It is known that when x belongs to (1,2), the inequality loga (A / x) - x ^ 2 + 2x greater than or equal to 0 holds, and the range of a is obtained

It is known that when x belongs to (1,2), the inequality loga (A / x) - x ^ 2 + 2x greater than or equal to 0 holds, and the range of a is obtained

When the inequality is reduced to logax = 1, f (x) > = 0
So if we want to make the inequality hold, then f (2) = 2
Finally, the range of a is (0,1) and [2, positive infinity]

F (x) = loga 1 + X / 1-x (a > 0, a ≠ 1) to solve the inequality f (x) > o

f(x)=loga 1+1/x-x
=0+1/x-x>0
1/x>x
x

F (x) = | loga (x) | where 0f (1 / 3)
B f(2)>f(1/3)>f(1/4)
Cf(1/4)>f(1/3)>f(2)
Df(1/3)>f(2)>f(1/4）

Draw the image, choose C

Let the image of exponential function y = f (x) pass through the point (2,1 / 2), then the inequality f (x)

y=f（x）=a^x
Passing through point (2,1 / 2),
1/2=a^2
a=√2/2
f（x）