Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

$p(x^2+1)={p\left(x\right)}^2+1;p\left(0\right)=0$

$p\left(1\right)=p(0{)}^2+1$

$p(1{)}^2+1=p(1^2+1)$

$=2$

for no : $p\left(n\right)=n$

$\therefore$ we can say $p\left(x\right)=x$

$p^{\prime }\left(x\right)=1$

$p^{\prime }\left(0\right)=1$