Let $p(x)$ be a polynomial of degree 4 having extremum at $x = 1, 2$ and $\lim_{x \to 0}\left(1 + \frac{p(x)}{x^{2}}\right) = 2$. Then the value of $p(2)$ is

Let p(x) be a polynomial of° 4 having extremum at x = 1, 2 and lim_(x → 0)(1 + p(x)/x²) = 2. Then the value of p(2) is

$p(x) \text{ degree } 4,\ \text{ extrema at } x=1,\ 2$

$\lim_{x\to0}\left(1+\dfrac{p(x)}{x^2}\right)=2$

$\text{To find: } p(2)$

$\text{use: limit condition, then } p'(1)=p'(2)=0$

$p(x)=ax^4+bx^3+cx^2+dx+e$ $\lim_{x\to0}\left(1+\dfrac{p(x)}{x^2}\right)=2 \Rightarrow \dfrac{p(x)}{x^2}\to 1$ $\dfrac{p(x)}{x^2}=ax^2+bx+c+\dfrac dx+\dfrac e{x^2} \to 1$ $\Rightarrow d=e=0,\quad c=1$ $p(x)=ax^4+bx^3+x^2$ $1.\ p'(1)=p'(2)=0$ $2.\ \text{solve for } a, b$ $3.\ p(2)$ $p'(x)=4ax^3+3bx^2+2x$ $p'(1)=0:\ 4a+3b+2=0$ $p'(2)=0:\ 8a+3b+1=0$ $4a=1 \Rightarrow a=\tfrac14$ $3b=-3 \Rightarrow b=-1$ $p(x)=\tfrac14 x^4 - x^3 + x^2$ $p'(x)=x^3-3x^2+2x=x(x-1)(x-2)$ $p(2)=\tfrac14(16)-8+4 = 4-8+4 = 0$ $\boxed{p(2)=0}$