Suppose $x:[0,\infty)\to [0,\infty)$ be continuous function and $x(0)=0$. If $\displaystyle (x(t))^2 \le 2+\int_0^t x(s)\,ds$ for all $t\ge 0$ then which of the following is true?
(A) $x(\sqrt 2)\in [0,2]$.
(B) $x(\sqrt 2)\in [0,3/\sqrt 2]$.
(C) $x(\sqrt 2)\in [5/\sqrt 2,7/\sqrt 2]$.
(D) $x(\sqrt 2)\in [10,\infty]$.
As it is given an inequality we can not differentiate it. How can I proceed to get the proper interval of solution ?
Take the square root at both sides and then divide by the RHS and multiply by $1/2 $ to obtain
$$ \frac{x(t)}{ 2\sqrt{ 2 + \int _0^t x(t)dt }}\leq \frac{1}{2} $$ Therefore, $$\left (\sqrt{ 2 + \int _0^t x(t)dt }\right )'\leq \frac{1}{2} $$ so by integrating $$\sqrt{ 2 + \int _0^t x(t)dt }\leq \frac{t}{2} +\sqrt{2} $$ Therefore, $$x(t)\leq \frac{t}{2} +\sqrt{2} \Longrightarrow x(\sqrt2) \leq \frac{3}{\sqrt{2}}$$
Also, if you solve $$ \frac{x(t)}{ 2\sqrt{ 2 + \int _0^t x(t)dt }}= \frac{1}{2} $$ you find that $x(\sqrt{2}) = \frac{3}{\sqrt{2}}$ Therefore, (b) is correct.