If π ≤ x ≤ 2π and y = cos–1 (cos x), find


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)


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