Let `f:R -> ( 0,(2pi)/2]` defined as `f(x) = cot^-1 (x^2-4x + alpha)` Then the smallest integral value of `alpha` such that, `f(x)` is into function is

Options

A. 2
B. 4
C. 6
D. 8

Answer

B

Explanation

Clearly
`cot^(-1)(x^(2)-4x+alpha) gt (2pi)/(3), AA x in R`.
`therefore" "x^(2)-4x+alpha gt (-1)/(sqrt3), AA x in R`
`rArr" "x^(2)-4x+alpha+(1)/(sqrt3)lt0 AA x in R`
So, discriminant `lt 0 rArr 16-4(alpha+(1)/(sqrt3))lt0`
`4-alpha-(1)(sqrt3)lt0 rArr alpha gt 4-(1)/(sqrt3)`
`therefore" "alpha in (4-(1)/(sqrt3),oo)`
Hence, minimum integral `alpha` is 4.