Cho hàm số \(f(x)\) có đạo hàm liên tục thỏa mãn \...

D

- Hướng dẫn giải

Phương pháp giải:

Sử dụng công thức tích phân từng phần: \(\int\limits_{a}^{b}{udv}=\left. uv \right|_{a}^{b}-\int\limits_{a}^{b}{vdu}\).

Giải chi tiết:

\(\begin{array}{l}
\int\limits_{\frac{\pi }{2}}^\pi {\cos x.f(x)dx} = \int\limits_{\frac{\pi }{2}}^\pi {f(x)d\left( {\sin x} \right)} = \left. {f(x)\sin x} \right|_{\frac{\pi }{2}}^\pi - \int\limits_{\frac{\pi }{2}}^\pi {\sin x.f'(x)dx} = \frac{\pi }{4}\\
\Rightarrow 0 - f\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{2}} \right) - \int\limits_{\frac{\pi }{2}}^\pi {\sin x.f'(x)dx} = \frac{\pi }{4}\\
\Rightarrow \int\limits_{\frac{\pi }{2}}^\pi {\sin x.f'(x)dx} = - \frac{\pi }{4}\left( {dof\left( {\frac{\pi }{2}} \right) = 0} \right)
\end{array}\)

Xét 

\(\begin{array}{l}
\int\limits_{\frac{\pi }{2}}^\pi {{{\left[ {f'(x) + k\sin x} \right]}^2}dx} = \int\limits_{\frac{\pi }{2}}^\pi {{{\left[ {f'(x)} \right]}^2}dx} + 2k\int\limits_{\frac{\pi }{2}}^\pi {f'\left( x \right).\sin xdx} + {k^2}\int\limits_{\frac{\pi }{2}}^\pi {{{\sin }^2}xdx} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{\pi }{4} + 2k.\left( { - \frac{\pi }{4}} \right) + {k^2}\int\limits_{\frac{\pi }{2}}^\pi {\frac{{1 - \cos 2x}}{2}dx}
\end{array}\)

\(=\frac{\pi }{4}-\frac{k\pi }{2}+\frac{1}{2}{{k}^{2}}.\left. \left( x-\frac{1}{2}\sin 2x \right) \right|_{\frac{\pi }{2}}^{\pi }=\frac{\pi }{4}-\frac{k\pi }{2}+\frac{1}{2}{{k}^{2}}.\frac{\pi }{2}=\frac{\pi }{4}{{\left( 1-k \right)}^{2}}=0\Rightarrow k=1\)

Khi đó \(\int\limits_{\frac{\pi }{2}}^{\pi }{{{\left[ f'(x)+\sin x \right]}^{2}}dx}=0\Rightarrow f'(x)+\sin x=0\Rightarrow f'(x)=-\sin x\Rightarrow f(x)=\cos \,x+C\)

Mà \(f\left( \frac{\pi }{2} \right)=0\Rightarrow 0+C=0\Rightarrow C=0\Rightarrow f(x)=\cos x\)

\(\Rightarrow f\left( 2018\pi  \right)=\cos 2018\pi =1\)

Chọn: D