TITLE: $P$ is a monic polynomial of degree $n$ , then which are correct?
QUESTION [7 upvotes]: Suppose that $P$ is a monic polynomial of degree $n$ in one variable with real coefficients and $K$ is a real number. Then which of the following statements are necessarily correct ?

If $n$ is even and $K>0$ then there exists $x_0\in \mathbb R$ such that $P(x_0)=Ke^{x_0}$.
If $n$ is odd and $K<0$ , then there exists $x_0 \in \mathbb{R}$ such that $P(x_0)=Ke^{x_0}$.
For any natural number $n$, and $0<K<1$ then there exists $x_0 \in \mathbb{R}$ such that $P(x_0)=Ke^{x_0}$.
If $n$ is odd and $K \in \mathbb R$ , then there exists $x_0 \in \mathbb{R}$ such that $P(x_0)=Ke^{x_0}$.


For (3), consider, $P(x)=x^2+5$. Then , there does not exist $x_0\in \mathbb R$ such that $0<e^{-x_0}P(x_0)<1$. So it is FALSE.
But what about the others ?

REPLY [10 votes]: You need to use the intermediate value theorem:
1) Since $n$ is even, $\lim\limits_{x\to-\infty} P=\infty$. From that, we know that $\lim\limits_{x\to-\infty}\frac{P(x)}{e^x}=\infty$. We also know that $\lim\limits_{x\to\infty}\frac{P(x)}{e^x}=0$. Therefore, there is a value $x_0$ verifying $\frac{P(x_0)}{e^{x_0}}=K$ (Since $0<K$).
2) Since $n$ is odd, $\lim\limits_{x\to-\infty} P=-\infty$. From that, we know that $\lim\limits_{x\to-\infty}\frac{P(x)}{e^x}=-\infty$. We also know that $\lim\limits_{x\to\infty}\frac{P(x)}{e^x}=0$. Therefore, there is a value $x_0$ verifying $\frac{P(x_0)}{e^{x_0}}=K$ (Since $0>K$).  
3) Your justification seems to be false, since $\lim\limits_{x_0\to\infty} e^{-x_0}P(x_0)=0^+$. Instead consider $K=\frac{1}{2}$, and $P(x)=x-1$.  
4) Consider $K=1$, and $P(x)=x-1$.