If ^n-1Cr=(k^2-3)^n C(r+1),then k in...

If `^n-1C_r=(k^2-3)^n C_(r+1),then k in `

`(-infty, -2]`

`[2,infty)`

`[-sqrt(3),sqrt(3)]`

`(sqrt(3), 2]`

Text Solution

Verified by Experts

The correct Answer is:

`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]`

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