^ n-1 C_r= (k^2-8 ) ^n C_ r+1 if and only if: - Vidyadip

MCQ

${ }^{n-1} C_r=\left(k^2-8\right){ }^n C_{r+1}$ if and only if:

A
$2 \sqrt{2}< k \leq 3$
B
$2 \sqrt{3}< k \leq 3 \sqrt{2}$
C
$2 \sqrt{3}< k <3 \sqrt{3}$
D
$2 \sqrt{2}< k <2 \sqrt{3}$

✓

Answer

${ }^{n-1} C_r=\left(k^2-8\right)^n C_{r+1}$
$\underbrace{r+1 \geq 0, r \geq 0}_{r \geq 0}$
$\frac{{ }^{n-1} C_r}{{ }^n C_{r+1}}=k^2-8$
$\frac{r+1}{n}=k^2-8$
$\Rightarrow k^2-8>0$
$(k-2 \sqrt{2})(k+2 \sqrt{2})>0$
$k \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty......(I)$
$\therefore n \geq r+1, \frac{r+1}{n} \leq 1$
$\Rightarrow k^2-8 \leq 1$
$k^2-9 \leq 0$
$-3 \leq k \leq 3......(II)$
From equation $(I)$ and $(II)$ we get
$k \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$

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