukushla
    • 3 years ago
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    thats right :)

  47. mukushla
    • 3 years ago
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    good job saura :)

  48. sauravshakya
    • 3 years ago
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    But I still have a question for u

  49. mukushla
    • 3 years ago
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    guys @sauravshakya @nipunmalhotra93 u cooked the problem :D

  50. mukushla
    • 3 years ago
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    yeah what ?

  51. sauravshakya
    • 3 years ago
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    (x-1)^1/3 - (x-3)^1/3 =1...........i (x-1)^2/3 +{(x-1)(x-3)}^1/3 + (x-3)^2/3=2........ii

  52. sauravshakya
    • 3 years ago
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    There I solved for x and got the value

  53. sauravshakya
    • 3 years ago
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    Similarly if I solved this (x)^1/2 + y^1/2 =2........i and x^1/2 - y^1/2 =4.......ii

  54. sauravshakya
    • 3 years ago
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    Will I get the values of x and y

  55. mukushla
    • 3 years ago
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    sorry can u plz elaborate ? where these 2 equations come from?

  56. sauravshakya
    • 3 years ago
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    Which one: (x-1)^1/3 - (x-3)^1/3 =1...........i (x-1)^2/3 +{(x-1)(x-3)}^1/3 + (x-3)^2/3=2........ii

  57. sauravshakya
    • 3 years ago
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    this one

  58. mukushla
    • 3 years ago
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    no second

  59. sauravshakya
    • 3 years ago
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    Second one just came from my mind

  60. mukushla
    • 3 years ago
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    ahh ok

  61. sauravshakya
    • 3 years ago
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    While I gave a thought when u said that x=2 only

  62. mukushla
    • 3 years ago
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    nope there is no answer for that system

  63. mukushla
    • 3 years ago
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    that gives\[\sqrt{y}=-1\]

  64. sauravshakya
    • 3 years ago
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    Yep

  65. sauravshakya
    • 3 years ago
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    So, Actually my approach was not really good............ As it may give somrtime solution when there should be no solution

  66. mukushla
    • 3 years ago
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    yeah maybe :)

  67. mukushla
    • 3 years ago
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    ok i really enjoyed guys....thank u

  68. sauravshakya
    • 3 years ago
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    Welcome........ And thanx for such a nice question

  69. nipunmalhotra93
    • 3 years ago
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    @sauravshakya haha yea we both had the same answer :D @mukushla I enjoyed it too man... thanks !!

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