Question 5

Let $g(x) = (x^2 - 1)^{10}$. Using the binomial expansion: $g(x) = \sum_{r=0}^{10} \binom{10}{r} (x^2)^r (-1)^{10-r} = \sum_{r=0}^{10} \binom{10}{r} (-1)^{10-r} x^{2r}$ The function $f(x)$ is defined as the $10^{th}$ derivative of $g(x)$: $f(x) = \frac{d^{10}}{dx^{10}} \left( \sum_{r=0}^{10} \binom{10}{r} (-1)^{10-r} x^{2r} \right) = \sum_{r=5}^{10} \binom{10}{r} (-1)^{10-r} \frac{(2r)!}{(2r-10)!} x^{2r-10}$

Analysis of Options:

• Option (C): The highest power in $g(x)$ is $x^{20}$ (for $r=10$). After differentiating 10 times, the highest power becomes $x^{20-10} = x^{10}$. Thus, the degree of $f(x)$ is 10. Correct.

• Option (A): To find the coefficient of $x^8$, we set the exponent $2r - 10 = 8$, which gives $r = 9$. The coefficient is: $\text{Coeff} = \binom{10}{9} (-1)^{10-9} \frac{(2 \cdot 9)!}{(18-10)!} = 10 \cdot (-1) \cdot \frac{18!}{8!} = -10 \left( \frac{18!}{8!} \right)$ So, Option (A) is correct.

• Option (B): Let $g(x) = (x-1)^{10}(x+1)^{10}$. Using the Leibniz rule for the $n^{th}$ derivative of a product: $f(x) = g^{(10)}(x) = \sum_{k=0}^{10} \binom{10}{k} \frac{d^k}{dx^k}(x-1)^{10} \frac{d^{10-k}}{dx^{10-k}}(x+1)^{10}$ At $x = 1$, all terms $\frac{d^k}{dx^k}(x-1)^{10}$ are zero for $k < 10$. Only the $k=10$ term survives: $f(1) = \binom{10}{10} (10!) (1+1)^{10} = 10! \cdot 2^{10}$ Similarly, at $x = -1$, only the $k=0$ term survives because $\frac{d^{10-k}}{dx^{10-k}}(x+1)^{10}$ is zero for $k > 0$: $f(-1) = \binom{10}{0} (-1-1)^{10} (10!) = (-2)^{10} \cdot 10! = 2^{10} \cdot 10!$ Therefore, $f(1) + f(-1) = 10! \cdot 2^{10} + 10! \cdot 2^{10} = 2 \cdot 10! \cdot 2^{10} = 10! \cdot 2^{11}$. Correct.

• Option (D): The constant term of $f(x)$ is $f(0)$. In the summation, this occurs when $2r - 10 = 0$, or $r = 5$: $f(0) = \binom{10}{5} (-1)^{10-5} \frac{10!}{0!} = \frac{10!}{5! 5!} (-1) (10!) = -\left( \frac{10!}{5!} \right)^2$ This is not equal to $-(\frac{10!}{5!})$. Incorrect.