The value of `k(k >0)` such that the length of the longest interval in which the function `f(x)=sin^(-1)|sink x|+cos^(-1)(cosk x)` is constant is `pi/

Correct Answer - B
`f(x) = sin^(-1) |sin kx| + cos^(-1) (cos kx)`
Let `g(x) = sin^(-1) |sin x| + cos^(-1) (cos x)`
`g(x){(2x,0 le x le(pi)/(2)),(pi,(pi)/(2) lt x le (3pi)/(2)),(4pi - 2x,(3pi)/(2) lt x le 2pi):}`
`g(x)` is periodic with period `2pi` and is constant in the continuous interval `[2n pi + (pi)/(2), 2n pi + (3pi)/(2)] ("where " n in I) and f(x) = g(kx)`.
So, `f(x)` is constant in the interval
`[(2npi)/(k) + (pi)/(2k), (2n pi)/(k) + (3pi)/(2k)]`
Thus, `(pi)/(4) = (3pi)/(2k) -(pi)/(2k)`
or `(pi)/(k) = (pi)/(4)`
or `k = 4`