(C).解 AC段: Q(x)=-qx(0≤x<a) B M(x)= (0≤x≤a) CB段: Q(x)=-2ga (a<x<2a) qe 2 ga MA C al x ga(x-a) q =-2qx+qa2(a≤x<2a) 2 XT7TU2 22 0 2 0 2 2 2 2 3 2 2 2 2 ( ). ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) c AC Q x qx x a M x q x x a CB Q x qa a x a M x qa x a qa x a qax qa a x a 解： 段： 段： = −   = −   = −   = − −       − − = − +   q qa x x a a Q A C B M qa 2 2 5 2 2 qa qa 2qa XT7TU2