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

Options

A. 3
B. 2
C. 4
D. 8

Answer

A, C

Explanation

We have
`cot^(-1) ((n^(2) -10n + 21.6)/(pi)) gt (pi)/(6)`
`rArr (n^(2) - 10n + 21.6)/(pi) lt cot.(pi)/(6)` (as cot x decreasing for `0 lt x lt pi`)
or `n^(2) - 10n + 21.6 lt pisqrt3`
or `n^(2) - 10n + 25 + 21.6 - 25 lt pi sqrt3`
or `(n-5)^(2) lt pi sqrt3 + 3.4`
or `-sqrt(sqrt3 po + 3.4) lt n - 5 lt sqrt(sqrt3 pi + 3.4)`
or `5 -x sqrt(sqrt5 pi + 3.4) lt n - 5 lt sqrt(sqrt3 pi + 3.4)` ...(i)
Since `sqrt3 pi = 5.5`, nearly , `sqrt(sqrt3 pi + 3.4) ~ sqrt(8.9) ~ 2.9`
`rArr 2.1 lt n lt 7.9`
`:. n = 3, 4, 5, 6, 7 " " {"as " n in N}`