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Consider the function y = cos (kx), where k > 0. The value of k has been chosen
so that a circle can be drawn, centred at the origin, which has exactly two points
of intersection with the graph of the function and so that the circle is never
above the graph of the function. The point P(a, b) is the point of intersection in
the first quadrant, so a > 0 and b > 0.

The vector joining the origin to the point P(a, b) is perpendicular to the tangent
to the graph of the function at that point.

Show that k > 1.

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