Question

${}^{n-1}{C}_r=(K^2-3){.}^n{C}_{r+1}$, If $K\in$

• A. $[-1\sqrt{3},\sqrt{3}]$
• B. $(-\infty ,-2)$
• C. $(2,\infty )$
• D. $(\sqrt{3},2]$

Answer 1

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

Given,

${}^{n-1}{C}_r=\left(K^2-3\right){.}^n{C}_{r+1}$

Now,

$\begin{array}{c}{}^{n-1}{C}_r=\left(K^2-3\right){.}^n{C}_{r+1}\\ \Rightarrow \frac{\left(n-1\right)!}{|!r!\left(n-r-1\right)!}=\left(K^2-3\right)\frac{n!}{\left(r!\right)\left(n-r-1\right)!}\\ =k^2-3=\frac{\left(n-1\right)!\left(r+1\right)!}{r!n!}=\frac{r+1}{n}\end{array}$

But, $\frac{r+1}{n}\in \left[0,2\right]$

$\begin{array}{c}\therefore k^2-3\in \left[0,1\right]\\ \Rightarrow 0<k^2-3\le 1\\ \Rightarrow 3<k^2-3+3\le 1+3\\ \Rightarrow 3<k^2\le 4\\ \Rightarrow \sqrt{3}<k\le 2or-\sqrt{3}>k\ge -2\\ \therefore k\in \left[\sqrt{3},2\right]ork\in \left[-2,-\sqrt{3}\right]\end{array}$

Hence,

$therefore$ Option $D$ is correct answer.