Let $$f(x)$$ be a polynomial of degree 4 having extreme values at $$x = 1$$ and $$x = 2$$. If $$\lim_{x \to 0}\left(\frac{f(x)}{x^2} + 1\right) = 3$$, then $$f(-1)$$ is equal to:

We know that a point of extreme value of a differentiable function occurs where its first derivative vanishes. Because the given quartic $$f(x)$$ has extreme values at $$x = 1$$ and $$x = 2$$, we must have

$$f'(1)=0 \quad\text{and}\quad f'(2)=0.$$

Therefore $$(x-1)$$ and $$(x-2)$$ are factors of $$f'(x).$$ Since $$f(x)$$ is of degree 4, its derivative $$f'(x)$$ is of degree 3, so it can be written as

$$f'(x)=k\,(x-1)(x-2)(x-a),$$

where $$k\neq0$$ is a constant multiplier and $$a$$ is the third root (still to be determined).

Next, we expand the cubic inside the bracket. First multiply the first two linear factors:

$$(x-1)(x-2)=x^{2}-3x+2.$$

Now multiply by the third factor $$(x-a):$$

$$\begin{aligned} (x^{2}-3x+2)(x-a) &=x(x^{2}-3x+2)-a(x^{2}-3x+2)\\ &=x^{3}-3x^{2}+2x-a x^{2}+3a x-2a\\ &=x^{3}-(a+3)x^{2}+(3a+2)x-2a. \end{aligned}$$

Thus

$$f'(x)=k\Bigl[x^{3}-(a+3)x^{2}+(3a+2)x-2a\Bigr].$$

To obtain $$f(x)$$ itself, we integrate term by term. We first recall the basic power‐rule formula

$$\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C,$$

where $$C$$ is the constant of integration. Applying that to each term, we have

$$\begin{aligned} f(x) &=\int f'(x)\,dx\\ &=k\left[\frac{x^{4}}{4}-\frac{(a+3)x^{3}}{3}+\frac{(3a+2)x^{2}}{2}-2a x\right]+C. \end{aligned}$$

The problem now supplies the limit

$$\lim_{x\to0}\left(\frac{f(x)}{x^{2}}+1\right)=3.$$

Rewriting, we need

$$\lim_{x\to0}\frac{f(x)}{x^{2}}=2.$$

Hence $$\frac{f(x)}{x^{2}}$$ must approach a finite number (namely 2) as $$x\to0$$, which means no negative powers of $$x$$ may remain in that quotient. Let us therefore divide our current expression for $$f(x)$$ by $$x^{2}$$ and inspect all terms:

$$\frac{f(x)}{x^{2}} =k\left[\frac{x^{2}}{4}-\frac{(a+3)x}{3}+\frac{(3a+2)}{2}-\frac{2a}{x}\right]+\frac{C}{x^{2}}.$$

For the limit as $$x\to0$$ to be finite, the coefficients of the troublesome terms $$\frac{1}{x}$$ and $$\frac{1}{x^{2}}$$ must be zero. That forces the two independent conditions

$$-2a k = 0 \quad\text{and}\quad C = 0.$$

Because $$k\neq0$$, the first equation immediately gives $$a = 0.$$ Substituting $$a = 0$$ and $$C = 0$$ back, the derivative simplifies neatly to

$$f'(x)=k\,x(x-1)(x-2).$$

Expanding the right-hand side once more:

$$x(x-1)(x-2)=x\bigl(x^{2}-3x+2\bigr)=x^{3}-3x^{2}+2x.$$

Thus

$$f'(x)=k\left(x^{3}-3x^{2}+2x\right).$$

We integrate again (now the integrand is simpler):

$$\begin{aligned} f(x) &=k\left[\frac{x^{4}}{4}-x^{3}+x^{2}\right]+0\\ &=k\left(\frac{x^{4}}{4}-x^{3}+x^{2}\right). \end{aligned}$$

We still need the constant $$k$$. Divide by $$x^{2}$$ once more:

$$\frac{f(x)}{x^{2}}=k\left(\frac{x^{2}}{4}-x+1\right).$$

Now take $$x\to 0$$:

$$\lim_{x\to0}\frac{f(x)}{x^{2}}=k\left(0-0+1\right)=k.$$

But we already required that this limit equal 2, so $$k=2.$$ Hence the explicit polynomial becomes

$$\begin{aligned} f(x) &=2\left(\frac{x^{4}}{4}-x^{3}+x^{2}\right)\\[4pt] &=\frac{x^{4}}{2}-2x^{3}+2x^{2}. \end{aligned}$$

Finally, we evaluate $$f(-1):$$

$$\begin{aligned} f(-1) &=\frac{(-1)^{4}}{2}-2(-1)^{3}+2(-1)^{2}\\ &=\frac{1}{2}-2(-1)+2(1)\\ &=\frac{1}{2}+2+2\\ &=\frac{9}{2}. \end{aligned}$$

Hence, the correct answer is Option D.