Cho \(f(x) = \dfrac{{4m}}{\pi } + {\sin ^2}x\). Tì...

A

- Hướng dẫn giải

Ta có:

\(\int {\left( {\dfrac{{4m}}{\pi } + {{\sin }^2}x} \right)\,dx}  \)

\(= \int {\left( {\dfrac{{4m}}{\pi } + \dfrac{{1 - \cos 2x}}{2}} \right)} \,dx \)

\(= \int {\left( {\dfrac{{8m + \pi }}{{2\pi }} - \dfrac{{\cos 2x}}{2}} \right)\,dx} \)

\( = \left( {\dfrac{{8m + \pi }}{{2\pi }}} \right)x - \dfrac{1}{4}\int {\cos 2x\,d\left( {2x} \right)}\)

\(  = \left( {\dfrac{{8m + \pi }}{{2\pi }}} \right)x - \dfrac{{\sin 2x}}{4} + C\)

Theo giả thiết ta có:

+ \(F\left( 0 \right) = 1 \Rightarrow C = 1\)

+ \(F\left( {\dfrac{\pi }{4}} \right) = \dfrac{\pi }{8}\)

\(\Rightarrow \left( {\dfrac{{8m + \pi }}{{2\pi }}} \right).\dfrac{\pi }{4} - \dfrac{1}{4} + 1 = \dfrac{\pi }{8}\)

\( \Leftrightarrow \dfrac{{8m + \pi }}{8} = \dfrac{\pi }{8} - \dfrac{3}{4} = \dfrac{{\pi  - 6}}{8} \)

\(\Leftrightarrow 8m =  - 6 \Rightarrow m =  - \dfrac{3}{4}\).