Solve the set c first,
F(-x)+f(x)≤2|x|, bringing in simplification:
2x? ≤2|x|
(1) when x≤0, that is, x? +x≤0, and the solution is:-1≤x≤0.
(2) when x > 0, namely x? -x≤0, the solution is: 0≤x≤ 1.
So C=[- 1, 1]
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Let m = a x (m > 0), then f (a x)-a (x+ 1) = 5 is arranged as follows:
m? +( 1-a)m-5=0, let g(m)=m? +( 1-a)m-5
① when 0 < a < 1,
On the interval [- 1, 1], a ≤ a x ≤1/a.
Equation m? +( 1-a)m-5=0 interval [a, 1/a] has a root and must meet the following requirements:
G(a)≤0 and g( 1/a)≥0,
Solution: 0 < a ≤ 1/2.
② when a > 1,
On the interval [- 1, 1],1/a ≤ a x ≤ a.
Equation m? +( 1-a)m-5=0 interval [a, 1/a] has a root and must meet the following requirements:
G( 1/a)≤0 and g(a)≥0,
Solution: a≥5
So the range of real number A is (0, 1/2]∩[5, +∞).
⒉
<1> What courses do preventive medicine take?
Main course:
Pathological anatomy, physiology, psychiatry, infectious disease theory, labor hygiene endogen