Let f(x) be a non-negative differentiable fun | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $f(x) \geq 0$ and $f^{\prime}(x) \leq 2f(x)$.Consider the function $g(x) = f(x) e^{-2x}$.Then $g^{\prime}(x) = f^{\prime}(x) e^{-2x} -

Vedclass · Accepted Answer

$f(x) = 0$ for all $x \geq 0$

Vedclass · Answer

(A) Given that $f(x) \geq 0$ and $f^{\prime}(x) \leq 2f(x)$.Consider the function $g(x) = f(x) e^{-2x}$.Then $g^{\prime}(x) = f^{\prime}(x) e^{-2x} - 2f(x) e^{-2x} = e^{-2x} (f^{\prime}(x) - 2f(x))$.Since $f^{\prime}(x) \leq 2f(x)$,we have $f^{\prime}(x) - 2f(x) \leq 0$.Thus,$g^{\prime}(x) \leq 0$ for all $x > 0$,which means $g(x)$ is a non-increasing function.Since $g(0) = f(0) e^0 = 0 	imes 1 = 0$ and $g(x)$ is non-increasing for $x \geq 0$,we must have $g(x) \leq g(0) = 0$ for all $x \geq 0$.However,$f(x) \geq 0$ and $e^{-2x} > 0$,so $g(x) = f(x) e^{-2x} \geq 0$ for all $x \geq 0$.Therefore,$g(x) = 0$ for all $x \geq 0$,which implies $f(x) = 0$ for all $x \geq 0$.

Let $f(x)$ be a non-negative differentiable function on $[0, \infty)$ such that $f(0)=0$ and $f^{\prime}(x) \leq 2 f(x)$ for all $x>0$. Then,on $[0, \infty)$:

Similar Questions

Let $f(x) = ax^2 - b|x|$,where $a$ and $b$ are constants. Then at $x = 0$,$f(x)$ has

$A$ function $f(x) = \begin{cases} 1 + x, & x \le 2 \\ 5 - x, & x > 2 \end{cases}$ is

The number of points,where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$,$x \in R$ cuts the $x$-axis,is equal to

The value of $f(4)-f(3)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module