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2017年高考新课标卷Ⅱ文科第21题:设函数f(x)=(1-x~2)e~x。(1)讨论f(x)的单调性;(2)当x≥0时,f(x)≤ax+1,求a的取值范围。解析(1)由于f′(x)=(-x~2-2x+1)e~x,令f’(x)>0有-1-2~(1/2)-1+2~(1/2),故f(x)在(-1-2~(1/2),-1+2~(1/2))上单调递增,在(-∞,-1-2~(1/2))和(-1+2~(1/2),+∞)上单调递减。(2)解法1分类讨论、假设反证法构造函数g(x)=(1-x~2)e~x-ax-1,则g’(x)=(-x~2-2x+1)e~x-a,g“(x)=(-x~2-4x-1)e~x,
2017 New Curriculum Entrance Examination Ⅱ Liberal Arts 21st Problem: Let f (x) = (1-x ~ 2) e ~ x. (1) to discuss the monotonicity of f (x); (2) when x≥0, f (x) ≤ax + 1, find the range of a. (1) Since f ’(x) = (- x ~ 2-2x + 1) e ~ x, let f’ (x)> 0 be -1-2 ~ (1/2) -1 + 2 ~ (1/2) (-1 ~ 2 ~ (1/2)), (-1 + 2 ~ (1 ~ 2) / 2), + ∞) monotonically decreasing. (2) Solution 1 According to the category discussion, assuming that the inverse syndrome method constructor g (x) = (1-x ~ 2) e ~ x-ax- 1, g ’(x) = (- x ~ 2-2x + 1) e ~ xa, g ”(x) = (- x ~ 2-4x-1) e ~ x,