Question

If ${}^{n-1}{C}_r={\left(K^2-3\right)}^n{C}_{r+1^{\prime }}$, then $k$ belongs to

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

Answer 1

The correct option is A $\left[-\sqrt{3},\sqrt{3}\right]$

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

$\frac{(n-1)!}{(n-1-r)!r!}=(k^2-3)\frac{n!}{(n-r-1)!(r+1)!}$

$\frac{(n-1)!}{(n-1-r)!r!}=(k^2-3)\frac{n(n-1)!}{(n-r-1)!(r+1)r!}$

or $1=(k^2-3)\frac{n}{r+1}$

or $\frac{r+1}{n}=(k^2-3)$

or $r⩽n-1$

and $r+1⩽n$

$0<\frac{r+1}{n}⩽1$

$0⩽(k^2-3)⩽1$

$k^2⩾3$ and $k^2⩽4$

$\therefore$ common region

$\left[-\sqrt{3},\sqrt{3}\right]$