Differentiation

Concept:The function $f(x)$ is the $10^{\text{th}}$ derivative of $(x^2-1)^{10}$. Use binomial expansion and differentiate term‑wise.Explanation:Expand $(x^2-1)^{10}$ using the binomial theorem:$$(x^2-1)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^{20-2k}.$$Differentiating $x^{20-2k}$ ten times gives $\frac{(20-2k)!}{(10-2k)!} x^{10-2k}$, provided $20-2k \ge 10$ (i.e., $k \le 5$). For $k>5$ the term vanishes. Hence,$$f(x) = \sum_{k=0}^{5} \binom{10}{k} (-1)^k \frac{(20-2k)!}{(10-2k)!} x^{10-2k}.$$Check each option:(A) The $x^8$ term corresponds to $k=1$ (since $10-2k=8$). Coefficient = $\binom{10}{1}(-1)^1\frac{18!}{8!} = -10\frac{18!}{8!}$. Option A is correct.(B) $(x^2-1)^{10}$ is an even function, so its $10^{\text{th}}$ derivative $f(x)$ is also even. Hence $f(1)=f(-1)$ and $f(1)+f(-1)=2f(1)$. Computing $f(1)$ from the sum gives $10! \cdot 2^{10}$, so $2f(1)=10! \cdot 2^{11}$. Option B is correct.(C) The highest power in $f(x)$ is $x^{10}$ (from $k=0$). So the degree of $f(x)$ is $10$. Option C is correct.(D) The constant term comes from $k=5$ (since $10-2k=0$). Constant term = $\binom{10}{5}(-1)^5\frac{10!}{0!} = -\frac{(10!)^2}{5!\,5!}$, not $-\frac{10!}{5!}$. Option D is false.Answer:Options A, B, and C are true.