Solution :
Let f(x) = cos x – 1, Clearly f(x) is continous on \([{\pi\over 2}, {3\pi\over 2}]\) and differentiable on \(({\pi\over 2}, {3\pi\over 2})\).
Also, f\((\pi\over 2)\) = \(cos {\pi\over 2}\) – 1 = -1 = f\((3\pi\over 2)\).
Thus, all the conditions of rolle’s theorem are satisfied. Consequently,there exist at least one point c \(\in\) \(({\pi\over 2}, {3\pi\over 2})\) for which f'(c) = 0. But,
f'(c) = 0 \(\implies\) -sin c = 0 \(\implies\) c = \(\pi\)
\(\therefore\) 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.
Similar Questions
Find the approximate value of f(3.02), where f(x) = \(3x^2 + 5x + 3\).
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