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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm đến cấp 2 liên tục trên [1; 3], $f\left( 1 \right)={f}'\left( 1 \right)=1$ và $f\left( x \right)>0, f\left( x \right){f}''\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}}-{{\left[ xf\left( x \right) \right]}^{2}}, \forall x\in \left[ 1; 3 \right]$. Tính $\ln f\left( 3 \right)$.
A. -4
B. -3
C. 4
D. 3
A. -4
B. -3
C. 4
D. 3
Ta có
$f\left( x \right){f}''\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}}-{{\left[ xf\left( x \right) \right]}^{2}}\Leftrightarrow f\left( x \right){f}''\left( x \right)-{{\left[ {f}'\left( x \right) \right]}^{2}}=-{{\left[ xf\left( x \right) \right]}^{2}}$
$\Rightarrow \dfrac{f\left( x \right){f}''\left( x \right)-{{\left[ {f}'\left( x \right) \right]}^{2}}}{{{f}^{2}}\left( x \right)}=-{{x}^{2}}\Rightarrow {{\left[ \dfrac{{f}'\left( x \right)}{f\left( x \right)} \right]}^{'}}=-{{x}^{2}}\Rightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=-\dfrac{{{x}^{3}}}{3}+C$
Do $f\left( 1 \right)={f}'\left( 1 \right)=1$ nên $C=\dfrac{4}{3}\Rightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=-\dfrac{{{x}^{3}}}{3}+\dfrac{4}{3}\Rightarrow \ln f\left( x \right)=-\dfrac{{{x}^{4}}}{12}+\dfrac{4}{3}x+{C}'$
Vì $f\left( 1 \right)=1$ nên ${C}'=-\dfrac{5}{4}$
Do đó $\ln f\left( x \right)=-\dfrac{{{x}^{4}}}{12}+\dfrac{4}{3}x-\dfrac{5}{4}\Rightarrow \ln f\left( 3 \right)=-4$
$f\left( x \right){f}''\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}}-{{\left[ xf\left( x \right) \right]}^{2}}\Leftrightarrow f\left( x \right){f}''\left( x \right)-{{\left[ {f}'\left( x \right) \right]}^{2}}=-{{\left[ xf\left( x \right) \right]}^{2}}$
$\Rightarrow \dfrac{f\left( x \right){f}''\left( x \right)-{{\left[ {f}'\left( x \right) \right]}^{2}}}{{{f}^{2}}\left( x \right)}=-{{x}^{2}}\Rightarrow {{\left[ \dfrac{{f}'\left( x \right)}{f\left( x \right)} \right]}^{'}}=-{{x}^{2}}\Rightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=-\dfrac{{{x}^{3}}}{3}+C$
Do $f\left( 1 \right)={f}'\left( 1 \right)=1$ nên $C=\dfrac{4}{3}\Rightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=-\dfrac{{{x}^{3}}}{3}+\dfrac{4}{3}\Rightarrow \ln f\left( x \right)=-\dfrac{{{x}^{4}}}{12}+\dfrac{4}{3}x+{C}'$
Vì $f\left( 1 \right)=1$ nên ${C}'=-\dfrac{5}{4}$
Do đó $\ln f\left( x \right)=-\dfrac{{{x}^{4}}}{12}+\dfrac{4}{3}x-\dfrac{5}{4}\Rightarrow \ln f\left( 3 \right)=-4$
Đáp án A.