[tex]A^2= x^2 .(99+\sqrt{101-x^2})^2 \leq x^2 .(\sqrt{99}.\sqrt{99}+\sqrt{101-x^2})^2 \displaystyle \leq_ {(Bunyakovski)}x^2(99+101-x^2)(99+1)=100x^2.(200-x^2)\leq 100.\frac{(x^2+200-x^2)^2}{4}=1000000\\\Leftrightarrow A^2\leq 1000000\Leftrightarrow -1000\leq A\leq 1000[/tex]