Explanation:
Let \( g(x)=(x^{2}-1)^{10} \)
\( = {}^{10}C_{0}x^{20}-{}^{10}C_{1}x^{18}+{}^{10}C_{2}x^{16}\ldots {}^{10}C_{10}1 \)
\( (B) \) is correct because on differentiating a polynomial, its degree reduces by \( 1 \) and the same holds for power of every term in the polynomial.
\( (A) \) correct because \( x^{18} \) term in \( g(x) \) will give the \( x^{8} \) term in \( f(x) \) on being differentiated \( 10 \) times.
Coefficient of \( x^{18} \) in \( g(x) \) is \( -{}^{10}C_{1} \)
\( \dfrac{d}{dx}(x^{18})=18x^{17} \)
\( \dfrac{d^{2}}{dx^{2}}(x^{18})=18\times17\,x^{16} \)
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\( \dfrac{d^{10}}{dx^{10}}(x^{18})=18\times17\times16\ldots 9 \times x^{8} \)
Hence, we get \( -10\times18\times17\times16\ldots 9=-\dfrac{10(18!)}{8!} \)

(D) is incorrect because: $g'(x) = 10(x^2 - 1)^9 \times 2x$

$\text{For } x = \pm 1 : g'(x) = 0$

$g''(x) = 10 \times 9(x^2 - 1)^8 \times (2x)^2 + 10(x^2 - 1)^9$

$\text{For } x = \pm 1 \cdot g''(x) = 0$

Note: Till the power of the $(x^2 - 1)$ term doesn't become 0, $\frac{d^n}{dx^n} g(x) = 0$ at $x = \pm 1$.

Since, at each differentiation step, the power of $(x^2 - 1)$ reduces exactly once and a factor of $(2x)$ in multiplied simultaneously,

The term that'll become non-zero for the first time at $x = \pm 1$, when degree of $(x^2 - 1)$ will become zero.

That term will b

$10 \times 9 \times 8 \times 7 \dots 1(2x)^{10} = 2^{10}10!$

Of course: $\frac{d^{10}}{dx^{10}} g(x)$ will have other terms with higher degrees of $x^2 - 1$ but they'll all vanish at $x = \pm 1$.

$\Rightarrow \quad f(1) = f(-1) = 2^{10}10! \quad \Rightarrow \quad f(1) + f(-1) = 2^{11} \times 10!$

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