If $P(x)=2013x^{2012}-2012x^{2011}-16x+8$,then $P(x)=0$ for $x\in\left[0,8^{\frac{1}{2011}}\right]$ has 
If $P(x)=2013x^{2012}-2012x^{2011}-16x+8$, then $P(x)=0$ for $x\in\left[0,8^{\frac{1}{2011}}\right]$ has
  
  
*
  
*exactly one real root.
  
*no real root.
  
*at least one and at most two real roots.
  
*at least two real roots.
  


$$\begin{align}P(x)&=2013x^{2012}-2012x^{2011}-16x+8  \\
P'(x)&=2013\cdot 2012x^{2011}-2012\cdot 2011x^{2010}-16\end{align}$$
$P'(0)$ is negative and $P'\left(8^{\frac{1}{2011}}\right)$ is positive. So $P'(x)$ has at least one real root. What should be the correct choice. In my previous question, it has been proved by answerers that if $f'(x)$ has at least one real root then $f(x)$ has not necessarily at least two real roots.
 A: We have $p(0)=8,p(1)=-7$ so there is at least one root in [0,1]. Put $\alpha=8^{1/2011}$. Then $p(\alpha)$ is approx 0. So we may have another root just inside the interval $[0,\alpha]$.
Rather than do any more work, look carefully at the options given. (2) is definitely wrong. (1) might be correct, and (4) might be correct, but (3) is certainly correct.
Well, not quite, because something weird might be happening so that we got 3 roots just below $\alpha$. So we check $p'(x)=2013\cdot2012x^{2011}-2012\cdot2011x^{2010}-16$. That is $-16$ at $x=0$, and 4008 and rapidly increasing at $x=1$, so clearly there is only one root near $\alpha$.
Hence pick (3).
[In fact, more careful checking shows that there are exactly 2 roots in the interval, so (4) is also true]
A: Let $Q(x) = x^{2013} - x^{2012} -8x^2 +8x + c $. Note that $Q'(x)=P(x)$. Since $Q(0)=Q(1)=Q(8^\frac{1}{2011})$, we apply Rolle's Theorem to each of the intervals $[0,1]$ and $[1,8^\frac{1}{2011}]$ to conclude that $Q'(x)=P(x)$ has at least a root in each of the intervals $(0,1)$ and $(1,8^\frac{1}{2011})$. Hence $P(x)$ has at least two roots.
Remark: This solution was basically what was suggested by the OP in a repeated asking of the question.
