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If cot^(-1)((n^2-10 n+21. 6)/pi)>pi/6, w...

If `cot^(-1)((n^2-10 n+21. 6)/pi)>pi/6,` where `x y<0` then the possible values of `z` is (are) 3 (b) 2 (c) 4 (d) 8

Text Solution

Verified by Experts

`cot^-1((n^2-10n+21.6)/pi) gt pi/6`
`=>(n^2-10n+21.6)/pi gt cot (pi/6)`
`=>(n^2-10n+25 -25 + 21.6)/pi gt sqrt3`
`=>(n-5)^2 - 3.4 gt sqrt3pi`
`=>(n-5)^2 gt sqrt3pi+3.4`
`=>-sqrt(sqrt3pi+3.4) lt n-5 lt sqrt(sqrt3pi+3.4)`
`=>2.1 lt n lt 7.9`
So, integral values of `n` can be `3,4,5,6,7`.
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