Set 0

Set 0

1 / M + 8 / (1-2m) ≥ K,
Then K ≤ [1 / M + 8 / (1-2m)] min
∵0

Let 0 ＜ m ＜ 1 / 3, if (1 / M) + (3 / 1-3m) ≥ K is constant, then the maximum value of K is

Solution [(1 / M) + (3 / 1-3m)]
=[(3/3m)+(3/1-3m)]*1
=[(3/3m)+(3/1-3m)]*[（3m）+（1-3m）]
=(3/3m)*3m+3/（1-3m）*3m+3（1-3m）/3m+（3/1-3m）*（1-3m）
=3+3+9m/（1-3m）+3（1-3m）/3m
≥6+2√9m/（1-3m）*3（1-3m）/3m
=6+2√27
=6+6√3
That is, K ≤ 6 + 6 √ 3
The maximum value of K is 6 + 6 √ 3

If f (a) = (3m-1) a + b-2m, when m ∈ [0.1], f (a) ≤ 1 holds, then the maximum value of (a + b) is
If f (a) = (3m-1) a + b-2m, when m ∈ [0.1], f (a) ≤ 1 holds, then the maximum value of (a + b) is
A 1/3 B 2/3 C 5/3 D 7/3
Answer D7 / 3 please help me write out the calculation process

M ∈ [0,1],
F (a) ≤ 1,
We get b-a

If f (a) = (3m-1) a + b-2m, when m ∈ [0, 1], f (a) ≤ 1 holds, then the maximum value of a + B is () a.13b.23c.53d.73

Let g (m) = f (a) = (3a-2) m + B-A, since g (m) = f (a) = (3a-2) m + B-A ≤ 1 holds when m ∈ [0, 1], then G (0) ≤ 1g (1) ≤ 1, that is, B-A ≤ 1b + 2A ≤ 1, the points (a, b) satisfying this inequality group constitute the shadow part in the graph, where a (23, 53), let a + B = t, obviously when the straight line a + B = t passes through point a, t gets the maximum value 73