Q. If $P(x)$ is a polynomial such that $P\left( x^{2} + 1 \right) = \left\{ P(x) \right\}^{2} + 1$ and $P(0) = 0$, then $P^{'}(0)$ is equal to
A) $- 1$
B) 0
C) 1
D) None of these
Explanation
(c)
Polynomial $P(x)$, satisfying the given relation can be taken as $x$
$ie,\ P(x) = x$
$\therefore P^{'}(x) = 1$
$\Rightarrow P^{'}(0) = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.