Question

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

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

Answer 1

Answer: (B), (D)

The correct options are
B $[-3,-2\sqrt{2})$
D $(2\sqrt{2},3]$

Given,

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

We have $r-1\ge 0,r\le n+1$

Therefore $1\le r\le n+1\Rightarrow \frac{1}{n+1}\le \frac{r}{n+1}\le 1$

Also, $k^2-8=\frac{n!}{(n-r)!(n-r+1)!}=\frac{r}{n+1}$

$\Rightarrow \frac{1}{n+1}\ge k^2-8\le 1$

$\Rightarrow 8<\frac{1}{n+1}+8\le k^2\le 9$

$\Rightarrow 8<k^2\le 9$

so,

$k\in [-3,-2\sqrt{2}]andk\in [2\sqrt{2},3]$