If the inequality `cot^(2)x + (k +1) cot x - (k-3) < 0` is true for at least one `x in (0, pi//2)`, then `k in `.

Options

A. `(-oo,3-2sqrt(5))`
B. `(3,oo)`
C. `(-1, oo)`
D. `(-oo,3)`

Answer

1,2

Explanation

Let cot x = t.
Since `x in (0, pi//2), t in (0 ,oo)`
Therefore, `f(t) = t^(2) +(k + 1) t- (k - 3) lt 0` for at least one `t in (0,oo)` .
We have following possibilities for the graph of f(t).
Case I :

`f(0) lt 0 rArr k gt 3`
Case II :

`D gt 0 `
`rArr (k +1)^(2) + 4(k - 3) gt 0`
`rArr k^(2) + 6k - 11 gt 0`
`rArr (k + 3)^(2) gt 20`
`rArr k + 3 lt - 2sqrt(5) or k + 3 gt 2 sqrt(5)`
`rArr k lt - 3 - 2 - sqrt(2) or k gt -3 + 2sqrt(5)`
`(-b)/(2a) gt 0`
`rArr - (k+1) gt 0`
`rArr k lt - 1" .....(2)`
`rArr k(0) ge 0 `
`rArr - k + 3 ge 0`
`rArr k le 3" ....(3)`
From (1),(2) and (3) we get
`k in (-oo, -3-2sqrt(5))`
So, from the both cases, `k in (-oo, -3 -2sqrt(5)) uu (3,oo)`