Question

求解题，谢谢！

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Answer 1

(8)
y=|2^x -1|
= 2^x -1 ; x≥0
= -2^x +1 ; x<0

y=m
case 1: x≥0
2^x -1 = m
x= log<2> (m+1) ≥0
=>
m+1>0 and m+1≥1
m> -1 and m≥0
m≥0

case 2: x<0
-2^x +1 = m
x= log<2> (1-m)
=>
1-m>0 and 1-m < 1
m<1 and m>0
=> 0<m<1

ie
case 1 or case 2
m≥0 or 0<m<1
=> m≥0

(11)
f(a+x) > f(x^2)
a+x <x^2
x^2-x-a >0
(x-1/2)^2 -a -1/4 >0

=>
-a-1/4 >0
a< -1/4

(14)
f(x)= √(3-ax)/(a-1) ( a≠1)

3-ax >0
a< 3/x
=> a< 3 (1) ;x∈(0,1]

f'(x) = -a/[2(a-1)√(3-ax)]

f'(x) <0 for x∈(0,1]
-a/[2(a-1)√(3-ax)] <0

-a/(a-1) <0
a/(a-1)>0
a>1 or a<0

(1) and (2)
a< 3 and (a>1 or a<0)
1<a<3 or a<0