Cho hàm số $\Large f(x)$ thỏa mãn $\Large {{\left[ {f}'(x) \right]}^{2

Cho hàm số $\Large f(x)$ thỏa mãn $\Large {{\left[ {f}'(x) \right]}^{2}}+f(x).{f}''(x)=2{{x}^{2}}-x+1,\forall x\in R$ và $\Large f(0)={f}'(0)=3$ . Giá trị của $\Large {{\left[ f(1) \right]}^{2}}$ bằng.

• A.

  $\Large 28$

• B.

  $\Large 22$

• C.

  $\Large \dfrac{19}{2}$

• D.

  $\Large 10$

Ta có 

$\Large {{\left[ {f}'(x) \right]}^{2}}+f(x).{f}''(x)=2{{x}^{2}}-x+1$

$\Large \Leftrightarrow {{\left[ f(x).{f}'(x) \right]}^{\prime }}=2{{x}^{2}}-x+1$

$\Large \Leftrightarrow f(x).{f}'(x)=\int{(2{{x}^{2}}-x+1)dx}$

$\Large \Leftrightarrow f(x).{f}'(x)=\dfrac{2}{3}{{x}^{3}}-\dfrac{1}{2}{{x}^{2}}+x+C$

Thay $\Large x=0$ ta được $\Large f(0).{f}'(0)=C\Leftrightarrow C=9$

Khi đó

$\Large f(x).{f}'(x)=\dfrac{2}{3}{{x}^{3}}-\dfrac{1}{2}{{x}^{2}}+x+9$

$\Large \Leftrightarrow \int{f(x).{f}'(x)dx=\int{\left( \dfrac{2}{3}{{x}^{3}}-\dfrac{1}{2}{{x}^{2}}+x+9 \right)dx}}$

$\Large \Leftrightarrow \int{f(x)d\left[ f(x) \right]=\dfrac{1}{6}{{x}^{4}}-\dfrac{1}{6}{{x}^{3}}+\dfrac{1}{2}{{x}^{2}}+9x+{{C}_{1}}}$

$\Large \Leftrightarrow \dfrac{1}{2}{{f}^{2}}(x)=\dfrac{1}{6}{{x}^{4}}-\dfrac{1}{6}{{x}^{3}}+\dfrac{1}{2}{{x}^{2}}+9x+{{C}_{1}}$

$\Large \Leftrightarrow {{f}^{2}}(x)=\dfrac{1}{3}{{x}^{4}}-\dfrac{1}{3}{{x}^{3}}+{{x}^{2}}+18x+2{{C}_{1}}$

Thay $\Large x=0$ ta được $\Large {{f}^{2}}(0)=2{{C}_{1}}\Leftrightarrow {{C}_{1}}=\dfrac{9}{2}$

Vậy $\Large {{f}^{2}}(x)=\dfrac{1}{3}{{x}^{4}}-\dfrac{1}{3}{{x}^{3}}+{{x}^{2}}+18x+9$ nên $\Large {{f}^{2}}(1)=28$

Chọn đáp án A