Question

If ${ }^n C_{r-1}=\left(k^2-8\right)\left({ }^{n+1} C_r\right)$, then $k$ belongs to

• A. $[-3,-2 \sqrt{2})$
• B. $(-3,3)$
• C. $[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$
• D. $[-2 \sqrt{2}, 2 \sqrt{2}]$

Answer 1

Answer: (C)

We have, $r-1 \geq 0, r \leq n+1$
$\Rightarrow 1 \leq r \leq n +1 \Rightarrow \frac{1}{n+1} \leq \frac{r}{n+1} \leq 1 $
$\text { Also, } k^2-8 =\frac{n !}{(r-1) !(n-r+1) !} \cdot \frac{r !(n+1-r) !}{(n+1) !} $
$ =\frac{r}{n+1}$
Thus, $\frac{1}{n+1} \leq k^2-8 \leq 1$
$\Rightarrow 8<\frac{1}{n+1}+8 \leq k^2 \leq 9 $
$\Rightarrow 8 Hence, $k \in[-3,-2 \sqrt{2})$