Evaluating f(θ) at the boundary points:
For θ=6π, the first and second rows of the determinant are identical, so f(6π)=0.
For θ=3π, the first and third rows of the determinant are identical, so f(3π)=0.
Since f(θ) is continuous on [6π,3π] and differentiable on (6π,3π), by Rolle's Theorem, there exists at least one θ∈(6π,3π) such that:
f′(θ)=0 Thus, statement (I) is TRUE.
Expanding the determinant along the first row gives a non-constant function of the form f(θ)=c1sinθ+c2cosθ+c3anθ. Since f(θ) is not constant over the interval, its derivative f′(θ) is not identically zero everywhere on (6π,3π). Therefore, there must exist some θ∈(6π,3π) such that:
f′(θ)=0 Thus, statement (II) is also TRUE.
Hence, both statements (I) and (II) are TRUE, matching option C.