Let $\mathbb{R}$ denote the set of all real numbers. Consider the polynomial function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=\frac{d^{10}}{dx^{10}}\left((x^{2}-1)^{10}\right)$, for all $x\in\mathbb{R}$. Here $\frac{d^{10}}{dx^{10}}\left((x^{2}-1)^{10}\right)$ is the $10^{\text{th}}$ order derivative of the function $(x^{2}-1)^{10}$. Then which of the following statements is (are) TRUE?

$f(x)=\dfrac{d^{10}}{dx^{10}}\left((x^{2}-1)^{10}\right)$

$g(x)=(x^{2}-1)^{10}$

$\text{Check: coeff of } x^{8},\ f(1)+f(-1),\ \deg f,\ \text{constant term}$

$g(x)=\sum_{k=0}^{10}\binom{10}{k}(-1)^{10-k}x^{2k}$

$[x^{m}]f=a_{m+10}\cdot\dfrac{(m+10)!}{m!}$

$\deg g=20 \Rightarrow \deg f=20-10=10$

$\Rightarrow (3) \text{ TRUE}$

$a_{18}=\binom{10}{9}(-1)^{1}=-10$

$[x^{8}]f=a_{18}\cdot\dfrac{18!}{8!}=(-10)\dfrac{18!}{8!}$

$\Rightarrow (1) \text{ TRUE}$

$x=1+t:\ (x^{2}-1)^{10}=(2t+t^{2})^{10}=t^{10}(2+t)^{10}$

$f(1)=10!\,2^{10},\quad f(-1)=10!\,2^{10}$

$\Rightarrow f(1)+f(-1)=10!\,2^{11} \Rightarrow (2) \text{ TRUE}$

$a_{10}=\binom{10}{5}(-1)^{5}=-252$

$f(0)=10!\cdot(-252)=-\left(\dfrac{10!}{5!}\right)^{2}$

$\neq -\dfrac{10!}{5!} \Rightarrow (4) \text{ FALSE}$

$\text{TRUE: } (1),\ (2),\ (3)$