Câu hỏi
Cho hàm số \(f\left( x \right)\) liên tục trên đoạn \(\left[ {0;4} \right]\) thỏa mãn \(f''\left( x \right)f\left( x \right) + \frac{{{{\left[ {f\left( x \right)} \right]}^2}}}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}} = {\left[ {f'\left( x \right)} \right]^2}\) và \(f\left( x \right) > 0\) với mọi \(x \in \left[ {0;4} \right]\). Biết rằng \(f'\left( 0 \right) = f\left( 0 \right) = 1\), giá trị của \(f\left( 4 \right)\) bằng:
- A \({e^2}\)
- B \(2e\)
- C \({e^3}\)
- D \({e^2} + 1\)
Lời giải chi tiết:
\(\begin{array}{l}f''\left( x \right)f\left( x \right) + \frac{{{{\left[ {f\left( x \right)} \right]}^2}}}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}} = {\left[ {f'\left( x \right)} \right]^2}\\ \Leftrightarrow f''\left( x \right)f\left( x \right) - {\left[ {f'\left( x \right)} \right]^2} = - \frac{{{{\left[ {f\left( x \right)} \right]}^2}}}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}}\\ \Leftrightarrow \frac{{f''\left( x \right)f\left( x \right) - {{\left[ {f'\left( x \right)} \right]}^2}}}{{{{\left[ {f\left( x \right)} \right]}^2}}} = - \frac{1}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}}\\ \Leftrightarrow \left[ {\frac{{f'\left( x \right)}}{{f\left( x \right)}}} \right]' = - \frac{1}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}}\\ \Leftrightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} = - \int\limits_{}^{} {\frac{{dx}}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}}} = - \int\limits_{}^{} {{{\left( {2x + 1} \right)}^{ - \frac{3}{2}}}dx} \\ \Leftrightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} = - \frac{{{{\left( {2x + 1} \right)}^{ - \frac{1}{2}}}}}{{ - \frac{1}{2}.2}} + C = {\left( {2x + 1} \right)^{ - \frac{1}{2}}} + C\end{array}\)
Thay \(x = 0\) ta có: \(\frac{{f'\left( 0 \right)}}{{f\left( 0 \right)}} = 1 + C \Leftrightarrow 1 = 1 + C \Leftrightarrow C = 0 \Rightarrow \Leftrightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} = {\left( {2x + 1} \right)^{ - \frac{1}{2}}}\).
Lấy nguyên hàm 2 vế ta có:
\(\int\limits_{}^{} {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \int\limits_{}^{} {{{\left( {2x + 1} \right)}^{ - \frac{1}{2}}}dx} \Leftrightarrow \ln \left| {f\left( x \right)} \right| = \frac{{{{\left( {2x + 1} \right)}^{\frac{1}{2}}}}}{{\frac{1}{2}.2}} + C = {\left( {2x + 1} \right)^{\frac{1}{2}}} + C\).
Do \(f\left( x \right) > 0\,\,\forall x \in \left[ {0;4} \right] \Rightarrow \ln f\left( x \right) = {\left( {2x + 1} \right)^{\frac{1}{2}}} + C\).
Thay \(x = 0\) ta có \(\ln f\left( 0 \right) = 1 + C \Leftrightarrow \ln 1 = 1 + C \Leftrightarrow 1 + C = 0 \Leftrightarrow C = - 1\).
\( \Rightarrow \ln f\left( x \right) = {\left( {2x + 1} \right)^{\frac{1}{2}}} - 1 \Leftrightarrow f\left( x \right) = {e^{{{\left( {2x + 1} \right)}^{\frac{1}{2}}} - 1}} \Rightarrow f\left( 4 \right) = {e^{3 - 1}} = {e^2}\).
Chọn A.
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