Question

If ${}^{n-1}{C}_r=(k^2-3){}^n{C}_{r+1}$, then $kϵ$

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

Answer 1

The correct option is D $(\sqrt{3},2]$

Given, ${}^{n-1}{C}_r=(k^2-3{)}^n{C}_{r+1}$

$\Rightarrow$$\frac{{}^{n-1}{C}_r}{{}^n{C}_{r+1}}=(k^2-3)$ ....Using ${}^n{C}_x=\frac{n}{x}{.}^{n-1}{C}_{x-1}$

$\Rightarrow$$\frac{r+1}{n}=(k^2-3)$ ....As $n>r+1$

So, $0<\frac{r+1}{n}\le 1$

Therefore, $0<k^2-3\le 1$

$\Rightarrow 3<k^2\le 4$

$\Rightarrow \sqrt{3}<k\le 2$

So, k$\in$$(\sqrt{3},2]$