Let φ,ψ be independent solutions of y″+(2−cos x)y=0 on ℝ. Suppose α<β,φ(α)=φ(β)=0 and φ≠0 on (α,β). P:φ′(α)φ′(β)>0. Q:φ(x)ψ(x)≠0 throughout (α,β). Choose truth values.
Zeros of a nonzero solution are simple by uniqueness. Its constant sign between consecutive zeros makes its endpoint slopes opposite in sign, so their product is negative. Sturm separation says the independent solution ψ has one zero between consecutive zeros of φ, refuting Q. **C**.
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