Lời giải chi tiết:

Ta có: \(\int\limits_0^{\frac{\pi }{2}} {\left[ {{{\left( {f\left( x \right)} \right)}^2} - 2f\left( x \right)\left( {\sin \,x - \cos x} \right)} \right]dx}  = \int\limits_0^{\frac{\pi }{2}} {{{\left[ {f\left( x \right) - \left( {\sin \,x - \cos x} \right)} \right]}^2}dx}  - \int\limits_0^{\frac{\pi }{2}} {{{\left( {\sin \,x - \cos x} \right)}^2}dx} \)

\( = \int\limits_0^{\frac{\pi }{2}} {{{\left[ {f\left( x \right) - \left( {\sin \,x - \cos x} \right)} \right]}^2}dx}  - \int\limits_0^{\frac{\pi }{2}} {{{\left( {\sin \,x - \cos x} \right)}^2}dx} \)

Mà \(\int\limits_0^{\frac{\pi }{2}} {{{\left( {\sin \,x - \cos x} \right)}^2}dx}  = \int\limits_0^{\frac{\pi }{2}} {\left( {1 - \sin 2x} \right)dx}  = \left. {\left( {x + \frac{1}{2}\cos 2x} \right)} \right|_0^{\frac{\pi }{2}} = \left( {\frac{\pi }{2} - \frac{1}{2}} \right) - \left( {0 + \frac{1}{2}} \right) = \frac{\pi }{2} - 1\)

\( \Rightarrow \int\limits_0^{\frac{\pi }{2}} {{{\left[ {f\left( x \right) - \left( {\sin \,x - \cos x} \right)} \right]}^2}dx}  - \left( {\frac{\pi }{2} - 1} \right) = 1 - \frac{\pi }{2} \Rightarrow \int\limits_0^{\frac{\pi }{2}} {{{\left[ {f\left( x \right) - \left( {\sin \,x - \cos x} \right)} \right]}^2}dx = 0} \)

\( \Rightarrow f\left( x \right) = \sin \,x - \cos x\)\( \Rightarrow \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)dx}  = 0\).

Chọn: B