Question

Let \(\mathbb{R}\) denote the set of all real numbers. Consider the polynomial function \[ f : \mathbb{R} \to \mathbb{R} \] defined by \[ f(x) = \frac{d^{10}}{dx^{10}}\left((x^2 - 1)^{10}\right), \qquad \text{for all } x \in \mathbb{R}. \] 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?

• A. The coefficient of $x^8$ in the polynomial $f(x)$ is $(-10) \left( \frac{18!}{8!} \right)$
• B. The value of $f(1) + f(-1)$ is equal to $10! 2^{11}$
• C. The degree of the polynomial $f(x)$ is 10
• D. The constant term of the polynomial $f(x)$ is $- \left( \frac{10!}{5!} \right)$

Hint:

Recognizing the Rodrigues' formula structure ($n^{th}$ derivative of $(x^2-1)^n$) immediately simplifies problems involving specific values like $f(1)$ and $f(-1)$ due to the properties of Legendre polynomials.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks about properties of the $10^{th}$ derivative of a specific polynomial. This is closely related to Legendre polynomials.
Step 2: Key Formula or Approach:

Use binomial expansion for $(x^2 - 1)^{10}$.

Apply the power rule for derivatives: $\frac{d^n}{dx^n} x^m = \frac{m!}{(m-n)!} x^{m-n}$ for $m \ge n$.

Rodrigues' formula for Legendre polynomials: $P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2-1)^n$.

Step 3: Detailed Explanation:

Expanding $(x^2-1)^{10} = \sum_{r=0}^{10} \binom{10}{r} (x^2)^{10-r} (-1)^r = \sum_{r=0}^{10} (-1)^r \binom{10}{r} x^{20-2r}$.

$f(x) = \sum_{r=0}^{5} (-1)^r \binom{10}{r} \frac{(20-2r)!}{(10-2r)!} x^{10-2r}$ (since terms where $20-2r<10$ become zero).

Checking (C): The highest power comes from $r=0$, which is $x^{10}$. So the degree is 10. Statement (C) is TRUE.

Checking (A): Coefficient of $x^8$ corresponds to $10-2r = 8 \implies r=1$. \[ \text{Coeff} = (-1)^1 \binom{10}{1} \frac{(20-2)!}{(10-2)!} = -10 \cdot \frac{18!}{8!}. \] Statement (A) is TRUE.

Checking (B): Using Rodrigues' formula, $f(x) = 2^{10} 10! P_{10}(x)$. Since $P_{10}(1) = 1$ and $P_{10}(-1) = (-1)^{10} = 1$: \[ f(1) = 2^{10} 10!, f(-1) = 2^{10} 10! \implies f(1) + f(-1) = 2 \cdot 10! 2^{10} = 10! 2^{11}. \] Statement (B) is TRUE.

Checking (D): Constant term is for $10-2r=0 \implies r=5$. \[ \text{Constant} = (-1)^5 \binom{10}{5} \frac{10!}{0!} = - \frac{10!}{5! 5!} 10! = - \left(\frac{10!}{5!}\right)^2. \] Statement (D) is FALSE as it does not match the expression.

Step 4: Final Answer:
The true statements are (A), (B), and (C).