Binomial Theorem Question 3
Question 3 - 2024 (27 Jan Shift 1)
${ }^{n-1} C_{r}=\left(k^{2}-8\right)^{n} C_{r+1}$ if and only if :
(1) $2 \sqrt{2}<\mathrm{k} \leq 3$
(2) $2 \sqrt{3}<\mathrm{k} \leq 3 \sqrt{2}$
(3) $2 \sqrt{3}<\mathrm{k}<3 \sqrt{3}$
(4) $2 \sqrt{2}<\mathrm{k}<2 \sqrt{3}$
Show Answer
Answer (1)
Solution
$$ \begin{align*} & { }^{n-1} C_{r}=\left(k^{2}-8\right){ }^{n} C_{r+1} \ & \underbrace{r+1 \geq 0, r \geq 0}{r \geq 0} \ & \frac{{ }^{n} C{r}}{{ }^{n} C_{r+1}}=k^{2}-8 \ & \frac{r+1}{n}=k^{2}-8 \ & \Rightarrow k^{2}-8>0 \ & (k-2 \sqrt{2})(k+2 \sqrt{2})>0 \tag{I} \end{align*} $$
$\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)$
$\therefore \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1$
$\Rightarrow \mathrm{k}^{2}-8 \leq 1$
$$ \begin{equation*} \mathrm{k}^{2}-9 \leq 0 \tag{II} \end{equation*} $$
$-3 \leq \mathrm{k} \leq 3$
From equation (I) and (II) we get
$\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$
