by 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. \({e^2} + 1\)
Hướng dẫn
Chọn đáp án là A
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.
