Question

The number of polynomials $p\left(x\right)$ satisfying $\left(p\right(x){)}^2=1+xp(x+1)$ for all real values of $x$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is B $1$

$\left(p\right(x){)}^2=1+xp(x+1)\cdots (1)$
Let degree of $p\left(x\right)$ be $n.$
So, degree of $\left(p\right(x){)}^2$ is $2n$ and degree of $1+xp(x+1)$ is $n+1.$
$\Rightarrow 2n=n+1$
$\Rightarrow n=1$

So, $p\left(x\right)=ax+b$
Putting $p\left(x\right)=ax+b$ in $\left(1\right),$ we get
$(a^2-a)x^2+(2ab-a-b)x+(b^2-1)=0$
Since, the equation holds true for all real $x,$
$a^2-a=0,2ab-a-b=0,b^2-1=0$
$\Rightarrow a=b=1$
$\therefore p\left(x\right)=x+1$