mukushla
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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sauravshakya
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f(x)=2 and 1
mukushla
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how?
mukushla
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@nipunmalhotra93 @sauravshakya
post complete solution ... i'll check it later
sauravshakya
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Take it to the form a^3+b^3
sauravshakya
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|dw:1347716779899:dw|
sauravshakya
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Wll that be enough
sauravshakya
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So, f(x) can only be either 1 or 2
sauravshakya
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So the hard part is to find x
sauravshakya
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Was that correct @mukushla
sauravshakya
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x=2
mukushla
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thats correct...complete it :)
can u find the exact range of f(x) ?
sauravshakya
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|dw:1347720322437:dw|
nipunmalhotra93
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well, here's my solution.
|dw:1347734655828:dw||dw:1347734803553:dw|
nipunmalhotra93
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|dw:1347736319687:dw|
nipunmalhotra93
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|dw:1347736404280:dw|
nipunmalhotra93
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so three answers in total.... :)
mukushla
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Fantastic work @nipunmalhotra93 :)
mukushla
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i think the answer is only x=2
sauravshakya
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|dw:1347769714174:dw|
sauravshakya
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I think that also works
mukushla
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that gives f(x)=1 ?
sauravshakya
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YEP
sauravshakya
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And is solved the equation
sauravshakya
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WHAT! so my solution was wrong
mukushla
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because the range of that function is\[[\sqrt[3]{2},2]\]
sauravshakya
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Cant it be zero as x---->infinity
sauravshakya
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Oh I got it there is a problem with my solution
sauravshakya
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So if (x)^1/2 + y^1/2 =2........i
and x^1/2 - y^1/2 =4.......ii
sauravshakya
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It has no solution........ though we get the value of x and y..... if we solve those equation
sauravshakya
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Same thing happened to my equations
sauravshakya
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right? @mukushla
mukushla
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where that equations come from?
sauravshakya
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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
sauravshakya
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Solved those two equations
sauravshakya
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Let x-1=A and x-3=B
sauravshakya
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Square first equation
sauravshakya
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|dw:1347770764883:dw|
mukushla
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yeah thats right...have u checked ur solutions?
nipunmalhotra93
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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.
nipunmalhotra93
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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.
sauravshakya
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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
sauravshakya
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@mukushla can u PLZ CHECK it
sauravshakya
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I think the range of f(x) is|dw:1347775870157:dw|
mukushla
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thats right :)
mukushla
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good job saura :)
sauravshakya
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But I still have a question for u
mukushla
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guys @sauravshakya @nipunmalhotra93 u cooked the problem :D
mukushla
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yeah what ?
sauravshakya
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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
sauravshakya
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There I solved for x and got the value
sauravshakya
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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
sauravshakya
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Will I get the values of x and y
mukushla
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sorry can u plz elaborate ? where these 2 equations come from?
sauravshakya
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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
sauravshakya
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this one
mukushla
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no second
sauravshakya
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Second one just came from my mind
mukushla
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ahh ok
sauravshakya
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While I gave a thought when u said that x=2 only
mukushla
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nope there is no answer for that system
mukushla
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that gives\[\sqrt{y}=-1\]
sauravshakya
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Yep
sauravshakya
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So, Actually my approach was not really good............ As it may give somrtime solution when there should be no solution
mukushla
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yeah maybe :)
mukushla
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ok i really enjoyed guys....thank u
sauravshakya
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Welcome........ And thanx for such a nice question
nipunmalhotra93
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@sauravshakya haha yea we both had the same answer :D
@mukushla I enjoyed it too man... thanks !!