Cho hàm số

B

- Hướng dẫn giải

Đáp án B.

Từ

$\begin{array}{l}f\left(x\right).f\text{'}\left(x\right)=2x\sqrt{f^2\left(x\right)+1}\\ \Rightarrow \frac{f\left(x\right).f\text{'}\left(x\right)}{\sqrt{f^2\left(x\right)+1}}=2x\\ \Rightarrow \int \frac{f\left(x\right).f\text{'}\left(x\right)}{\sqrt{f^2\left(x\right)+1}}dx=\int 2xdx\end{array}$

 (1)

Đặt  

$\begin{array}{l}\sqrt{f^2\left(x\right)+1}=t\\ \Rightarrow f^2\left(x\right)=t^2-1\\ \Rightarrow 2f\left(x\right).f\text{'}\left(x\right)dx=2tdt\\ \Rightarrow f\left(x\right).f\text{'}\left(x\right)dx=tdt\end{array}$

Suy ra  $\int \frac{f\left(x\right).f\text{'}\left(x\right)}{\sqrt{f^2\left(x\right)+1}}x=\int \frac{tdt}{t}$$=\int dt=t+C_1$$=\sqrt{f^2\left(x\right)+1}+C_1$và  $\int 2xdx=x^2+C_2$

Từ (1) ta suy ra $\sqrt{f^2\left(x\right)+1}+C_1=x^2+C_2$  . Do  $f\left(0\right)=0$nên $C_2-C_1=1$.

Như vậy  

$\begin{array}{l}\sqrt{f^2\left(x\right)+1}=x^2+\left(C_2-C_1\right)\\ =x^2+1\\ \Rightarrow f^2\left(x\right)={\left(x^2+1\right)}^2-1\\ =x^4+2x^2\end{array}$

$\begin{array}{l}\Rightarrow f\left(x\right)=\sqrt{x^4+2x^2}\\ =\left|x\right|\sqrt{x^2+2}\\ =x\sqrt{x^2+2}\end{array}$

 (do $x\in \left[1;3\right]$).

Ta có $f\text{'}\left(x\right)=\sqrt{x^2+2}+\frac{x^2}{\sqrt{x^2+2}}$$=\frac{2\left(x^2+1\right)}{\sqrt{x^2+2}}>\mathrm{0,}\forall x\in ℝ\Rightarrow$ Hàm số $f\left(x\right)=x\sqrt{x^2+2}$ đồng biến trên R nên $f\left(x\right)$ cũng đồng biến trên $\left[1;3\right]$  .

Khi đó $M=\underset{\left[1;3\right]}{\max }f\left(x\right)=f\left(3\right)=3\sqrt{11}$ và $m=\underset{\left[1;3\right]}{\min }f\left(x\right)=f\left(1\right)=\sqrt{3}$.

Vậy 

$\begin{array}{l}P=2M-m=6\sqrt{11}-\sqrt{3}\\ \Rightarrow a=6;b=1;c=0\\ \Rightarrow a+b+c=7\end{array}$