QJEEM24Ma4025: JEE Main 2024 Mathematics - Class 11 JEEM24Ma4025 Solution: if and only if :

${ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}}=(\mathrm{k}^2-8){ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}$

$\underbrace{\mathrm{r}+1 \geq 0, \quad \mathrm{r} \geq 0}_{\mathrm{r} \geq 0}$

$\begin{aligned}& \frac{{ }^{n-1} C_r}{{ }^n C_{r+1}}=k^2-8 \\& \frac{r+1}{n}=k^2-8 \\& \Rightarrow k^2-8>0 \\& (k-2 \sqrt{2})(k+2 \sqrt{2})>0\end{aligned}$

$\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)\quad \text{.... (I)}$

$\begin{aligned}\therefore & \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1 \\\Rightarrow & \mathrm{k}^2-8 \leq 1 \\& \mathrm{k}^2-9 \leq 0 \\& -3 \leq \mathrm{k} \leq 3 \quad \text{.... (II)}\end{aligned}$

From equation (I) and (II) we get

$\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$