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

The correct options are:

1. The coefficient of x⁸ in the polynomial f(x) is −10 $·\frac{18!}{8!}$

2. The value of f(1) + f(−1) is equal to $10!·2^{11}$

3. The degree of the polynomial f(x) is 10

Step 1: Understand the definition of f(x)

We are given the polynomial function $f\left(x\right)=\frac{d^{10}}{d{x}^{10}}{(x^2-1)}^{10}$.

Let $g\left(x\right)={(x^2-1)}^{10}$. Expanding $g\left(x\right)$ using the binomial theorem gives:

$g\left(x\right)=\sum _{k=0}^{10}\left(\genfrac{}{}{0ex}{}{10}{k}\right){\left(x^2\right)}^{10-k}$

Here, $g\left(x\right)$ is a polynomial of degree 20 containing only even powers of $x$.

Step 2: Determine the degree of f(x)

Taking the 10^th derivative of a polynomial of degree 20 reduces its degree by 10.

$\text{Degree of}f\left(x\right)=20-10=10$

Therefore, the degree of the polynomial $f\left(x\right)$ is 10. This confirms that the third option is correct.

Step 3: Find the coefficient of x⁸ in f(x)

A general term in $g\left(x\right)$ is given by:

$T_k={(-1)}^k\left(\genfrac{}{}{0ex}{}{10}{k}\right)x^{20-2k}$

After taking the 10^th derivative, the power of $x$ in this term becomes $(20-2k)-10=10-2k$.

We want the coefficient of $x^8$ in $f\left(x\right)$, so we set:

$10-2k=8\Rightarrow 2k=2\Rightarrow k=1$

For $k=1$, the corresponding term in $g\left(x\right)$ is:

${(-1)}^1\left(\genfrac{}{}{0ex}{}{10}{1}\right)x^{18}=-10x^{18}$

Differentiating $-10x^{18}$ ten times with respect to $x$ yields:

$\frac{d^{10}}{d{x}^{10}}(-10x^{18})=-10·\frac{18!}{(18-10)!}{x}^8=-10·\frac{18!}{8!}{x}^8$

Hence, the coefficient of $x^8$ in $f\left(x\right)$ is $-10·\frac{18!}{8!}$. This confirms that the first option is correct.

Step 4: Calculate f(1) + f(−1)

Notice that $f\left(x\right)$ is an even function because $g\left(x\right)={(x^2-1)}^{10}$ contains only even powers, so taking an even number of derivatives (10 derivatives) preserves the even symmetry, i.e., $f(-1)=f\left(1\right)$.

Thus, $f\left(1\right)+f(-1)=2f\left(1\right)$.

By Rodrigues' formula for Legendre polynomials $P_n\left(x\right)$:

${\mathrm{Paccess}}_n\left(x\right)=\frac{1}{2^{n}n!}\frac{d^n}{d{x}^n}{(x^2-1)}^n$

For $n=10$, we have:

$f\left(x\right)=\frac{d^{10}}{d{x}^{10}}{(x^2-1)}^{10}=2^{10}·10!·P_{10}\left(x\right)$

Since $P_n\left(1\right)=1$ for all $n$, we get:

$f\left(1\right)=2^{10}·10!·P_{10}\left(1\right)=2^{10}·10!$

Therefore:

$f\left(1\right)+f(-1)=2f\left(1\right)=2·(2^{10}·10!)=10!·2^{11}$

This confirms that the second option is also correct.