Explanation

(d)
$\because\ \left( k^{2} - 3 \right) = \frac{\ ^{n - 1}C_{r}}{\ ^{n - 1}C_{r + 1}} = \frac{\ ^{n - 1}C_{r}}{\left( \frac{n}{r + 1} \right)\ ^{n - 1}C_{r}} = \left( \frac{r + 1}{n} \right)$ ...(i)
$\because\ 0 \leq r \leq n - 1$
$\Rightarrow \ 1 \leq r + 1 \leq n$
$\Rightarrow \ \frac{1}{n} \leq \frac{r + 1}{n} \leq 1$
$\Rightarrow \ \frac{1}{n} \leq k^{2} - 3 \leq 1$ [from Eq. (i)]
$\Rightarrow 3 + \frac{1}{n} \leq k^{2} \leq 4$
When $n \rightarrow \infty,\ 3 \leq k^{2} \leq 4$
or $k \in \left\lbrack - 2, - \sqrt{3} \right\rbrack \cup (\sqrt{3},2\rbrack$