Cho hàm số (fleft( x right)) thỏa mãn sau ({left[ {fleft( x right)} right]^2} + fleft( x right).

Ta  có: (VT = left[ {fleft( x right).f’left( x right)} right]’ = f’left( x right).f’left( x right) + fleft( x right).f”left( x right) = {left[ {f’left( x right)} right]^2} + fleft( x right).f”left( x right))

( Rightarrow left[ {f’left( x right).fleft( x right)} right]’ = {x^3} – 2x;;;left( * right))

Nguyên hàm hai vế của (left( * right)) ta được: (f’left( x right).fleft( x right) = dfrac{{{x^4}}}{4} – {x^2} + C.;;left( 1 right))

Lại có: (f’left( 0 right) = fleft( 0 right) = 2 Rightarrow C = 2.2 = 4.) 

(begin{array}{l} Rightarrow left( 1 right) Leftrightarrow fleft( x right).f’left( x right) = dfrac{{{x^4}}}{4} – {x^2} + 4\ Rightarrow int {fleft( x right)f’left( x right)dx = int {left( {dfrac{{{x^4}}}{4} – {x^2} + 4} right)} } dx Leftrightarrow int {fleft( x right)dfleft( x right) = dfrac{{{x^5}}}{{20}} – dfrac{{{x^3}}}{3} + 4x + A} \ Leftrightarrow dfrac{{{f^2}left( x right)}}{2} = dfrac{{{x^5}}}{{20}} – dfrac{{{x^3}}}{3} + 4x + A Leftrightarrow {f^2}left( x right) = dfrac{{{x^5}}}{{10}} – dfrac{{2{x^3}}}{3} + 8x + 2A.end{array})

Có (fleft( 0 right) = 2 Rightarrow 4 = 2A Leftrightarrow A = 2)

(begin{array}{l} Rightarrow {f^2}left( x right) = dfrac{{{x^5}}}{{10}} – dfrac{{2{x^3}}}{3} + 8x + 4\ Rightarrow {f^2}left( 2 right) = dfrac{{{2^5}}}{{10}} – dfrac{{{{2.2}^3}}}{3} + 8.2 + 4 = dfrac{{268}}{{15}}.end{array})

Chọn A.