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man111

(1) let P(x) be polynomials such that P(x²+2)=x¹⁷-3x⁵+x³-3 than

P(x) = ?

(2) P(x)=a₀+a₁x+a₂x²............a_nxⁿ, a_n # 0 be such that P(x²)=(P(x))^2. than value of P(x) =

phyana

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

phyana

are yar............i think the only solution to ur 2nd question is

a0=1
a1=a2=a3=a3=...........=an=0

i got it mathematically......

animal

hey phayana man it is clearly given inthe ques that a_n#0.

animal

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+.......+a_n)=(a₀+......+a_n)²

which again gives (a₀+ .......+a_n)=0 or (a₀+.........+a_n)=1.

spideyunlimited

(2) P(x)=a₀+a₁x+a₂x²............a_nxⁿ, a_n # 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

hsbhatt

spideyunlimited

ok sir i understood.. but where did i go wrong? :(

anchitsaini

excellent sir :)

anandghegde

same prob posted by the legendary rushil Mr. Green

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42881 Smile

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