Courses
Courses for Kids
Free study material
Offline Centres
More Last updated date: 30th Nov 2023
Total views: 280.2k
Views today: 4.80k

# If $P(1)=0$ and $\dfrac{d\text{P}\left( \text{x} \right)}{dx}>\text{P}\left( \text{x} \right)$ for all $x\ge 1$, then prove $\text{P}\left( \text{x} \right)>0$ for all $x>1$ Verified
280.2k+ views
Hint: Type of question is based on the function, as it is asked to prove that $\text{P}\left( \text{x} \right)$ is greater than zero when x is greater than 1. As the given condition we have is $\text{P}\left( 1 \right)$ is equal to zero, and its differentiation is greater than $\text{P}\left( \text{x} \right)$. When x is greater than 1.

Complete step-by-step solution:
So moving ahead with the question; i.e.
$\dfrac{d\text{P}\left( \text{x} \right)}{dx}>\text{P}\left( \text{x} \right)$
Bring both to same side, then we will get;
$\dfrac{d\text{P}\left( \text{x} \right)}{dx}-\text{P}\left( \text{x} \right)>0$
Try to make it as a one single function, rather than having two function, as we have i.e. $\dfrac{d\text{P}\left( \text{x} \right)}{dx}$and$\text{P}\left( \text{x} \right)$, as it will be easy to find the value of function. So Multiplying both side by ${{e}^{-x}}$, as it will look like we had a differentiation of $\left[ {{e}^{-x}}\text{P}\left( \text{x} \right) \right]$; so we will get;
\begin{align} & {{e}^{-x}}\dfrac{d\text{P}\left( \text{x} \right)}{dx}-{{e}^{-x}}\text{P}\left( \text{x} \right)>0 \\ & \dfrac{d\left[ {{e}^{-x}}\text{P}\left( \text{x} \right) \right]}{dx}>0 \\ \end{align}
As we know that if differentiation of function is increasing than the function is increasing, so we can write it as; ${{e}^{-x}}\text{P}\left( \text{x} \right)$ is increasing function.
So we can say that ${{e}^{-x}}\text{P}\left( \text{x} \right)$ will give a minimum at its minimum value of ‘x’ and then it will increase. According to question 1 is the minimum value we have in domain, so we can write ity as;
${{e}^{-x}}\text{P}\left( \text{x} \right)>{{e}^{-1}}\text{P}\left( 1 \right)$in which $x>1$
So on solving it; we will get
\begin{align} & {{e}^{-x}}\text{P}\left( \text{x} \right)>{{e}^{-1}}\text{P}\left( 1 \right) \\ & {{e}^{-x}}\text{P}\left( \text{x} \right)>{{e}^{-1}}(0) \\ & {{e}^{-x}}\text{P}\left( \text{x} \right)>0 \\ \end{align}
As we know that for ${{e}^{-x}}\text{P}\left( \text{x} \right)$ to be greater than zero, than both ${{e}^{-x}}$ and $\text{P}\left( \text{x} \right)$ must be greater than zero or less than zero, but as we know that ${{e}^{-x}}$is always greater than zero then $\text{P}\left( \text{x} \right)$ must be greater than zero.
Hence we had proved that $\text{P}\left( \text{x} \right)>0$

Note: Rather than multiplying by ${{e}^{-x}}$ you can also multiply by some variable which gives the same answer after differentiation. Then also you will get the same answer. Moreover, in the way we had reduce the two function into one $\dfrac{d\text{P}\left( \text{x} \right)}{dx}$ and $\text{P}\left( \text{x} \right)$, through the multiplication of ${{e}^{-x}}$ is the common way how we deal with the type of question.