The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all trigonometric functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval. Now, you need to check if `f(pi)...

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The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all trigonometric functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval. Now, you need to check if `f(pi) = f(2pi).`

`f(pi) =sec (pi)= 1/(cos pi) = 1/(-1) = -1`

`f(2pi) =sec 2pi = 1/(cos 2pi) = 1/1 = 1`

**Since one of all the three conditions is not valid, `f(pi) != f(2pi)` , you cannot apply Rolle's theorem.**