Lời giải:Ta có:
\(\begin{array}{l}\left[ {f'\left( x \right).f\left( x \right)} \right]'\\ = f''\left( x \right).f\left( x \right) + f'\left( x \right).f'\left( x \right)\\ = {\left( {f'\left( x \right)} \right)^2} + f\left( x \right).f''\left( x \right)\end{array}\)
Do đó: \(\left[ {f'\left( x \right).f\left( x \right)} \right]' = 15{x^4} + 12x,\,\,\forall x \in \mathbb{R}\).
Lấy nguyên hàm hai vế ta được:
\(\begin{array}{l}\int {\left[ {f'\left( x \right).f\left( x \right)} \right]'dx} = \int {\left( {15{x^4} + 12x} \right)dx} \\ \Leftrightarrow f'\left( x \right).f\left( x \right) = 3{x^5} + 6{x^2} + C\end{array}\)
Thay \(x = 0\) ta có: \(f'\left( 0 \right).f\left( 0 \right) = C \Leftrightarrow C = 1\).
\( \Rightarrow f'\left( x \right).f\left( x \right) = 3{x^5} + 6{x^2} + 1\)
Tiếp tục lấy nguyên hàm hai vế ta được:
\(\begin{array}{l}\int {f'\left( x \right)f\left( x \right)dx} = \int {\left( {3{x^5} + 6{x^2} + 1} \right)dx} \\ \Leftrightarrow \frac{{{f^2}\left( x \right)}}{2} = \frac{1}{2}{x^6} + 2{x^3} + x + C'\end{array}\)
Thay \(x = 0\) ta có: \(\frac{{{f^2}\left( 0 \right)}}{2} = C' \Leftrightarrow C' = \frac{1}{2}\).
\(\begin{array}{l} \Rightarrow \frac{{{f^2}\left( x \right)}}{2} = \frac{1}{2}{x^6} + 2{x^3} + x + \frac{1}{2}\\ \Leftrightarrow {f^2}\left( x \right) = {x^6} + 4{x^3} + 2x + 1\end{array}\)
Vậy \({f^2}\left( 1 \right) = 1 + 4 + 2 + 1 = 8\).
