Let g(x) = f(x) sin x, where f(x) is a twice differentiable function on -∞,∞ such that f'-π=1. The value of g''-π equals _____

We have g(x) = f(x) sin x                                                (1)On differentiating equation (1) w.r.t. x, we getg'x=fxcosx+f'xsinx                                              (2)Again differentiating equation (2) w.r.t. x, we get g''x=fx-sinx+f'xcosx+f'xcosx+f''xsinx    (3)or g''-π=2f'-πcos-π=2×1×-1=-2Hence, g''-π=-2