Given f (x) = - 3x & # 178; + a (6-A) x, if the inequality f (x) > 0 holds on (1,3), the range of a is obtained -Is radical 6 + 3 ≤ a ≤ radical 6 + 3 to be discarded? The answer is a = 3

Given f (x) = - 3x & # 178; + a (6-A) x, if the inequality f (x) > 0 holds on (1,3), the range of a is obtained -Is radical 6 + 3 ≤ a ≤ radical 6 + 3 to be discarded? The answer is a = 3

First of all [1,3] is a closed interval, not an open interval, otherwise there is no answer
If - 3x & # 178; + a (6-A) x ＞ 0 holds on [1,3]
It is known that f (0) = 0, that is, f (x) passes through the origin
And we know that the image of F (x) is open downward
So the part of F (x) greater than 0 must be on the right side of the y-axis
There are two restrictions
The axis of symmetry of F (x) is greater than 0
-a（6-a）/(-6)＞0
The solution is 0