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Question
Verify Rolle’s theorem for the following functions : f(x) = `sin(x/2), x ∈ [0, 2pi]`
Solution
The function f(x) = `sin(x/2)` is continuous on [0, 2π] and differetiable on (0, 2π).
Now, f(0) = sin0 = 0
and f(2π) = sin π = 0
∴ f(0) = f(2π)
Thus, the functiontisfies all the conditions of Rolle's theorem.
∴ there exists c ∈ (0, 2π) such that f'(c) = 0.
Now, f(x) = `sin(x/2)`
∴ f'(x) = `d/dx[sin (x/2)]`
= `cos(x/2).d/dx(x/2)`
= `cos(x/2).(1)/(2) = (1)/(2) cos (x/2)`
∴ f'(c) = `(1)/(2)cos(c/2)`
∴ f'(c) = `0 "gives" (1)/(2)cos(c/2)` = 0
∴ `cos (c/2)` = 0
∴ `cos c/(2) = cos pi/(2) = cos (3pi)/(2) = cos (5pi)/(2)` = ...
∴ `c/(2)i/(2), (3pi)/(2), (5pi)/(2)`, ...
∴ c = π, 3π 5π, ...
But π ∈ (0, 2π)
∴ c = π
Hence, the Rolle’s theorem is verified.
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