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Since, both have at most one zero.
Hence, has at most one zero in (0,1).
Hence, (y_{k}(t)) has at most one zero at ([k,k+1]), which implies that (x t)) has at most one zero at ([k,k+1]).
More precisely, has at most one zero in the whole line.
We claim that has at most one zero in (0,1).
Hence, (bar {x}_{k}(t)) has at most one zero in ([k,k+1]), which implies (bar{x}(t)) has at most one zero in ([k,k+1]) for any integer k. □.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com