1i think the answer is (c)0
If P(x) be a polynomial such that P(x2+1) = {P(x)}2+1 and P(0) = 0 then P'(0) is equal to :
(a) 1 (b) -1 (c) 0 (d) none
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7 Answers
1By observation , I think " p ( x ) = x " is the reqd. function . So the answer would be " ( a ) " .
And Sagnik , " p ( x ) ≠x 2 + 1 " since " p ( 0 ) " is given to be " 0 " .
Jai , you made a mistake in differentiating in the first step only .
1Hey frnds..... correct ans is (a) 1
nd da msg is 4 nihal....
u r diff both sides.....
so i think u r wrong.....
u 've nt diff P(x) w.r.t. in RHS....
it will be...
P'(x2+1).2x = 2.P(x). P'(x)
am i correct!!!!
1I HAD DIFFERENTIATED SAME....
P'(x2+1).2x = 2.P(x). P'(x)
P'(X)=P'(x2+1).2x2.P(x)
PUT X=0
WE GET 00 FORM
applying l hospital rule
P'(X)= P"(x2+1).4x2.P'(x
[P'(X)]2=P"(x2+1).4x2
PUT X=0
[P'(0)]2={P"(1).0}/2
[P'(0)]2=0/2=0
SO, P'(0)=0
341Let {an} be the sequence defined by a_0=1; a_n = a_{n-1}^2+1 for n>0.
It is easy to see that P(a_n) = a_n \ \forall n \in \mathbb{Z} \ge 0
But, since P is of finite degree, we must have that P(x) is identically to x.
Hence P(x) = x and the rest is simple