1.BACCB BDCAD BA II。 13.2,14.,15.① ④16.④

3. 17. Solution: Let x 1 and x2 be any two real numbers in the interval [2,6], and x 1

f(x 1)-f(x2)= -

=

= .

by 2； 0，(x 1- 1)(x2- 1)>0,

So f (x 1)-f (x2) >: 0, that is, f (x1) >; f(x2)。

So the function y= is the decreasing function on the interval [2,6].

Therefore, the function y= obtains the maximum and minimum values at the two endpoints of the interval, that is, when x=2, ymax = 2；; When x=6, ymin=.

18. solution: let u= and choose x2 > x 1 > 1, then

u2-u 1=

=

= .

∵x 1> 1,x2> 1,∴x 1- 1>0,x2- 1>0.

And ∵ x 1 < x2, ∴ x 1-x2 < 0.

∴ < 0, namely U2 < u 1.

When a > 1, y=logax is increasing function, ∴ logau2 < logau 1,

That is f (x2) < f (x1);

When 0 < a < 1, y=logax is a decreasing function, ∴ logau2 > logau 1,

That is, f (x2) > f (x 1).

To sum up, when a > 1, f(x)=loga is a decreasing function at (1, +∞); When 0 < a < 1, f(x)=loga is a increasing function at (1, +∞).

This is mandatory.

too much

I suggest you go to the bookstore to buy a book or take a camera to take the answer!