Let $$f(x)$$ be a polynomial of degree 5, and have extrema at $$x = 1$$ and $$x = -1$$. If $$\lim_{x \to 0} \left(\frac{f(x)}{x^3}\right ) = -5$$, then $$f(2) - f(-2)$$ is equal to :

Since $$f(x)$$ is a degree-5 polynomial, its derivative $$f'(x)$$ is degree 4.
Given that $$f(x)$$ has extrema at $$x = 1$$ and $$x = -1$$, we must have $$f'(1)=0$$ and $$f'(-1)=0$$. Therefore $$(x-1)$$ and $$(x+1)$$ are factors of $$f'(x)$$, so:

$$f'(x)= (x-1)(x+1) \, q(x)= (x^2-1)\,q(x)$$ where $$q(x)$$ is a quadratic (degree 2) so the overall degree of $$f'(x)$$ remains 4.

Write $$q(x)=Ax^2+Bx+C$$, hence $$f'(x)=(x^2-1)(Ax^2+Bx+C)$$ $$-(1)$$

Integrating $$f'(x)$$ will give $$f(x)$$. First expand $$f'(x)$$:

$$f'(x)=A x^4 + B x^3 + C x^2 - A x^2 - B x - C$$ $$\;\;=A x^4 + B x^3 +(C-A)x^2 - B x - C$$

Integrate term by term:

$$f(x)=\int f'(x)\,dx$$ $$= \frac{A}{5}x^5 + \frac{B}{4}x^4 + \frac{C-A}{3}x^3 - \frac{B}{2}x^2 - Cx + D$$ $$-(2)$$

The limit condition is $$\lim_{x\to 0}\frac{f(x)}{x^3}=-5$$ $$-(3)$$

To make the limit finite, the terms $$x^0, x^1,$$ and $$x^2$$ must vanish in $$f(x)$$, i.e.

From $$(2)$$ we require $$f(0)=0 \;\Rightarrow\; D=0$$ $$f'(0)=0 \;\Rightarrow\; -C = 0 \;\Rightarrow\; C = 0$$ $$f''(0)=0 \;\Rightarrow\; -B = 0 \;\Rightarrow\; B = 0$$

Thus $$B=C=D=0$$. Equation $$(2)$$ simplifies to

$$f(x)=\frac{A}{5}x^5-\frac{A}{3}x^3$$ $$=A\left(\frac{x^5}{5}-\frac{x^3}{3}\right)$$ $$-(4)$$

Now use the limit $$(3)$$. Divide $$(4)$$ by $$x^3$$ and let $$x\to 0$$:

$$\frac{f(x)}{x^3}=A\left(\frac{x^2}{5}-\frac{1}{3}\right)\xrightarrow[x\to 0]{} -\frac{A}{3}$$

Set this equal to $$-5$$:

$$-\frac{A}{3}=-5\;\;\Longrightarrow\;\;A=15$$

Hence

$$f(x)=15\left(\frac{x^5}{5}-\frac{x^3}{3}\right)=3x^5-5x^3$$ $$-(5)$$

Finally compute $$f(2)-f(-2)$$:

$$f(2)=3(2)^5-5(2)^3=3\cdot32-5\cdot8=96-40=56$$ $$f(-2)=3(-2)^5-5(-2)^3=3(-32)-5(-8)=-96+40=-56$$

Therefore $$f(2)-f(-2)=56-(-56)=112$$

Option D which is: $$112$$