Cho hàm số \(f\left( x \right)\) có đạo hàm liên...

A

- Hướng dẫn giải

Phương pháp giải:

 Sử dụng bất đẳng thức Holder trong tích phân để tìm hàm số \({f}'\left( x \right)\) 

Giải chi tiết:

Đặt

\(\left\{ \begin{array}{l}
u = f\left( x \right)\\
{\rm{d}}v = \sin x\,{\rm{d}}x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\rm{d}}u = f'\left( x \right)\,{\rm{d}}x\\
v = - \,\cos x
\end{array} \right.,\)

khi đó \(\int\limits_{0}^{\frac{\pi }{2}}{\sin x.f\left( x \right)\,\text{d}x}=\left. -\,\cos x.f\left( x \right) \right|_{0}^{\frac{\pi }{2}}+\int\limits_{0}^{\frac{\pi }{2}}{\cos x.{f}'\left( x \right)\,\text{d}x}\) \(\Leftrightarrow \,\,\frac{\pi }{4}=-\,\cos \frac{\pi }{2}.f\left( \frac{\pi }{2} \right)+\cos 0.f\left( 0 \right)+\int\limits_{0}^{\frac{\pi }{2}}{\cos x.{f}'\left( x \right)\,\text{d}x}\Leftrightarrow \,\,\int\limits_{0}^{\frac{\pi }{2}}{\cos x.{f}'\left( x \right)\,\text{d}x}=\frac{\pi }{4}.\)

Xét \(\int\limits_{0}^{\frac{\pi }{2}}{{{\left[ {f}'\left( x \right)+k.\cos x \right]}^{\,2}}\,\text{d}x}=0\Leftrightarrow \int\limits_{0}^{\frac{\pi }{2}}{{{\left[ {f}'\left( x \right) \right]}^{\,2}}\,\text{d}x}+2k.\int\limits_{0}^{\frac{\pi }{2}}{\cos x.{f}'\left( x \right)\,\text{d}x}+{{k}^{2}}.\int\limits_{0}^{\frac{\pi }{2}}{{{\cos }^{2}}x\,\text{d}x}=0\) \(\Leftrightarrow \,\,\frac{\pi }{4}+2k.\frac{\pi }{4}+{{k}^{2}}.\frac{\pi }{4}=0\Leftrightarrow \,\,k=-\,1.\)

Khi đó \(\int\limits_{0}^{\frac{\pi }{2}}{{{\left[ {f}'\left( x \right)-\cos x \right]}^{\,2}}\,\text{d}x}=0\Leftrightarrow \,\,{f}'\left( x \right)=\cos x.\)

Suy ra \(f\left( x \right)=\int{{f}'\left( x \right)\,\text{d}x}=\int{\cos x\,\text{d}x}=\sin x+C\) mà \(f\left( 0 \right)=0\Rightarrow \,\,C=0.\) Vậy \(f\left( x \right)=\sin x\,\,\xrightarrow{{}}\,\,\int\limits_{0}^{\frac{\pi }{2}}{\sin x\,\text{d}x}=1.\)

Chọn A