cho hàm số f(x) xác định trên R /(-1,1) thỏa mãn f'(x) =1/(x ∧2-1) ,f(-3)+f(3)=0/ f(-1/2) + f(1/2) =2 tính f(-2) +f(0) + f(4) A 3+ ㏑(3/5) B 5_ ㏑

Đáp án: D

Giải thích các bước giải:

$\begin{array}{l}
f’\left( x \right) = \frac{1}{{{x^2} – 1}} = \frac{1}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
 = \frac{1}{2}.\frac{{\left( {x + 1} \right) – \left( {x – 1} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = \frac{1}{2}\left( {\frac{1}{{x – 1}} – \frac{1}{{x + 1}}} \right)\\
 \Rightarrow f\left( x \right) = \int {f’\left( x \right)dx} \\
 \Rightarrow f\left( x \right) = \int {\frac{1}{2}\left( {\frac{1}{{x – 1}} – \frac{1}{{x + 1}}} \right)dx} \\
 \Rightarrow f\left( x \right) = \frac{1}{2}\ln \left| {x – 1} \right| – \frac{1}{2}\ln \left| {x + 1} \right| + C\\
 \Rightarrow \left\{ \begin{array}{l}
khi:x \in \left( { – 1;1} \right) \Rightarrow f\left( x \right) = \frac{1}{2}\ln \left| {x – 1} \right| – \frac{1}{2}\ln \left| {x + 1} \right| + a\\
khi:x \notin \left( { – 1;1} \right) \Rightarrow f\left( x \right) = \frac{1}{2}\ln \left| {x – 1} \right| – \frac{1}{2}\ln \left| {x + 1} \right| + b
\end{array} \right.\\
Do:\left\{ \begin{array}{l}
f\left( { – 3} \right) + f\left( 3 \right) = 0\\
f\left( { – \frac{1}{2}} \right) + f\left( {\frac{1}{2}} \right) = 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
\frac{1}{2}.\ln \left( 4 \right) – \frac{1}{2}\ln 2 + \frac{1}{2}\ln 2 – \frac{1}{2}\ln 4 + 2a = 0\\
\frac{1}{2}\left( {\ln \frac{3}{2} – \ln \frac{1}{2} + \ln \frac{1}{2} – \ln \frac{3}{2}} \right) + 2b = 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
a = 0\\
b = 1
\end{array} \right.\\
 \Rightarrow f\left( { – 2} \right) + f\left( 0 \right) + f\left( 4 \right)\\
 = \frac{1}{2}\ln \left| {\frac{{ – 2 – 1}}{{ – 2 + 1}}} \right| + 1 + \frac{1}{2}\ln \left| {\frac{{0 – 1}}{{0 + 1}}} \right| + 0 + \frac{1}{2}\ln \left| {\frac{{4 – 1}}{{4 + 1}}} \right| + 1\\
 = \frac{1}{2}\ln 3 + \frac{1}{2}\ln \frac{3}{5} + 2\\
 = \ln 3 – \ln \sqrt 5  + 2\\
 = 2 + \ln \frac{3}{{\sqrt 5 }}
\end{array}$