Cho hàm số \(f\left( x \right) = A\sin \pi x + B\) với \(A,B\) là hằng số.

a) Đ, b) S, c) S, d) S

a) Khi \(A = \pi ,B = 1\) thì \(f\left( x \right) = \pi \sin \pi x + 1\).

Khi đó \(\int\limits_0^2 {f\left( x \right)dx} = \int\limits_0^2 {\left( {\pi \sin \pi x + 1} \right)dx} \)\( = \left. {\left( { - \cos \pi x + x} \right)} \right|_0^2\)\( = 2\).

b) \(\int\limits_0^2 {f\left( x \right)dx} \)\( = \int\limits_0^2 {\left( {A\sin \pi x + B} \right)dx} \)\( = \left. {\left( { - \frac{A}{\pi }\cos \pi x + Bx} \right)} \right|_0^2\)\( = \left. {\left( { - \frac{A}{\pi }\cos \pi x + Bx} \right)} \right|_0^2 = 2B\).

c) Vì \(\int\limits_0^2 {f\left( x \right)dx} = 4\) nên \(2B = 4 \Leftrightarrow B = 2\).

d) Ta có \(f'\left( x \right) = A\pi \cos \pi x\).

Vì \(f'\left( 1 \right) = 2\) nên \(A\pi \cos \pi = 2 \Rightarrow A = - \frac{2}{\pi }\).

Theo câu c, ta có \(B = 2\).