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f(x)=2 and 1

how?

@nipunmalhotra93 @sauravshakya
post complete solution ... i'll check it later

Take it to the form a^3+b^3

|dw:1347716779899:dw|

Wll that be enough

So, f(x) can only be either 1 or 2

So the hard part is to find x

x=2

thats correct...complete it :)
can u find the exact range of f(x) ?

|dw:1347720322437:dw|

well, here's my solution.
|dw:1347734655828:dw||dw:1347734803553:dw|

|dw:1347736319687:dw|

|dw:1347736404280:dw|

so three answers in total.... :)

Fantastic work @nipunmalhotra93 :)

i think the answer is only x=2

|dw:1347769714174:dw|

I think that also works

that gives f(x)=1 ?

YEP

And is solved the equation

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

WHAT! so my solution was wrong

because the range of that function is\[[\sqrt[3]{2},2]\]

Cant it be zero as x---->infinity

Oh I got it there is a problem with my solution

So if (x)^1/2 + y^1/2 =2........i
and x^1/2 - y^1/2 =4.......ii

It has no solution........ though we get the value of x and y..... if we solve those equation

Same thing happened to my equations

where that equations come from?

Solved those two equations

Let x-1=A and x-3=B

Square first equation

|dw:1347770764883:dw|

yeah thats right...have u checked ur solutions?

I think the range of f(x) is|dw:1347775870157:dw|

thats right :)

good job saura :)

But I still have a question for u

guys @sauravshakya @nipunmalhotra93 u cooked the problem :D

yeah what ?

(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

There I solved for x and got the value

Similarly if I solved this (x)^1/2 + y^1/2 =2........i and x^1/2 - y^1/2 =4.......ii

Will I get the values of x and y

sorry can u plz elaborate ? where these 2 equations come from?

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

this one

no second

Second one just came from my mind

ahh ok

While I gave a thought when u said that x=2 only

nope there is no answer for that system

that gives\[\sqrt{y}=-1\]

Yep

yeah maybe :)

ok i really enjoyed guys....thank u

Welcome........ And thanx for such a nice question

@sauravshakya haha yea we both had the same answer :D
@mukushla I enjoyed it too man... thanks !!