Cho hàm số f ( x ) thỏa mãn f ( 0 ) = 1/3 và ( √ x + √ x + 2 ) f ′ ( x ) = 1 , ∀ x ≥ − 2 . Mỗi phát biểu sau đây là đúng hay sai?
Đáp án
Phát biểu | Đúng | Sai |
\(f\left( x \right) = \frac{1}{3}\left( {\sqrt {{{(x - 2)}^3}} - \sqrt {{x^3}} + 1} \right)\) | X | |
\(\int\limits_0^2 {f\left( x \right)dx = \frac{{54 - 16\sqrt 2 }}{{15}}} \) | X |
Giải thích
Ta có: \(\left( {\sqrt x + \sqrt {x + 2} } \right)f'\left( x \right) = 1 \Leftrightarrow f'\left( x \right) = \frac{1}{{\sqrt x + \sqrt {x + 2} }} \Leftrightarrow \mathop \smallint \nolimits^ f'\left( x \right)dx = \mathop \smallint \nolimits^ \frac{{dx}}{{\sqrt x + \sqrt {x + 2} }}\)
\( \Leftrightarrow f\left( x \right) = \frac{1}{2}\mathop \smallint \nolimits^ \left( {\sqrt {x + 2} - \sqrt x } \right)dx = \frac{1}{2}.\frac{2}{3}\left( {\sqrt {{{(x + 2)}^3}} - \sqrt {{x^3}} } \right) + C = \frac{1}{3}\left( {\sqrt {{{(x + 2)}^3}} - \sqrt {{x^3}} } \right) + C\)
\(f\left( 0 \right) = \frac{1}{3} \Leftrightarrow \frac{1}{3}.2\sqrt 2 + C = \frac{1}{3} \Leftrightarrow C = \frac{1}{3}\left( {1 - 2\sqrt 2 } \right)\)
\( \Rightarrow f\left( x \right) = \frac{1}{3}\left( {\sqrt {{{(x + 2)}^3}} - \sqrt {{x^3}} + 1 - 2\sqrt 2 } \right)\)
\[ \Rightarrow \mathop \smallint \nolimits^ _0^2f\left( x \right)dx = \frac{1}{3}\mathop \smallint \nolimits^ _0^2\left( {\sqrt {{{(x + 2)}^3}} - \sqrt {{x^3}} + 1 - 2\sqrt 2 } \right)dx = \left. {\frac{1}{3}.\left( {\frac{2}{5}\sqrt {{{(x + 2)}^5}} - \frac{2}{5}\sqrt {{x^5}} + (1 - 2\sqrt 2 )x} \right)} \right|_0^2 = \frac{{74 - 36\sqrt 2 }}{{15}}\]