Question

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

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

Answer 1

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

${}^{n-1}{C}_r=(K^2-3{)}^n{C}_{r+1}{\Rightarrow }^n{C}_r=\frac{n!}{(n-r)!r!}\Rightarrow \frac{\frac{(n-1)!}{(n-r-1)!r!}}{\frac{n!}{(n-r-1)!(r+1)!}}=K^2-3\Rightarrow \frac{r}{n}=K^2-3$

We know in the value of ${}^n{C}_r$

value of $n\ge r$

$\Rightarrow \frac{r}{n}$ value lies between $[0,1]$

$\Rightarrow K^2-3\in [0,1]\Rightarrow K^2\in [3,4]\Rightarrow K\in [\sqrt{3},2]$