JEE Main 2024 — 27 January, Shift 1 — Question 1

${ }^{n-1} C_{r}=\left(k^{2}-8\right){ }^{n} C_{r+1}$

$\underbrace{r+1 \geq 0, \quad r \geq 0}_{r \geq 0}$ $\frac{{ }^{n-1} C_{r}}{{ }^{n} C_{r+1}}=k^{2}-8$

$\frac{\mathrm{r}+1}{\mathrm{n}}=\mathrm{k}^{2}-8$

$\Rightarrow \mathrm{k}^{2}-8>0$

$(\mathrm{k}-2 \sqrt{2})(\mathrm{k}+2 \sqrt{2})>0$

\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)$$ …(1)

$\therefore \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1$

$\Rightarrow \mathrm{k}^{2}-8 \leq 1$

$\mathrm{k}^{2}-9 \leq 0$

$-3 \leq \mathrm{k} \leq 3$ $…(2)$

From equation (I) and (II) we get $\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$