Cho hàm số f(x) có \(f\left...

D

- Hướng dẫn giải

Do $f'\left( x \right) = x\sin x$ nên $f\left( x \right) = \int {f'\left( x \right){\rm{d}}x} = \int {x\sin x{\rm{d}}x = - \int {x{\rm{d}}\cos x} = } - x\cos x + \int {\cos x{\rm{d}}x} = - x\cos x + \sin x + C$.

Theo giả thiết $f\left( {\frac{\pi }{2}} \right) = 2 \Leftrightarrow 1 + C = 2 \Rightarrow C = 1$. Suy ra $f\left( x \right) = \sin x - x\cos x + 1$.

$\int\limits_0^{\frac{\pi }{2}} {\cos x.f\left( x \right){\rm{d}}x} = \int\limits_0^{\frac{\pi }{2}} {\cos x\left( {\sin x - x\cos x + 1} \right){\rm{d}}x} = \int\limits_0^{\frac{\pi }{2}} {\left( {\sin x\cos x - x{{\cos }^2}x + \cos x} \right){\rm{d}}x}$

$= \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2x{\rm{d}}x} - \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {x\left( {1 + \cos 2x} \right){\rm{d}}x} + \int\limits_0^{\frac{\pi }{2}} {\cos x{\rm{d}}x} = - \frac{1}{4}\cos 2x\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. + \sin x\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. - \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {x{\rm{d}}x} - \frac{1}{4}\int\limits_0^{\frac{\pi }{2}} {x{\rm{d}}\sin 2x}$

$= \frac{1}{2} + 1 - \frac{{{x^2}}}{4}\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. - \frac{1}{4}x\sin 2x\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. + \frac{1}{4}\int\limits_0^{\frac{\pi }{2}} {\sin 2x{\rm{d}}x} = \frac{3}{2} - \frac{{{\pi ^2}}}{{16}} - \frac{1}{8}\cos 2x\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. = \frac{7}{4} - \frac{{{\pi ^2}}}{{16}}$

Vậy a = 7,b = 4,c = 16. Suy ra a + b + c = 27.