设函数fx=2ln(x-1)-(x-1)^2
1.令2/(x-1)-2(x-1)≤0,解得x≥2"}}}'>

1个回答

  • (1)由f(x)=2ln(x-1)-(x-1)^2可知f'(x)=2/(x-1)-2(x-1)且x>1.

    令2/(x-1)-2(x-1)≤0,解得x≥2或x≤0,又x>1,综上,函数的单调减区间为[2,+∞)

    (2)令g(x)=f(x)+x^2-3x-a 整理得 g(x)=2ln(x-1)-x-a-1 (x>1)

    所以g'(x)=2/(x-1)-1 令g'(x)=2/(x-1)-1≥0得 g(x)的增区间为(1,3],减区间为(3,+∞)

    所以g(x)在[2,3]上单调递增,在[3,4]上单调递减

    又方程f(x)+x^2-3x-a=0在[2,4]内恰有两个相异实根,所以由图像及单调性可知g(2)≤0,g(3)>0,g(4)≤0.

    解得 2ln3-5 ≤ a < 2ln2-4