Question

Let $p\left(x\right)$ be a polynomial satisfying $p\left(x^3\right)=x^{2}p\left(x^2\right)$ and $p\left(0\right)=-3,$ $p^{\prime }\left(\frac{1}{3}\right)=0,$ $p\left(1\right)=2.$ Then the value of $p\left(3\right)$ is

• A. $101$
• B. $102$
• C. $103$
• D. $104$

Answer 1

The correct option is B $102$

Given : $p\left(x^3\right)=x^{2}p\left(x^2\right)$ and $p\left(0\right)=-3,p^{\prime }\left(\frac{1}{3}\right)=0,p\left(1\right)=2$
Let $p\left(x\right)$ be a polynomial of degree $n.$
$p\left(x^3\right)=x^{2}p\left(x^2\right)\Rightarrow 3n=2+2n$
$\Rightarrow n=2$
So, assuming $p\left(x\right)=a{x}^2+bx+c$
Using $p\left(0\right)=-3,$ we get
$c=-3$
Using $p^{\prime }\left(\frac{1}{3}\right)=0,$
$p^{\prime }\left(x\right)=2ax+b\Rightarrow \frac{2a}{3}+b=0\cdots \left(1\right)$
Using $p\left(1\right)=2,$
$a+b-3=2$
$\Rightarrow a+b=5\cdots \left(2\right)$
Solving equations $\left(1\right)$ and $\left(2\right),$ we get
$a=15$ and $b=-10$
$\Rightarrow p\left(x\right)=15x^2-10x-3\therefore p\left(3\right)=102$