Question

If P(x) is a polynomial such that $P(x^2+1)={\left\{P\left(x^2\right)\right\}}^2+1$ and $P\left(0\right)=0$ then $P^{{}^{\prime }}\left(0\right)$ is equal to

• A. $1$
• B. $0$
• C. $-$1
• D. none of these

Answer 1

The correct option is A $1$

Given, $P(x^2+1)={\left\{P\left(x^2\right)\right\}}^2+1$ ...(1)
Given $P\left(0\right)=0$
Put $x=0$ in (1)
$\Rightarrow P\left(1\right)={P\left(0\right)}^2+1$
$\Rightarrow P\left(1\right)=1$
Also, put $x=1$ in (1)
$\Rightarrow P\left(2\right)=2$
Also, since, P(x) is a polynomial so we have P(x)=x
$\Rightarrow P^{\prime }\left(x\right)=1$
$P^{\prime }\left(0\right)=1$