1. Use the condition on critical points

Since $f(x)$ is a polynomial of degree $5$, its derivative $f'(x)$ is a polynomial of degree $4$.

Given that $x=\pm 1$ are critical points, we must have $f'(1)=0,\qquad f'(-1)=0.$

A quartic derivative consistent with these critical points is of the form $f'(x)=k(x^2-1)^2,$ because this is degree $4$ and vanishes at $x=\pm 1$.

2. Integrate to get $f(x)$

Expand: $f'(x)=k(x^4-2x^2+1).$

Integrating, $f(x)=k\left(\frac{x^5}{5}-\frac{2x^3}{3}+x\right)+C.$

So $f(x)=\frac{k}{5}x^5-\frac{2k}{3}x^3+kx+C.$

3. Use the limit condition

Given $\lim_{x\to 0}\left(2+\frac{f(x)}{x^3}\right)=4,$ we get $\lim_{x\to 0}\frac{f(x)}{x^3}=2.$

Now, $\frac{f(x)}{x^3}=\frac{k}{5}x^2-\frac{2k}{3}+\frac{k}{x^2}+\frac{C}{x^3}.$ This form suggests divergence unless lower-order terms vanish appropriately. So for the limit to exist and be finite, we must have $C=0,\qquad kx\text{ term must be absent.}$

Hence we should instead infer directly from the limit that $f(x)$ must behave like $2x^3$ near $x=0$, so coefficients of constant, $x$, and $x^2$ must be zero.

Thus let $f(x)=ax^5+bx^4+cx^3+dx^2+ex+g.$ Then $\frac{f(x)}{x^3}=ax^2+bx+c+\frac{d}{x}+\frac{e}{x^2}+\frac{g}{x^3}.$

For the limit to exist finitely, we need $d=e=g=0.$ And since $\lim_{x\to 0}\left(2+\frac{f(x)}{x^3}\right)=4,$ we get $2+c=4 \implies c=2.$

So $f(x)=ax^5+bx^4+2x^3.$

4. Use the critical point conditions

Differentiate: $f'(x)=5ax^4+4bx^3+6x^2=x^2(5ax^2+4bx+6).$

Since $x=1$ and $x=-1$ are critical points, $f'(1)=0 \implies 5a+4b+6=0,$ $f'(-1)=0 \implies 5a-4b+6=0.$

Adding, $10a+12=0 \implies a=-\frac65.$

Subtracting, $8b=0 \implies b=0.$

Hence $f(x)=-\frac65x^5+2x^3.$

So $f(x)=x^3\left(2-\frac65x^2\right).$ This is clearly an odd function.

Therefore option D is true.

5. Check nature of critical points

Compute second derivative: $f'(x)=-6x^4+6x^2=6x^2(1-x^2),$ $f''(x)=-24x^3+12x=12x(1-2x^2).$

Now, $f''(1)=12(1)(1-2)=-12<0,$ so $x=1$ is a point of maximum.

And $f''(-1)=12(-1)(1-2)=12>0,$ so $x=-1$ is a point of minimum.

Therefore option C is true, and option B is false.

6. Check option A

Compute values: $f(1)=-\frac65+2=\frac45,$ $f(-1)=\frac65-2=-\frac45.$

Then $f(1)-4f(-1)=\frac45-4\left(-\frac45\right)=\frac45+\frac{16}{5}=\frac{20}{5}=4.$

So option A is true.

7. Conclusion

• A is true
• B is false
• C is true
• D is true

Hence the statement which is not true is: $\boxed{\text{B}}$