The known function f (x) = (1-x) / (AX) + LNX (1) If a ≥ 1, it is proved that the function f (x) is an increasing function on [1, + ∞) (2) When a = 1, we prove that ln (n / (n-1)) > 1 / N for positive integer n greater than 1 There is no need to answer the first question The second question is that I change the left side into: lnn-ln (n-1), shift term, lnn-ln (n-1) - 1 / n It is proved that lnn-ln (n-1) - 1 / N > 0, but the left derivative is less than zero, as if it can not be proved,

The known function f (x) = (1-x) / (AX) + LNX (1) If a ≥ 1, it is proved that the function f (x) is an increasing function on [1, + ∞) (2) When a = 1, we prove that ln (n / (n-1)) > 1 / N for positive integer n greater than 1 There is no need to answer the first question The second question is that I change the left side into: lnn-ln (n-1), shift term, lnn-ln (n-1) - 1 / n It is proved that lnn-ln (n-1) - 1 / N > 0, but the left derivative is less than zero, as if it can not be proved,

Use the result of the first question
a=1,f(x)=(1-x)/x + lnx
For positive integer n greater than 1, where n / (n-1) > 1, the function is an increasing function on [1, + ∞)
f(n/(n-1))=ln(n/(n-1))-1/n
However, f (1) = 0, f (n / (n-1)) > 0
Ln (n / (n-1)) - 1 / N > 0
ln(n/(n-1))>1/n

If the function f (x) = x power of a * (x power of A-3 * 2 power of A-1) (a > 0 and a is not equal to 1) is an increasing function in the interval [0, positive infinity], then the value range of real number a is ()
A. (0,2 / 3) B. (radical 3 / 3,1) C. (1, radical 3) d. (2 / 3, positive infinity)

B
Let a ^ x = t
f(t)=t^2-(3a^2+1)t
When 0 1, y = a ^ x is an increasing function
And T > 1
The problem is transformed into that y = f (T) is an increasing function when t > 1
So the axis of symmetry

A very simple high school math problem about function
It is known that f (x) is an odd function defined on R, and f (π / 3 + x) = f (π / 3 - x). When x ∈ [0, π / 3], f (x) = SiNx
(1) Find the value (2) of F (3 π) and solve the inequality f (x) < 1 / 2

Since f is an odd function, we have f (- pi / 3-x) = - f (PI / 3 + x) = - f (PI / 3-x), let y = - pi / 3-x, and substitute f (y) = - f (y + 2 / 3 * PI) into the above formula. Furthermore, we have f (y) = - f (y + 2 / 3 * PI) = f (y + 4 / 3 * PI). F is a function with a period of 4 / 3 * PI, f (3 * PI) = f (3 * Pi-4 / 3 * pi * 2) = f (PI / 3) = sin (1)