5.已知函数f(x)=(x−1)(ax+b)f(x)=(x-1)(ax+b)f(x)=(x−1)(ax+b)为偶函数,且在(0,+∞)(0,+\infty )(0,+∞)上单调递增,则f(x)<0f(x)<0f(x)<0的解集为((( )))