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Suppose $x:[0,\infty) \to [0,\infty)$ is continuous and $x(0)=0 $ If $$(x(t))^2 \leq 2+\int_{0}^{t}x\left(s\right){\rm d}s . $$ for all $t \geq 0$, Then which of the following is true?

$a$) $x(\sqrt2)\in[0,2]$

$b$) $x(\sqrt2)\in [0,\frac3{\sqrt2}]$

$c$) $x(\sqrt2) \in [\frac5{\sqrt2},\frac7{\sqrt2}]$

$d$) $x(\sqrt2)\in [10,\infty)$

I tried function $x=0$ and $x=\frac{3t}2$. And I want to discard $a$ or $b$ ...(as I think $a$ should be wrong and $b$ is answer) not arguments like $a$ is true then $b$ should be true and only one answer can be correct so $b$ is correct.

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    $\begingroup$ Please, use math.meta.stackexchange.com/questions/5020/…, for a better formatting of your question $\endgroup$
    – Arnaldo
    Jul 21, 2017 at 16:13
  • $\begingroup$ Just tell me please what to use for greater then or equal $\endgroup$ Jul 21, 2017 at 16:14
  • $\begingroup$ Do you mean $X(t)=x(t)$? $\endgroup$
    – Arnaldo
    Jul 21, 2017 at 16:17
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    $\begingroup$ For future references \geq or \ge for greater or equal to $\endgroup$ Jul 21, 2017 at 16:17
  • $\begingroup$ Great I was about to wrote that but have to learn a lot in writing ques in this $\endgroup$ Jul 21, 2017 at 16:20

1 Answer 1

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I think the correct inequality is $x(t)^2\le2+\int_{0}^{t}x(s)\,ds$. In that case, let $y(t)=2+\int_{0}^{t}x(s)\,ds$. We have $x(t)=y'(t)\le\sqrt{y(t)}$, so $$\frac{y'(t)}{2\sqrt{y(t)}}\le\frac12$$ and thus $$x(t)\le\sqrt{y(t)}=\sqrt{y(0)}+\int^t_0\frac{y'(t)}{2\sqrt{y(t)}}\,dt\le\sqrt{y(0)}+\frac{t}2=\sqrt{2}+\frac{t}2.$$ So $x(\sqrt{2})\le\frac32\sqrt{2}=\frac3{\sqrt{2}},$ meaning it's b).

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