Question

If $P\left(2\right)=0$ and $P^{\prime }\left(x\right)+20P\left(x\right)<0$ for all $x>0.$Then the number of solution(s) for $P\left(x\right)=1$ for $x>2,$ is

Answer 1

Given : $P^{\prime }\left(x\right)+20P\left(x\right)<0$
$\Rightarrow \left[P^{\prime }\right(x)+20P(x\left)\right]e^{20x}<0$
$\Rightarrow \frac{d\left[e^{20x}P\right(x\left)\right]}{dx}<0$
So, $f\left(x\right)=e^{20x}P\left(x\right)$ is a decreasing function for all $x>0$
For $x>2:$
$\Rightarrow f\left(x\right)<f\left(2\right)$
$\Rightarrow e^{20x}P\left(x\right)<0(\because P(2)=0)$
$\therefore P\left(x\right)<0(\because e^{20x}>0\forall x\in R)$
So, there is no solution for $P\left(x\right)=1(\because P(x)<0)$