Differentiation 4

i think the answer is (c)0

P'(x²+1).2x = 2.P'(x)

P'(x² +1) .x = P'(x)

at x=0

P'(1).0 =P'(0)

P'(0)=0

By 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 .

oh yes.sorry!

Hey 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!!!!

I 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

Let {a_n} 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