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

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(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} -

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$f(x) = 0$ 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 $x \geq 0$.

Column $I$	Column $II$
$A$. $x\|x\|$	$I$. Strictly increasing and continuous in $(-1,1)$
$B$. $\sqrt{\|x\|}$	$II$. Continuous but not differentiable in $(-1,1)$
$C$. $x+[x]$	$III$. Differentiable in $(-1,1)$
$D$. $\|x-1\|+\|x+1\|+\|x\|$	$IV$. Differentiable in $(-1,0) \cup (0,1)$
	$V$. Strictly increasing and not differentiable in $(-1,1)$

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 $[0, \infty)$:

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