If π ≤ x ≤ 2π and

y = cos^-1 (cos x)

for x (π, 2π)

y = cos^-1(cos x)

= cos^-1(cos (π + (x - π)))

= cos^-1(-cos (x-π))

= π – (x - π)

= 2π - x

[Since, cos(π+x) = - cos x and cos^-1(-x) = π-x]

So,

For cos^-1(cos x), x = n are the ‘sharp corners’ where slope changes from 1 to -1 or vice versa, i.e., the points where the curves are not differentiable.

So, for x [π,2π]

(Ans)