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Pick out the true statements:

a. If $P$ is a polynomial in one variable with real coefficients which has all its roots real, then its derivative $P'$ has all its roots real as well.

b. The equation $\cos(\sin x) = x$ has exactly one solution in the interval $[0,\pi/2]$.

c. $\cos x > 1 −x^2/2$ for all $x > 0$.

My thoughts:

By Rolles theorem, (a) must be true, but I am not sure about the others.

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(a) Rolles Theorem, like you mentioned.

(b) Consider $f(x) = \cos (\sin x) - x$, then show that $f'(x) = \sin (\sin x) \sin x - 1 < 0$ (Hint: $\sin x < x$ in that range)

(c) The MacLaurin's Expansion of $\cos x$ is ... Use the remainder theorem.

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a) Rolle's Theorem is indeed useful here. It works very nicely if all the roots of $P(x)$ have multiplicity $1$. But we need also to deal with polynomials $P(x)$ like $x^3(x-1)^2 (x-2)$. The additional result that we need is that if $(x-a)^k$ divides $P(x)$, where $k\ge 2$, then $(x-a)^{k-1}$ divides $P'(x)$. This is easy to prove. Let $P(x)=(x-a)^kQ(x)$, and use the Product Rule.

Suppose $Q(x)$ has degree $d\ge 1$. Now you will be able to show that all of the $d-1$ complex roots of $P'(x)=0$ are real.

b) One standard approach to a question like this one is to let $f(x)=\cos(\sin x)-x$. Note that $f(0)$ is positive and $f(1)$ is negative, so by the Intermediate Value Theorem there is a root in between. Note that $f'(x)=(\cos x)(-\sin(\cos x))-1$, so $f$ is decreasing steadily on our interval.

c) Use the same basic strategy as in b). Let $f(x)=\cos x-(1-\frac{x^2}{2})$. We have $f(0)=0$. It is enough to show that $f$ is increasing on the interval $(0,\infty)$. We have $f'(x)=-\sin x+x$.

To show that this is $\gt 0$ on $(0,\infty)$, use the same basic strategy. Let $g(x)=x-\sin x$. Then $g(0)=0$ and $g'(x)=1-\cos x$.

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b. Try graphing both functions to get an idea of the answer. To prove that it's correct: what does the derivative of $\cos(\sin(x))$ look like on $[0, \pi /2]$?

c. Again, a graph can do wonders. Do you know the power series expansion $\cos(x)=1-x^2/2+x^4/4!-\dots$ ? That can help quite a bit in confirming your answer.

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