Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ia/paper-1/12e/solution

Rolle theorem states that if , is continuous on , differentiable on , and , then there is with .
If is constant, any interior point works. Otherwise the extreme value theorem gives an attained maximum and minimum on the interval, and at least one of them differs from the common endpoint value. That extremum must occur at an interior point . At an interior maximum, for small positive the difference quotient is nonpositive, while for negative it is nonnegative. Existence of the common derivative forces it to be both nonpositive and nonnegative, hence zero. The same argument with inequalities reversed applies to an interior minimum. This proves Rolle theorem, including the derivative-at-an-extremum step rather than assuming it.
Now let the nonconstant polynomial have distinct real roots , with multiplicities adding to its degree . If and , differentiation gives
The bracket is nonzero at , so its multiplicity as a root of is exactly . These roots already contribute real roots counted with multiplicity. Rolle theorem supplies an additional root of in each of the disjoint intervals . These additional roots are distinct from each other and from all the .
We have therefore exhibited at least real roots counted with multiplicity. The derivative has degree , so there is no room for any additional nonreal roots. Every root of is real. This is the Rolle root count with multiplicities; for the nonzero constant derivative has no roots, as expected.
For failure of the converse, take
The derivative has only the real root zero, of multiplicity two. But and the quadratic factor has roots , which are nonreal. Thus real-rootedness of the derivative does not imply real-rootedness of the original polynomial.

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