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

We are given: $$\lim_{x\to 0} \left[ 1 + \frac{f(x)}{x^2} \right] = 3 \implies \lim_{x\to 0} \frac{f(x)}{x^2} = 2$$

For this limit to exist and be equal to a finite non-zero value ($$2$$), the polynomial $$f(x)$$ must not contain any terms with a degree less than $$2$$.

Let the degree four polynomial be $$f(x) = ax^4 + bx^3 + cx^2 + dx + e$$

Since $$\lim_{x\to 0} \frac{f(x)}{x^2} = 2$$, we must have:

$$e = 0$$ and $$d = 0$$ (otherwise the limit would tend to infinity)

The coefficient of $$x^2$$ must be $$2$$, so $$c = 2$$

$$f(x) = ax^4 + bx^3 + 2x^2$$

$$f'(x) = 4ax^3 + 3bx^2 + 4x$$

Since $$x = 1$$ and $$x = 2$$ are points of extrema, they must be roots of the quadratic equation $$4ax^2 + 3bx + 4 = 0$$

$$\implies \mathbf{a = \frac{1}{2}}$$ , $$\mathbf{b = -2}$$

$$f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$$

$$f(2) = \frac{1}{2}(2)^4 - 2(2)^3 + 2(2)^2$$ 

$$f(2) = 0$$