If `[f((sqrt(3))/2)]` is [.] denotes the greatest integer function) 4 (b) 5 (c) 6 (d) `-7`

Options

A. `4`
B. `5`
C. `6`
D. `-7`

Answer

B

Explanation

`int_(cosx)^(1)t^(2)f(t)dt=1-cosx`
Differentiating both sides w.r.t `x`, we get
`d/(dx) int_(cosx)^(1)t^(2)f(t)dt=d/(dx)(1-cosx)`
or `-cos^(2)x f(cosx)(-sinx)=sinx`
or `cos^(2)xf(cosx)sinx=sinx`
or `f(cosx)=1/(cos^(2)x)`
Now `f((sqrt(3))/4)` is attained when `cosx=(sqrt(3))/4`
`:.f((sqrt(3))/4)=16/3=5.33`
`[f((sqrt(3))/4)]=5`