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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục và đồng biến trên đoạn $\left[ 1;4 \right],$ $f\left( 1 \right)=0$ và $x+2xf\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}},\forall x\in \left[ 1;3 \right].$ Đặt $I=\int\limits_{1}^{4}{f\left( x \right)dx}.$ Mệnh đề nào dưới đây đúng?
A. $1<I<4.$
B. $4<I<8.$
C. $8<I<12.$
D. $12<I<16.$
A. $1<I<4.$
B. $4<I<8.$
C. $8<I<12.$
D. $12<I<16.$
Ta có $f\left( x \right)$ đồng biến trên $\left[ 1;4 \right]$ nên $f\left( x \right)\ge f\left( 1 \right)=0$
$\Rightarrow x+2xf\left( x \right)\ge 0\forall x\in \left[ 1;4 \right]$
$x+2xf\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}}\Leftrightarrow \sqrt{x+2xf\left( x \right)}={f}'\left( x \right)\Leftrightarrow \dfrac{{f}'\left( x \right)}{\sqrt{1+2f\left( x \right)}}=\sqrt{x}$
$\Leftrightarrow {{\left[ \sqrt{1+2f\left( x \right)} \right]}^{\prime }}=\sqrt{x}\Leftrightarrow \sqrt{1+2f\left( x \right)}=\int{\sqrt{x}dx}\Leftrightarrow \sqrt{1+2f\left( x \right)}=\dfrac{2}{3}\sqrt{{{x}^{3}}}+C(1)$
Mà $f\left( 1 \right)=0\overset{(1)}{\mathop{\Rightarrow }} C=\dfrac{1}{3}\Rightarrow \sqrt{1+2f\left( x \right)}=\dfrac{2}{3}\sqrt{{{x}^{3}}}+\dfrac{1}{3}\Rightarrow f\left( x \right)=\dfrac{{{\left( \dfrac{2}{3}\sqrt{{{x}^{3}}}+\dfrac{1}{3} \right)}^{2}}-1}{2}$
Vậy $I=\int\limits_{1}^{4}{f\left( x \right)dx}=\int\limits_{1}^{4}{\dfrac{{{\left( \dfrac{2\sqrt{{{x}^{3}}}+1}{3} \right)}^{2}}-1}{2}}dx=\dfrac{1403}{90}\Rightarrow 12<I<16.$
$\Rightarrow x+2xf\left( x \right)\ge 0\forall x\in \left[ 1;4 \right]$
$x+2xf\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}}\Leftrightarrow \sqrt{x+2xf\left( x \right)}={f}'\left( x \right)\Leftrightarrow \dfrac{{f}'\left( x \right)}{\sqrt{1+2f\left( x \right)}}=\sqrt{x}$
$\Leftrightarrow {{\left[ \sqrt{1+2f\left( x \right)} \right]}^{\prime }}=\sqrt{x}\Leftrightarrow \sqrt{1+2f\left( x \right)}=\int{\sqrt{x}dx}\Leftrightarrow \sqrt{1+2f\left( x \right)}=\dfrac{2}{3}\sqrt{{{x}^{3}}}+C(1)$
Mà $f\left( 1 \right)=0\overset{(1)}{\mathop{\Rightarrow }} C=\dfrac{1}{3}\Rightarrow \sqrt{1+2f\left( x \right)}=\dfrac{2}{3}\sqrt{{{x}^{3}}}+\dfrac{1}{3}\Rightarrow f\left( x \right)=\dfrac{{{\left( \dfrac{2}{3}\sqrt{{{x}^{3}}}+\dfrac{1}{3} \right)}^{2}}-1}{2}$
Vậy $I=\int\limits_{1}^{4}{f\left( x \right)dx}=\int\limits_{1}^{4}{\dfrac{{{\left( \dfrac{2\sqrt{{{x}^{3}}}+1}{3} \right)}^{2}}-1}{2}}dx=\dfrac{1403}{90}\Rightarrow 12<I<16.$
Đáp án D.