Tenth Derivative Properties of Tenth Power Quadratic Polynomial

To determine which of the given statements are true for the polynomial function $f(x) = \frac{d^{10}}{dx^{10}} \left((x^2 - 1)^{10}\right)$, we can analyze the function using the Binomial Theorem and the Leibniz Rule for differentiation.

1. Binomial Expansion of $f(x)$

First, expand $g(x) = (x^2 - 1)^{10}$ using the binomial theorem: $g(x) = \sum_{k=0}^{10} \binom{10}{k} (x^2)^{10-k} (-1)^k = \sum_{k=0}^{10} (-1)^k \binom{10}{k} x^{20-2k}$

Differentiating $g(x)$ ten times with respect to $x$: $f(x) = \frac{d^{10}}{dx^{10}} g(x) = \sum_{k=0}^{10} (-1)^k \binom{10}{k} \frac{d^{10}}{dx^{10}} \left( x^{20-2k} \right)$

Recall that the $10^{\text{th}}$ derivative of $x^n$ is given by: $\frac{d^{10}}{dx^{10}} (x^n) = \begin{cases} \frac{n!}{(n-10)!} x^{n-10}, & \text{if } n \ge 10 \\ 0, & \text{if } n < 10 \end{cases}$

Since $20-2k < 10$ for $k > 5$, terms with $k > 5$ vanish after taking the $10^{\text{th}}$ derivative. Thus, $f(x)$ reduces to: $f(x) = \sum_{k=0}^{5} (-1)^k \binom{10}{k} \frac{(20-2k)!}{(10-2k)!} x^{10-2k}$

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2. Analysis of the Options

Checking Option (C):

The highest power of $x$ in $f(x)$ occurs for $k = 0$: $\text{Leading term} = (-1)^0 \binom{10}{0} \frac{20!}{10!} x^{10} = \frac{20!}{10!} x^{10}$ Since the coefficient $\frac{20!}{10!} \neq 0$, the degree of the polynomial $f(x)$ is $10$. Thus, Option (C) is TRUE.

Checking Option (A):

To find the coefficient of $x^8$ in $f(x)$, set the exponent $10 - 2k = 8$, which gives $k = 1$. Substituting $k = 1$ into the general term of $f(x)$: $\text{Term for } x^8 = (-1)^1 \binom{10}{1} \frac{(20 - 2)!}{(10 - 2)!} x^8 = -10 \left(\frac{18!}{8!}\right) x^8$ Hence, the coefficient of $x^8$ is $(-10)\left(\frac{18!}{8!}\right)$. Thus, Option (A) is TRUE.

Checking Option (D):

The constant term of $f(x)$ corresponds to the exponent $10 - 2k = 0$, which gives $k = 5$. Substituting $k = 5$: $\text{Constant term} = (-1)^5 \binom{10}{5} \frac{(20 - 10)!}{(10 - 10)!} = -1 \cdot \frac{10!}{5! 5!} \cdot \frac{10!}{0!} = -\left(\frac{10!}{5!}\right)^2$ Since $-\left(\frac{10!}{5!}\right)^2 \neq -\left(\frac{10!}{5!}\right)$, Option (D) is incorrect. Thus, Option (D) is FALSE.

Checking Option (B):

We express $g(x) = (x^2 - 1)^{10} = (x-1)^{10} (x+1)^{10}$. By Leibniz's Rule for the $n^{\text{th}}$ derivative of a product, we have: $f(x) = \frac{d^{10}}{dx^{10}} \left[ (x-1)^{10} (x+1)^{10} \right] = \sum_{j=0}^{10} \binom{10}{j} \frac{d^{10-j}}{dx^{10-j}} \left((x-1)^{10}\right) \cdot \frac{d^j}{dx^j} \left((x+1)^{10}\right)$

Evaluating $f(x)$ at $x = 1$: Notice that $\frac{d^{10-j}}{dx^{10-j}} \left((x-1)^{10}\right) = \frac{10!}{j!} (x-1)^{j}$. When evaluated at $x = 1$, $(x-1)^j = 0$ for all $j > 0$. Therefore, only the term corresponding to $j = 0$ survives: $f(1) = \binom{10}{0} \left[ \frac{d^{10}}{dx^{10}} (x-1)^{10} \right]_{x=1} \cdot \left[ (x+1)^{10} \right]_{x=1}$ $f(1) = 1 \cdot 10! \cdot (1+1)^{10} = 10! \cdot 2^{10}$

Since $f(x)$ contains only even powers of $x$ (i.e., $x^{10-2k}$), $f(x)$ is an even function, which implies: $f(-1) = f(1) = 10! \cdot 2^{10}$

Now, calculating $f(1) + f(-1)$: $f(1) + f(-1) = 2 \cdot f(1) = 2 \cdot (10! \cdot 2^{10}) = 10! \cdot 2^{11}$ Thus, Option (B) is TRUE.

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Conclusion

The correct statements are A, B, and C.