If `.^(n-1)C_r=(k^2-3).^nC_(r+1)`, then `k` belong to

Options

A. `(-infty, -2]`
B. `[2,infty)`
C. `[-sqrt(3),sqrt(3)]`
D. `(sqrt(3), 2]`

Answer

D

Explanation

`because ""^(n-1)C_(r) =- (k^(2)-3)cdot ^(n)C_(r+1_`
`rArr k^(2) - 3 = (""^(n-1)C_(r))/(""^(n)C_(r+1))=(r+1)/n` ...(i)
`rArr 0le rle n-1`
` rArr 1le r + 1 le n`
` rArr 1/nle (r+1)/nle 1`
`rArr 1/n le(k^(2)-3) le1 `
`rArr 3+1/n le k^(2) le 4 or 3lt k^(2) le4 [ "here,"nge2]`
`therefore k in[-2, sqrt(3) ) cup (sqrt(3),2]`