$P(x)$ is a polynomial function such that $P(1)=0$, $P^{\prime}(x)P(x), \forall x1$. $\forall x1

Question

$P(x)$ is a polynomial function such that $P(1)=0$, $P^{\prime}(x)>P(x), \forall x>1$.  $\forall x>1$, which of the following statements is true?

Professor PC Thomas Classes · Answer

$P^{\prime}(x)-P(x)>0, \forall x>1$\ $\Rightarrow \quad e^{-x} P^{\prime}(x)-e^{-x} P(x)>0, \forall x>1$\ $\Rightarrow \quad \frac{d}{d x}\left(e^{-x} \cdot P(x)ight)>0, \forall x>1$\ $\Rightarrow \quad e^{-x} P(x)$ is an increasing function of\ $x \forall x \in[1, \infty]$\ (as $P(x)$ is a continuous function)\ Thus $x>1$\ $\Rightarrow \quad e^{-x} P(x)>e^{-1} P(1)$\ $\Rightarrow \quad e^{-x} P(x)>0$, as $P(1)=0$\ $\Rightarrow \quad P(x)>0$\