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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định, có đạo hàm, liên tục và đồng biến trên $\left[ 1;4 \right]$ thỏa mãn $x+2xf\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}},\forall x\in \left[ 1;4 \right],f\left( 1 \right)=\dfrac{3}{2}.$ Giá trị $f\left( 4 \right)$ bằng:
A. $\dfrac{391}{18}.$
B. $\dfrac{361}{18}.$
C. $\dfrac{381}{18}.$
D. $\dfrac{371}{18}.$
A. $\dfrac{391}{18}.$
B. $\dfrac{361}{18}.$
C. $\dfrac{381}{18}.$
D. $\dfrac{371}{18}.$
HD: Vì $y=f\left( x \right)$ là hàm số đồng biến trên $\left[ 1;4 \right]\Rightarrow f\left( x \right)\ge f\left( 1 \right)=\dfrac{3}{2}>0.$
Khi đó $x+2x.f\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}}\Leftrightarrow \sqrt{x.\left[ 2f\left( x \right)+1 \right]}={f}'\left( x \right)\Leftrightarrow \dfrac{{f}'\left( x \right)}{\sqrt{2f\left( x \right)+1}}=\sqrt{x}\text{ }\left( * \right).$
Lấy nguyên hàm hai vế của $\left( * \right),$ ta được $\int{\dfrac{{f}'\left( x \right)}{\sqrt{2f\left( x \right)+1}}dx}=\int{\sqrt{x}dx}=\dfrac{2}{3}x\sqrt{x}+C\text{ }(1).$
Đặt $t=\sqrt{2f\left( x \right)+1}\Leftrightarrow dt=\dfrac{{f}'\left( x \right)}{\sqrt{2f\left( x \right)+1}}dx\Rightarrow \int{\dfrac{{f}'\left( x \right)}{\sqrt{2f\left( x \right)+1}}dx=\int{dt}=t\text{ }(2)}.$
Từ (1), (2) suy ra $\sqrt{2f\left( x \right)+1}=\dfrac{2}{3}x\sqrt{x}+C$ mà $f\left( 1 \right)=\dfrac{3}{2}\Rightarrow \sqrt{2.\dfrac{3}{2}+1}=C+\dfrac{2}{3}\Leftrightarrow C=\dfrac{4}{3}.$
Do đó $\sqrt{2f\left( x \right)+1}=\dfrac{2}{3}x\sqrt{x}+\dfrac{4}{3}\Leftrightarrow f\left( x \right)=\dfrac{1}{2}\left[ {{\left( \dfrac{2}{3}x\sqrt{x}+\dfrac{4}{3} \right)}^{2}}-1 \right].$ Vậy $f\left( 4 \right)=\dfrac{391}{18}.$
Khi đó $x+2x.f\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}}\Leftrightarrow \sqrt{x.\left[ 2f\left( x \right)+1 \right]}={f}'\left( x \right)\Leftrightarrow \dfrac{{f}'\left( x \right)}{\sqrt{2f\left( x \right)+1}}=\sqrt{x}\text{ }\left( * \right).$
Lấy nguyên hàm hai vế của $\left( * \right),$ ta được $\int{\dfrac{{f}'\left( x \right)}{\sqrt{2f\left( x \right)+1}}dx}=\int{\sqrt{x}dx}=\dfrac{2}{3}x\sqrt{x}+C\text{ }(1).$
Đặt $t=\sqrt{2f\left( x \right)+1}\Leftrightarrow dt=\dfrac{{f}'\left( x \right)}{\sqrt{2f\left( x \right)+1}}dx\Rightarrow \int{\dfrac{{f}'\left( x \right)}{\sqrt{2f\left( x \right)+1}}dx=\int{dt}=t\text{ }(2)}.$
Từ (1), (2) suy ra $\sqrt{2f\left( x \right)+1}=\dfrac{2}{3}x\sqrt{x}+C$ mà $f\left( 1 \right)=\dfrac{3}{2}\Rightarrow \sqrt{2.\dfrac{3}{2}+1}=C+\dfrac{2}{3}\Leftrightarrow C=\dfrac{4}{3}.$
Do đó $\sqrt{2f\left( x \right)+1}=\dfrac{2}{3}x\sqrt{x}+\dfrac{4}{3}\Leftrightarrow f\left( x \right)=\dfrac{1}{2}\left[ {{\left( \dfrac{2}{3}x\sqrt{x}+\dfrac{4}{3} \right)}^{2}}-1 \right].$ Vậy $f\left( 4 \right)=\dfrac{391}{18}.$
Đáp án A.