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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 May 2008 14:28:34 IST
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(1) let P(x) be polynomials such that P(x2+2)=x17-3x5+x3-3 than
P(x) = ?
(2) P(x)=a0+a1x+a2x2............anx n, an # 0 be such that P(x2)=(P(x))2. than value of P(x) =
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 May 2008 22:09:21 IST
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first question..........
we know p(x^2 +2)
put x^2+2 =y
so x=(y-2)^.5
now subtstitute for x in the given equation on both rhs and LHS.....
thus u have p(y) = so and so answer............
now replace y by x in the entire equation..
thus u get the answer for p(x)............
i think u can do the other question similarly......
p(x)=a0 +a1.x^2 +...........+an.x^2
so p(x^2)=a0 + a1.x^4 +....................+an.x^2n
since p(x^2)=(p(x) )^2
p(x) = (p(x^2) )^.5
thus expand it u will get the answer.............
if satisfied plz rate me
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 May 2008 22:25:09 IST
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are yar............i think the only solution to ur 2nd question is
a0=1 a1=a2=a3=a3=...........=an=0
i got it mathematically......
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 May 2008 11:05:24 IST
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hey phayana man it is clearly given inthe ques that an#0.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 May 2008 11:27:06 IST
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2nd ques
from where u get such ques man?
Tell me if I m going inthe right dir.
As I see p(x2)=(p(x))2 is possible only when x=0 and equating we get
a0 =a02 which gives a0=0 or a0=1.
it is also possible when x=1and on equating we get
(a0+a1+.......+an)=(a0+......+an)2
which again gives (a0+ .......+an)=0 or (a0+.........+an)=1.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 May 2008 11:31:52 IST
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(2) P(x)=a0+a1x+a2x2............anx n, an # 0 be such that P(x2)=(P(x))2. than value of P(x) =
P(12) = [ P(1) ]2
Let n = 1
a0 = a02
a0 = 1
LEt n = 2
a0 + a1 = a02 + a12 + 2.a0.a1
1 + a1 = 1 + a12 + 2a1
-a1 = a12
a1 = 0, -1
as an # 0 ... so only solution is a1 = -1
so P(1) for n = 2 is 1 - 1 = 0
Similarly by solving ahead, we observe that:
P(x) = 0
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---------------------------------------------------------------
* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 May 2008 19:23:26 IST
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 May 2008 20:27:42 IST
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ok sir i understood.. but where did i go wrong? :(
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---------------------------------------------------------------
* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
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