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A differentiable function $f :\mathbb R \to\mathbb R$ satisfies:

  1. $f(2) = 5$
  2. $f(3) = 8$
  3. $f'(x) \le 4 \sin^2(\pi x)$ for all $x \in\mathbb R$.

How many values of $s \ge 2$ satisfy $f(s) = s^2$ ?

I could conclude that $x < 3$ by the 3rd condition, and tried to use the Mean Value Theorem but found it hard to go on. Can anyone give some hints?

Thanks

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Let $g(x)=x^2$ and define a new function $h(x)=g(x)-f(x)$. The question asks for the number of solutions to $f(x)=g(x)$ or equivalently $h(x)=g(x)-f(x)=0$ where $x\ge2$.

All we have to do is note that when $x>2$, we have $$h'(x)=(g(x)-f(x))'=g'(x)-f'(x)=2x-f'(x)\ge 2x-4\sin^2(\pi x)\ge 2x-4>0$$

Now since $h(2)=g(2)-f(2)=-1<0$ while $h(3)=g(3)-f(3)=1>0$ we know by the intermediate value theorem that there is at least one solution to $h(x)=0$ and since $h'(x)>0$ for $x>2$, we know this solution is unique.

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