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Differential Calculus

Cool goIITian

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27 Mar 2009 09:33:00 IST

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{x+1}+2x=4[x+1]-6

None

{x+1}+2x=4[x+1]-6

{}=fractional part..[]=integral part...pls solvve..rates assured.

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saharsha kumar keshkar

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Joined: 7 Aug 2008

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27 Mar 2009 09:44:36 IST

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as we know...{x}=x-[x]

(x+1)-[x+1]+2x=4[x+1]-6

3x+1=5[x+1]-6

3x+1=5([x]+1)-6

3([x]+{x})=5[x]-2

3{x}=2[x]-2................(1)

as we know..0<={x}<1

0<=3{x}<3........................(2)

on substitutin...(1)

we get..

1<=[x]<5/2...so we conclude that...[x]=1 or 2.

hence from (1)..3{x}=2*1-2=0

or {x}=0 or 2/3.

x={x}+[x]...1+0 or 2+2/3 i.e

1,8/3 are solutions.....

nudge me if i m wrong....

ar rehman

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Joined: 24 Feb 2009

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27 Mar 2009 09:56:41 IST

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i think answer is correct....pls some one confirm the answer....

saharsha kumar keshkar

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Posts: 1104

27 Mar 2009 09:59:38 IST

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mr. rehman i think i clearly mention each and every step....kindly allow me any doubt abt any step...

ram kumar

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Joined: 15 Mar 2009

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27 Mar 2009 10:28:34 IST

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@rehman answer given by @saharsha...is correct....wats ur doubt just mention...

akhil

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27 Mar 2009 12:46:50 IST

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answer given by saharsha is correct..........

ar rehman

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Posts: 119

27 Mar 2009 16:28:47 IST

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sorry!

i was bit confused.......answer is correct....sorry saharsha...

sagar panda

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Joined: 5 Mar 2009

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31 Mar 2009 10:58:44 IST

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nice question.........

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