Đáp án:
\(\left\{ \begin{array}{l}
f{\left( x \right)_{\min }} = \frac{{2 + \sqrt 2 }}{2} \Leftrightarrow x = \frac{1}{2}\\
f{\left( x \right)_{\max }} = \frac{{2 + \sqrt 3 }}{2} \Leftrightarrow x = \frac{1}{4}
\end{array} \right.\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = \sqrt {2 - x + 2\sqrt {1 - x} } \\
= \sqrt {\left( {1 - x} \right) + 2\sqrt {1 - x} + 1} = \sqrt {{{\left( {\sqrt {1 - x} + 1} \right)}^2}} = \sqrt {1 - x} + 1\\
\frac{1}{4} \le x \le \frac{1}{2} \Rightarrow \frac{1}{2} \le 1 - x \le \frac{3}{4}\\
\Rightarrow \frac{{\sqrt 2 }}{2} \le \sqrt {1 - x} \le \frac{{\sqrt 3 }}{2}\\
\Leftrightarrow \frac{{2 + \sqrt 2 }}{2} \le f\left( x \right) \le \frac{{\sqrt 3 + 2}}{2}
\end{array}\)
