Maths F.4 Q 快!40分!

1a) Let y = ax + b√x where a and b are constants.

Then x = 9:

9a + 3b = 17 ... (1)

x = 1:

a + b = 5/3 ... (2)

Solving, we have a = 2 and b = -1/3

So y = 2x - (√x)/3

b) y = 2[x - (√x)/6]

= 2[(√x)2 - (√x)/6]

= 2[(√x)2 - (√x)/6 + 1/9] - 2/9

= 2[(√x) - 1/3]2 - 2/9

So when x = 1/9, y is min. = -2/9

2a) Let L = ks4/t2 where k is contant.

Then s = 2 and t = 20:

2200 = k x 16/400

k = 55000

So L = 55000s4/t2.

b) Totally there will be 18 small pillars, each with loading 55000 x (1/3)4/(1/2)2 = 55000 x 4/81

So total loading = 55000 x 16/81

So the ratio of total loading of small pillars : original pillar = 16 : 81

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