Đáp án:
$A.$
Giải thích các bước giải:
$y=x+\sqrt{1-x^2} \ \ \ \ D=[-1;1]\\ y'=1+\dfrac{-2x}{2\sqrt{1-x^2}}\\ =1-\dfrac{x}{\sqrt{1-x^2}}\\ =\dfrac{\sqrt{1-x^2}-x}{\sqrt{1-x^2}}\\ y'=0 \Leftrightarrow \sqrt{1-x^2}-x=0\\ \Leftrightarrow \sqrt{1-x^2}=x(x \ge 0)\\ \Leftrightarrow 1-x^2=x^2\\ \Leftrightarrow 2x^2=1\\ \Leftrightarrow x^2=\dfrac{1}{2}\\ \Leftrightarrow \left[\begin{array}{l} x=\dfrac{\sqrt{2}}{2}\\x=-\dfrac{\sqrt{2}}{2}(L)\end{array} \right.$
BBT:
\begin{array}{|c|ccccccccc|} \hline x&-1&&\dfrac{\sqrt{2}}{2}&&1\\\hline y'&&+&0&-&\\\hline &&&\sqrt{2}\\y&&\nearrow&&\searrow&\\&-1&&&&1\\\hline\end{array}
Dựa vào BBT, $\Rightarrow M=\sqrt{2},m=-1.$
