Solution :
Let f(x) = cos x – 1, Clearly f(x) is continous on [π2,3π2] and differentiable on (π2,3π2).
Also, f(π2) = cosπ2 – 1 = -1 = f(3π2).
Thus, all the conditions of rolle’s theorem are satisfied. Consequently,there exist at least one point c ∈ (π2,3π2) for which f'(c) = 0. But,
f'(c) = 0 ⟹ -sin c = 0 ⟹ c = π
∴ f(c) = cos \pi – 1 = -2
By the geometric interpretation of rolle’s theorem (\pi, -2) is the point on y = cos x – 1 where tangent is parallel to x-axis.
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