Question

Sol. [Key Concept: The limit fixes the low-order coefficients; $f' \propto \ x(x - 1)(x + 1)$ makes $f$ odd, so $f(2) - f( - 2) = 2f(2)$.]

Given $f(x)$ is a polynomial of degree 5

Also, $f^{'}(1) = 0;f^{'}( - 1) = 0$.

Also $\because\lim_{x \rightarrow 0}\mspace{2mu}\frac{f(x)}{x^{3}} = - 5$ (fixed and finite)

$${\therefore f(0) = 0;f^{'}(0) = 0;f^{'}(0) = 0 }{\frac{f^{'''}(0)}{6} = - 5 \Rightarrow f^{'''}(0) = - 30 }$$Hence, let $f^{'}(x) = (ax + b)(x - 0)(x - 1)(x + 1)$

$${\Rightarrow f^{'}(x) = {ax}^{4} + {bx}^{3} - {ax}^{2} - bx }{f^{''}(x) = 4{ax}^{3} + 3{bx}^{2} - 2ax - b }{\because f^{'}(0) = 0 \Rightarrow \text{ }b = 0 }{f^{'''}(x) = 12ax^{2} + 6bx - 2a }{\because f^{''}(0) = - 2a = - 30 }{\Rightarrow a = 15 }{\therefore f^{'}(x) = 15x.x.(x - 1)(x + 1) }{\Rightarrow f^{'}(x) = 15x^{4} - 15x^{2} }{\Rightarrow f(x) = 3x^{5} - 5x^{3} + C }{\because f(0) = 0 \Rightarrow C = 0 }{\therefore f(x) = 3x^{5} - 5x^{3} }{\therefore f(2) - f( - 2) = 112 }$$