Đáp án đúng: A
Phương pháp giải:
Tính công thức tổng quát của \(f'\left( x \right)\)
Đạo hàm của hàm số \(y = \ln g\left( x \right)\) là \(y' = \dfrac{{g'\left( x \right)}}{{g\left( x \right)}}\)
Giải chi tiết:Ta có :
\(\begin{array}{l}f\left( x \right) = \ln \left( {1 - \dfrac{1}{{{x^2}}}} \right) = \ln \left( {\dfrac{{{x^2} - 1}}{{{x^2}}}} \right)\\\left( {\dfrac{{{x^2} - 1}}{{{x^2}}}} \right)' = \dfrac{{2x.{x^2} - 2x.\left( {{x^2} - 1} \right)}}{{{x^4}}} = \dfrac{{2x}}{{{x^4}}} = \dfrac{2}{{{x^3}}}\end{array}\)
\(\begin{array}{l} \Rightarrow f'\left( x \right) = \dfrac{{\dfrac{2}{{{x^3}}}}}{{\dfrac{{{x^2} - 1}}{{{x^2}}}}} = \dfrac{2}{{{x^3}}}.\dfrac{{{x^2}}}{{{x^2} - 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{2}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{1}{x}\left( {\dfrac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{x}\left( {\dfrac{1}{{x - 1}} - \dfrac{1}{{x + 1}}} \right) = \dfrac{1}{x}.\dfrac{1}{{\left( {x - 1} \right)}} - \dfrac{1}{{x + 1}}.\dfrac{1}{x}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\dfrac{1}{{x - 1}} - \dfrac{1}{x}} \right) - \left( {\dfrac{1}{x} - \dfrac{1}{{x + 1}}} \right) = \dfrac{1}{{x - 1}} + \dfrac{1}{{x + 1}} - \dfrac{2}{x}\end{array}\)
Do đó ta có :
\(f'\left( 2 \right) + f'\left( 3 \right) + f'\left( 4 \right) + .... + f'\left( {2019} \right)\)
\( = \left( {\dfrac{1}{1} + \dfrac{1}{3} - \dfrac{2}{2}} \right) + \left( {\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{2}{3}} \right) + \left( {\dfrac{1}{3} + \dfrac{1}{5} - \dfrac{2}{4}} \right)\)\( + ..... + \left( {\dfrac{1}{{2018}} + \dfrac{1}{{2020}} - \dfrac{2}{{2019}}} \right)\)
\( = \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + .... + \dfrac{1}{{2018}}} \right)\)\( + \left( {\dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + .... + \dfrac{1}{{2020}}} \right)\)\( - 2\left( {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + .... + \dfrac{1}{{2019}}} \right)\)
\( = 1 + \dfrac{1}{{2020}} - \dfrac{1}{{2019}} - \dfrac{1}{2}\)
\( = \dfrac{1}{2} - \dfrac{1}{{2019.2020}} = \dfrac{{2019.2020 - 1}}{{2.2019.2020}}\)
Do đó \(b = 2a\)
Chọn A.
