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If .^(n-1)C1= (k^2-3) ^nC(r+1) then K b...

If `.^(n-1)C_1= (k^2-3) ^nC_(r+1) ` then K belongs to

A

`(- oo, - 2)`

B

`[2 oo]`

C

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

D

`(sqrt(3), 2]`

Text Solution

Verified by Experts

Given, `.^(n-1)C_r=(k^2-3).^Nc_(r+1)`
`rArr .^(n-1)C_(r)=(k^2-3)(n)/(r+1).^(n-1)C_(r)`
`rArr k^2 -3 =(r+1)/(n)`
`[" since ", n ge r rArr (r+1)/(n) le 1 and n r gt 0]`
`rArr 0lt k^2-3 le 1 rArr 3lt k^2 le 4`.
`rArr k in [ -2, -sqrt(3))cup (sqrt(3),2]`.
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